{"id":9260,"date":"2019-11-30T14:42:38","date_gmt":"2019-11-30T09:12:38","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=9260"},"modified":"2021-07-02T15:44:23","modified_gmt":"2021-07-02T10:14:23","slug":"1st-puc-physics-question-bank-chapter-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-physics-question-bank-chapter-3\/","title":{"rendered":"1st PUC Physics Question Bank Chapter 3 Motion in a Straight Line"},"content":{"rendered":"

You can Download Chapter 3 Motion in a Straight Line Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Physics Question Bank Chapter 3 Motion in a Straight Line<\/h2>\n

1st PUC Physics Motion in a Straight Line TextBook Questions and Answers<\/h3>\n

Question 1.
\nIn which of the following examples of motion, can the body be considered approximately a point object:<\/p>\n

    \n
  1. a railway carriage moving without jerks between two stations.<\/li>\n
  2. a monkey sitting on top of a man cycling smoothly on a circular track.<\/li>\n
  3. a spinning cricket ball that turns sharply on hitting the ground.<\/li>\n
  4. a tumbling beaker that has slipped off the edge of a table.<\/li>\n<\/ol>\n

    Answer:<\/p>\n

      \n
    1. This is an example of uniform linear motion. Hence the carriage can be considered as a point object.<\/li>\n
    2. The monkey also undergoes motion smoothly on the circular track. Hence it can be heated as a point object.<\/li>\n
    3. The ball is spinning and undergoes changes in the plane of its motion when it hits the ground. Various parts of the ball experience different forces when the ball hits the ground. It is a rigid body and cannot be treated as a point object.<\/li>\n
    4. Different parts of the beaker experience different magnitude of forces during its motion. Hence it cannot be treated as a point object.<\/li>\n<\/ol>\n

      Question 2.
      \nThe position-time (x-i) graphs for two children A and B returning from their school! O to their homes P and Q respectively are shown in Fig. 3.19. Choose the correct entries in the brackets below;<\/p>\n

        \n
      1. (A\/B) lives closer to the school than (B\/A)<\/li>\n
      2. (A\/B) starts from the school earlier than (B\/A)<\/li>\n
      3. (A\/B) walks faster than (B\/A)<\/li>\n
      4. A and B reach home at the (same\/ different) time.<\/li>\n
      5. (A\/B) overtakes (B\/A) on the road (once\/twice).<\/li>\n<\/ol>\n

        \"1st<\/p>\n

        Answer:
        \n1. The distance between P and O (the school) is less then the distance between Q and O. (from the graph)
        \n\u21d2 A lives closer to the school than B.
        \n2. From the graph, we see that the line for A starts before the line for B.
        \n\u21d2 A starts from the school earlier than B.
        \n3. Velocity = Slope of the x-t graph. Clearly, the slope for B > slope for A.
        \n\u21d2 B walks faster than A
        \n4. From the graph, we see that both lines end at the same time.
        \n\u21d2 A and B reach home at the same time.
        \n5. From the graph, we see that the lines intersect once. Also, B walks faster.
        \n\u21d2 B overtakes A on the road once.<\/p>\n

        \"KSEEB<\/p>\n

        Question 3.
        \nA woman starts from her home at 9.00 am, walks with a speed of 5 km h-1<\/sup> on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h-1<\/sup>. Choose suitable scales and plot the x-t graph of her motion.
        \nAnswer:
        \nTime for the woman to reach her office = \\(\\frac{2.5 \\mathrm{km}}{5 \\mathrm{km} \\mathrm{h}^{-1}}\\)
        \n= 30 min.
        \n\u2234 She reaches her office at 9.30 am. Time for the woman to return home = \\(\\frac{2.5 \\mathrm{km}}{25 \\mathrm{km} \\mathrm{h}^{-1}}\\)
        \n= 0.1 h = 6 min.
        \n\u2234 She reaches her home at 5.06 pm
        \n\"1st<\/p>\n

        Question 4.
        \nA drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x – t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
        \nAnswer:
        \n\"1st
        \nFrom the above table, see that every 8 s the drunkard moves 2 m forward.
        \n\"1st
        \nBy looking at the shape of the graph, we can see that the graph for entire motion will be like this:-
        \n\"1st
        \nFrom the graph, we see that the drunkard will fall into the pit at 137 s.
        \nAnalytically:-
        \nThe drunkard covers 2 m every 8 s.
        \n\u2234 He will cover 8 m in 32 s. When he takes 5 steps forwards, he will fall into the pit.
        \n\u2234 Total time = 32 + 5 = 37 s.<\/p>\n

        Question 5.
        \nA jet airplane travelling at the speed of 500 km h-1<\/sup> ejects its products of combustion at the speed of 1500 km h-1<\/sup> relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?
        \nAnswer:
        \nSpeed of the exhaust with respect to an observer on the ground = Speed of exhaust with respect to the plane – Speed of plane with respect to the ground, (minus sign because the plane and exhaust move in opposite directions)
        \n= (1500 – 500) km h-1<\/sup>
        \n= 1000 km h-1<\/sup><\/p>\n

        Question 6.
        \nA car moving along a straight highway with a speed of 126 km h-1<\/sup> is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
        \nAnswer: v0<\/sub> = 126 km h-1<\/sup>
        \n= 126 \u00d7 \\(\\frac{5}{18}\\) m s-1<\/sup>
        \n= 35 m s-1<\/sup>
        \nStopping distance = (x – x0<\/sub>) = 200 m
        \nSince the car is stopped , v = 0.
        \nv2<\/sub> = v0<\/sub>2<\/sup> + 2 a (x – x0<\/sub>)
        \n0 = 35\u00b2 + 2a (200)
        \n\u21d2 a = – 3.06 m s-2<\/sup>
        \nRetardation = 3.06 m s-2<\/sup>
        \nv = v0<\/sub> + at
        \n\u21d2 0 = 35 – 3.06 t
        \n\u21d2 t = 11.4 s
        \nTime for car to stop = 11.4 s.<\/p>\n

        \"KSEEB<\/p>\n

        Question 7.
        \nTwo trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h-1<\/sup> in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s-2<\/sup>. If after 50 s, the guard of B Just brushes past the driver of A, what was the original distance between them?
        \nAnswer:
        \nInitially,
        \n\"1st
        \nRelative Acceleration a = 1 ms-2<\/sup>
        \nv0<\/sub> =72 km h-1<\/sup> = 72 \u00d7 \\(\\frac{5}{18}\\) m s-1<\/sup>
        \n= 20 ms-1<\/sup>
        \nRelative displacement = x
        \nRelative velocity initially = V0<\/sub> – V0<\/sub> = 0
        \nX = 1\/2 at\u00b2
        \n= 1\/2 \u00d7 1 \u00d7 50\u00b2 = 1250 m
        \nOriginal distance between the trains = 1250 m.<\/p>\n

        Question 8.
        \nOn a two-lane road, car A Is travelling with a speed of 36 km h-1<\/sup>. Two cars B and C approach car A in opposite directions with a speed of 54 km h-1<\/sup> each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
        \nAnswer:
        \n\"1st
        \nVA<\/sub>= 36 km h-1<\/sup> = 10 m s-1<\/sup>
        \nVB = Vc = 54 km h-1<\/sup> = 15 m s-1<\/sup>
        \nVelocity Of B with respect to A
        \n= VBA<\/sub> = VB<\/sub> – VA<\/sub>
        \n= 15m -10 = 5 m s-1<\/sup>
        \nVelocity of C with respect to A .
        \nVCA<\/sub> = VC<\/sub> – VA<\/sub>
        \n= 15 – ( -10) = 25 m s-1<\/sup>
        \nS = VCA<\/sub> t
        \n1000 m = 25 \u00d7 t
        \n\u21d2 t = 40 s
        \nC will take 40 s to cover AC. In that time, B must cover AB to overtake and avoid collisions.
        \ns = VBA<\/sub> t + 1\/2 at\u00b2
        \n1000 = 5 \u00d7 40 + 1\/2 \u00d7 a \u00d7 40\u00b2
        \n\u21d2 a = 1 ms-2<\/sup>
        \nAcceleration of B = 1 ms-2<\/sup><\/p>\n

        Question 9.
        \nTwo towns A and B are connected by a regular bus service with a bus leaving In either direction every T minutes. A man cycling with a speed of 20 km h-1<\/sup> In the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period Tof the bus service and with what speed (assumed constant) do the buses ply on the road?
        \nAnswer:
        \nLet the speed of the bus be v km h-1<\/sup>
        \nDistance covered by one bus before the next one leaves the same town is vT For buses going in the same direction as the cyclist, frequency of the bus is 18 min i.e.,
        \n\\(\\frac{v T}{\\text { relative velocity of the bus w.r.t. the cyclist }}\\) = 18 min
        \n\\(\\frac{v T}{v-20}\\) = 18……………………… (1)
        \nSimilarly, for buses going in the opposite direction, frequency of the bus = 6 min
        \n\\(\\frac{v T}{\\text { relative velocity of the bus w.r.t. the cyclist }}\\) = 6 min
        \n\\(\\frac{v T}{v+20}\\) = 6 ……………………… (2)
        \n\\((1) \\div(2)\\)
        \n\u21d2 \\(\\frac{v+20}{v-20}\\)
        \nSubstitute for v in (1)
        \n\\(\\frac{40 \\times T}{40-20}\\) =18 \u21d2 T = 9 min.<\/p>\n

        Question 10.
        \nA player throws a ball upwards with an initial speed of 29.4 ms-1<\/sup>.<\/p>\n

          \n
        1. What is the direction of acceleration during the upward motion of the ball?<\/li>\n
        2. What are the velocity and acceleration of the ball at the highest point of its motion?<\/li>\n
        3. Choose the X = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of X-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.<\/li>\n
        4. To what height does the ball rise and after how long does the ball return to the player\u2019s hands? (Take g = 9.8 m s-2<\/sup> and neglect air resistance).<\/li>\n<\/ol>\n

          Answer:
          \n1. The direction of acceleration is vertically downwards (towards the Earth)
          \n2. At the highest point, velocity = 0 m s-1<\/sup> acceleration = g = 9.8 m s-2<\/sup> vertically downwards.
          \n3. During upward motion:<\/p>\n

            \n
          • position – positive<\/li>\n
          • velocity – negative<\/li>\n
          • acceleration – positive<\/li>\n<\/ul>\n

            During downward motion:<\/p>\n

              \n
            • position – positive<\/li>\n
            • velocity – positive<\/li>\n
            • acceleration – positive<\/li>\n<\/ul>\n

              4. v = 0 at the highest point
              \nv0<\/sub> = 29.4 m s-1<\/sup>
              \ng = -9.8 m s-2<\/sup>
              \nv\u00b2 = v0<\/sub>2<\/sup> + 2 g h
              \n0 = 29.4\u00b2 = 2 \u00d7 (-9.8) \u00d7 4
              \n\u21d2 h = 44.1m
              \nHeight of ascension = 44.1 m
              \nv = v0<\/sub> + g t
              \n0 = 29.4 – 9.8 t
              \nt = 3 s
              \nTime of ascension = Time of descension
              \n\u2234 Time for the ball to return = 6 s<\/p>\n

              \"KSEEB<\/p>\n

              Question 11.
              \nRead each statement below carefully and state with reasons and examples, if it is true or false; A particle in one-dimensional motion<\/p>\n

                \n
              1. with zero speed at an instant may have non-zero acceleration at that instant<\/li>\n
              2. with zero speed may have non-zero velocity,<\/li>\n
              3. with constant speed must have zero acceleration,<\/li>\n
              4. with positive value of acceleration must be speeding up.<\/li>\n<\/ol>\n

                Answer:
                \n1. True:
                \nConsider a ball thrown vertically, upward. At the highest point, the speed is zero but the acceleration of the ball is non-zero\u00a0 \u00a0 \u00a0 \u00a0( 9.8ms-2<\/sup> vertically downwards). Acceleration does not depend on instantaneous speed.<\/p>\n

                2. False:
                \nSince the magnitude of velocity is speed, a body with zero speed must have zero velocity.<\/p>\n

                3. True:
                \nIn the case of a body rebounding with the same speed, the acceleration at the time of impact is infinite, which is not practical physically.<\/p>\n

                4. False:
                \nThis depends on the chosen positive direction. The statement is true when the direction of motion and acceleration are along the chosen positive direction.<\/p>\n

                Question 12.
                \nA ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one-tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
                \nAnswer:
                \nVelocity of the ball when it hits the ground = v
                \nv0<\/sub> = 0, h = 90 m, g = 9.8m s-2<\/sup>
                \nv\u00b2 = v0<\/sub>2<\/sup> + 2 g h
                \nv\u00b2 = 02<\/sup> + 2 \u00d7 9.8 \u00d7 90
                \n\u21d2 v = 42 ms-1<\/sup>
                \nTime for descent = t
                \nv = v0<\/sub> + g t
                \n42 = 0 + 9.8 t
                \nt = 4.3 s
                \nDuring the rebound, initial speed of the ball = 9\/10 of v
                \n= 9\/10 \u00d7 42
                \n= 37.8 ms-1<\/sup>
                \nTotal time for ascent and descent = t1<\/sub>
                \nNet displacement = 0
                \n0 = 37.8 t – 1\/2 gt\u00b2
                \n37.8 t = 1\/2 \u00d7 9.8 \u00d7 t\u00b2 \u21d2 t = 7.7 s
                \nTime at which maximum height is achieved = 4.3 + 1\/2 7.7
                \n= 8.15 s
                \n\"1st<\/p>\n

                Question 13.
                \nExplain clearly, with examples, the distinction between:
                \n(a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;<\/p>\n

                (b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show In both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider one-dimensional motion only].
                \nAnswer:
                \n(a) Consider a person who leaves home for work in the morning and returns home in the evening.
                \nThe magnitude of the displacement of the person during the interval is zero but the total length of path covered = 2 \u00d7 distance between the home and the workplace.
                \nMagnitude of displacement is the magnitude of the shortest length between the two points. The second quantity is clearly always greater than or equal to the first quantity. The equality holds if the body has not moved at all (path length = magnitude of displacement = 0)<\/p>\n

                (b) Average velocity = \\(\\frac{\\Delta x}{\\Delta t}\\). If \u0394 X is zero (like in the example explained in (a)) then average velocity = 0.
                \nAverage speed = \\(\\frac{\\text { Total path length }}{\\text { Time taken }}\\)
                \n2 \u00d7 \\(\\frac{\\text { Distance between the home and the work place }}{\\text { Time taken to traverse this distance }}\\)
                \n= Clearly average speed \u2265 magnitude of average velocity. Equality exists when body does not undergo any motion.<\/p>\n

                Question 14.
                \nA man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h-1<\/sup>. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h-1<\/sup>. What is the<\/p>\n

                  \n
                1. magnitude of average velocity, and<\/li>\n
                2. average speed of the man over the interval of time<\/li>\n<\/ol>\n

                  (i) 0 to 30 min,
                  \n(ii) 0 to 50 min,
                  \n(iii) 0 tp 40 min?
                  \n[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]
                  \nAnswer:
                  \nv1<\/sub> = 5 km h-1<\/sup>, v2<\/sub> = 7.5 km h-1<\/sup>, d = 2.5 km
                  \nTime for the man to reach the market
                  \n= t1<\/sub>\u00a0= d\/v1<\/sub>= 2.5\/5 = 0.5 h = 30 min
                  \nTime for the man to return from the market
                  \n= t2<\/sub>= d\/v2<\/sub> = 2.5\/7.5 = 1\/3 h= 20 min<\/p>\n

                  1. To find the average velocity,
                  \nAverage velocity = \\(\\frac{\\text { Total path length }}{\\text { Total time taken }}\\)<\/p>\n

                    \n
                  • During the interval between 0 to 30 min,
                    \nDisplacement = 2.5 km
                    \ntime = 30 minutes
                    \n\u2234 Average velocity = \\(\\frac{2.5 \\mathrm{km}}{30 \\mathrm{min}}\\) = 5 km h-1<\/sup><\/li>\n
                  • interval between 0 and 50 minutes,
                    \nDisplacement = 0
                    \n\u2234 Average velocity = 0<\/li>\n
                  • During between 0 to 40 minutes,
                    \nDisplacement = 2.5 km – 7.5 kmh-1<\/sup> \u00d7 10 min = 1.25 km
                    \ntime = 40 minutes
                    \n\u2234 Average velocity = \\(\\frac{1.25 \\mathrm{km}}{40 \\mathrm{min}}\\) = 1.875 km h-1
                    \n<\/sup><\/li>\n<\/ul>\n

                    2. To find the average speed:
                    \nAverage speed = \\(\\frac{\\text { Total path length }}{\\text { Total time taken }}\\)<\/p>\n

                      \n
                    • During the interval between 0 to 30 min
                      \nPath length = 2.5 km
                      \ntime = 30 minutes
                      \n\u2234 Average speed = \\(\\frac{2.5 \\mathrm{km}}{30 \\mathrm{min}}\\) = 5 km h-1<\/sup><\/li>\n
                    • During the interval between 0 and 50 minutes,
                      \nPath length = 2.5 km + 2.5 km
                      \nTime taken = 50 minutes
                      \n\u2234 Average speed = \\(\\frac{2.5 \\mathrm{km}+2.5 \\mathrm{km}}{50 \\mathrm{min}}\\) = 6 km h-1<\/sup><\/li>\n
                    • During 0 to 40 minutes
                      \nPath length = 5 kmh-1<\/sup> \u00d7 30 min + 7.5 kmh-1<\/sup> \u00d7 10 min
                      \n2.5 km + 1.25 km = 3.75 km
                      \ntime = 40 minutes
                      \n\u2234 Average velocity = \\(\\frac{3.75 \\mathrm{km}}{40 \\mathrm{min}}\\) = 5.625 km h-1<\/sup><\/li>\n<\/ul>\n

                      \"KSEEB<\/p>\n

                      Question 15.
                      \nIn Exercises 3.13 and 3.14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
                      \nAnswer:
                      \nThe instantaneous speed is always equal to the magnitude of the instantaneous velocity because for very small instants of time the length of the path is equal to the magnitude of displacement.<\/p>\n

                      Question 16.
                      \nLook at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
                      \n\"1st
                      \nAnswer:
                      \n(a) From the graph, we see that for certain instants of time, the particle has 2 positions, which is not possible. Hence this graph cannot possibly represent one-dimensional motion of a particle.<\/p>\n

                      (b) From the graph, we see that for certain instants of time, the particle has 2 velocities, which is not possible. Hence this graph cannot possibly represent one-dimensional motion of a particle.<\/p>\n

                      (c) From the graph, we see that for some instants of time, the particle has negative speed. But speed is always a non-negative quantity. Hence this graph cannot possibly represent one-dimensional motion of a particle.<\/p>\n

                      (d) From the graph, we see that for a time interval the total path length is decreasing. But total path length is always a nondecreasing quantity. Hence this graph cannot possibly represent one-dimensional motion of a particle.<\/p>\n

                      Question 17.
                      \nFigure 3.21 shows the X- t plot of onedimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph.
                      \n\"1st
                      \nAnswer:
                      \nNo, it is wrong to make such statements about the trajectory of the particle because an x-t graph does not show the particle\u2019s trajectory.
                      \nContext:
                      \nConsider a body dropped from the top of a tower at time t = 0. If the vertically downward direction is chosen as the positive direction, then the body\u2019s x -t graph would resemble the one given in the question.<\/p>\n

                      Question 18.
                      \nA police van moving on a highway with a speed of 30 km h-1<\/sup> fires a bullet at a thief\u2019s car speeding away In the same direction with a speed of 192 km h-1<\/sup>. If the muzzle speed of the bullet is 150 m s-1<\/sup>, with what speed does the bullet hit the thief’s car?
                      \nAnswer:
                      \nSpeed of the police van = vp<\/sub>
                      \n= 30 km h-1<\/sup> = 25\/3 ms-1<\/sup>.
                      \nSpeed of the thief\u2019s car is
                      \nvt<\/sub> = 192 km h-1<\/sup> = 160\/3 ms-1<\/sup>
                      \nSpeed of the bullet = vb<\/sub> = 150 ms-1<\/sup>
                      \nSpeed of the bullet with respect to the police car = vb<\/sub> + vp<\/sub> = 150 + 25\/3
                      \n= 475\/3 msv-1<\/sup>
                      \nSpeed with which the bullet hits the thief\u2019s car = Speed of the bullet with respect to the thief\u2019s car
                      \n= 475\/3 – vt<\/sub> = 475\/3 – 160\/3 =105 ms-1<\/sup><\/p>\n

                      Question 19.
                      \nSuggest a suitable physical situation for each of the following graphs (Fig 3.22):
                      \n\"1st
                      \n\"1st
                      \nAnswer:
                      \n(a) consider a ball which is pushed at some time t < 0 towards a wall. Upon rebounding from this wall it hits the opposite wall and comes to a stop. If x = 0 for the initial position, then this context may have the given x -t graph.<\/p>\n

                      (b) Consider a ball thrown vertically upwards, with the vertically upward direction chosen as the positive direction. Each time it hits the ground, it loses a fraction of its velocity and finally comes to rest. The ball may be represented the given v -t graph in this context.<\/p>\n

                      (c) Consider a cricket ball moving with a uniform velocity which is hit by the bat and then turns back. In this case, the a – t graph for the ball may be similar to the one given above.<\/p>\n

                      Question 20.
                      \nFigure 3.23 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position, velocity and acceleration variables of the particle at t= 0.3 s, 1.2 s, -1.2 s.
                      \n\"1st
                      \nAnswer:
                      \nLet the maximum amplitude of the sine wave be A, where A is positive. We can see that the particle obeys the
                      \nequation x = – A sin \\(\\left(\\frac{2 \\pi \\mathrm{t}}{\\mathrm{T}}\\right)\\)
                      \nWhere T = 2 s = period of the sine wave
                      \n\u2234 position = x = – A sin (\u03c0 t)
                      \nvelocity – v = \\(\\frac{dx}{dt}\\) = – A\u03c0<\/sub> cos (\u03c0 t)
                      \nacceleration = a = \\(\\frac{dv}{dt}\\) = A\u03c0<\/sub>\u00b2 sin(\u03c0 t)
                      \nAt t = 0.3 s
                      \nx = – A sin (0.3\u03c0<\/sub>) = negative
                      \nv = – A cos (0.3\u03c0<\/sub>) = negative
                      \na = A \u03c0<\/sub>\u00b2 sin (0.3\u03c0<\/sub>) = positive
                      \nSince sin(0.3\u03c0<\/sub>) > 0 and cos(0.3\u03c0<\/sub>) > 0
                      \nAt t = 1.2 s
                      \nx = – A sin (1.2\u03c0<\/sub>) = – A sin (1.2\u03c0<\/sub>) = positive
                      \nv = – A \u03c0<\/sub> cos (1.2\u03c0<\/sub>) = – A cos (1.2\u03c0<\/sub>) = positive
                      \na = A\u03c0<\/sub>\u00b2 sin (1.2\u03c0<\/sub>) = negative
                      \nsince sin(1.2\u03c0<\/sub>)<0 and cos(1.2\u03c0<\/sub>) < 0
                      \nAt t = – 1.2 s
                      \nx = – A sin (- 1.2\u03c0<\/sub>) = A sin (1.2\u03c0<\/sub>)= negative
                      \nv = – A it cos(- 1.2\u03c0<\/sub>) = – A\u03c0<\/sub> cos (1.2\u03c0<\/sub>) = positive.
                      \na = A\u03c0<\/sub>\u00b2 sin (- 1.2\u03c0<\/sub>) = – A\u03c0<\/sub> 2sin(1.2\u03c0<\/sub>) = positive
                      \nSince sin(-\u03b8) = – sin\u03b8t cos(-\u03b8) = cos\u03b8 sin(1.2\u03c0<\/sub>) <0 and cos(1.2\u03c0<\/sub>) < 0<\/p>\n

                      Question 21.
                      \nFigure 3.24 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each Interval.
                      \n\"1st
                      \nAnswer:
                      \nThe magnitude of the slope of the x – t graph is highest in 3 and least in 2. Hence, the average speed is greatest in 3 and least in 2. Also, the sign of the slope of the x -t graph is positive for 1 and 2 and negative for 3. Therefore sign of the average velocity ‘V’ is:-
                      \nv > 0 for intervals 1 and 2
                      \nv < 0 for intervals 3.<\/p>\n

                      Question 22.
                      \nFigure 3.25 gives a speed-time graph of a particle In motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and an in the three Intervals. What are the accelerations at the points A, B, C and D?
                      \n\"1st
                      \nAnswer:
                      \nThe magnitude of the slope of the speed-time graph is greatest for interval 2. Hence average acceleration is greatest in magnitude in interval 2. The value of speeds in interval 3 is the highest. Hence the average speed in interval 3 is highest. Sign of a = sign of the slope of the speed-time graph, a > 0 for 1 and 3 a < 0 for 2. At points, A, B, C and D, the tangent to the curve will be parallel to the time axis. Hence slope at the points is zero,
                      \n\u2234 a = 0 at points A, B, C and D.<\/p>\n

                      \"KSEEB<\/p>\n

                      Question 23.
                      \nA three-wheeler starts from rest, accelerates uniformly with 1 m s-2<\/sup> on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the nth second (n = 1, 2, 3….) versus n. What do you expect this plot to be during accelerated motion: a straight line or a parabola?
                      \nAnswer:
                      \nInitial velocity u = 0.
                      \nAcceleration a = 1 ms-2<\/sup>.
                      \nAcceleration time = 10 s
                      \nLet sn<\/sub> be the total displacement (or distance in this case) in time \u2018n\u2019 seconds.
                      \nsn<\/sub> = ut + 1\/2 at\u00b2
                      \nsn<\/sub> = 0(n) + 1\/2 a(n\u00b2) = 1\/2 an\u00b2 for n \u2264 10
                      \nDistance covered in \u2018n\u2019th<\/sup> second dn<\/sub> = sn<\/sub> – sn<\/sub> – 1
                      \n= 1\/2 an\u00b2 – 1\/2 a(n-1)\u00b2
                      \n= an – 1\/2 a (for n \u2264 10)
                      \n= (n – 1\/2) m
                      \nAt the end of 10 seconds, velocity acquired v = u + at = 0 + a (10)
                      \n10 a = 10 ms-1<\/sup>
                      \nDistance covered in the \u2018n\u2019th<\/sup> second = 10m for n >10
                      \n\u2234 dn<\/sub> = {(n-1\/2)m n \u2264 10,
                      \n10m n > 10
                      \nThe plot is a straight line inclined to the time axis for uniformly accelerated motion. (It is a straight line parallel to the time axis for uniform motion).
                      \n\"1st<\/p>\n

                      Question 24.
                      \nA boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s-1<\/sup>. How much time does the bail take to return to his hands? if the lift starts moving up with a uniform speed of 5 m s-1<\/sup> and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?
                      \nAnswer:
                      \nInitial velocity n = 49 ms-1<\/sup>
                      \nAcceleration g = -9.8 ms\u00b2
                      \nDisplacement s = 0
                      \ns = ut + 1\/2 at\u00b2
                      \n0 = 49 t + 1\/2 (-9.8) t\u00b2
                      \n\u21d2 t = 0 or t = 10 s
                      \nt = 0 represents initial time. The required \u2018t\u2019 here is t = 10 s. In the case where the lift also moves, the relative velocity of the ball with respect to the boy remains the same, i.e, 49 ms-1<\/sup>. The relative displacement is also 0. Hence the time required in this case is also 10 s.<\/p>\n

                      Question 25.
                      \nOn a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed 9 km h-1<\/sup> (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h-1<\/sup>. For an observer on a stationary platform outside, what is the<\/p>\n

                        \n
                      1. speed of the child running in the direction of motion of the belt?.<\/li>\n
                      2. speed of the child running opposite to the direction of motion of the belt?<\/li>\n
                      3. time taken by the child in (1) and (2)?<\/li>\n<\/ol>\n

                        Which of the answers alter If motion is viewed by one of the parents?
                        \n\"1st
                        \nAnswer:
                        \n1. Speed of the child running in the direction of motion of the belt = Speed of the child with respect to the belt + Speed of the belt with respect to the ground
                        \n= 9 + 4 = 13 km h-1<\/sup>.
                        \n2. Speed of the child running opposite to the direction of motion of the belt = speed of the child with respect to the belt speed of the belt with respect to the ground
                        \n= 9 – 4 = 5 km h-1<\/sup>.
                        \n3. Time taken by the child in both cases is the same because the speed of the child with respect to the belt does not change.
                        \nSpeed v = 9 km h-1<\/sup> = 2.5 ms-1<\/sup>
                        \nDistance d = 50 m
                        \nTime t = \\(\\frac{\\text { Distance }}{\\text { Speed }}\\) = \\(\\frac{50}{2.5}\\) = 20 s
                        \nThe answer to (a) and (b) when the motion is viewed by one of the parents is 9 km h-1<\/sup>. This is because the parents are also on the belt and are moving with respect to the ground. They only see the motion of the child with respect to the belt. In (c), the answer remains unaltered because the speed of the child with respect to the belt does not depend on the speed of the parents.<\/p>\n

                        Question 26.
                        \nTwo stones are thrown up simultaneously from the edge of a cliff 200 m high with Initial speeds of 15 m s-1<\/sup> and 30 m s-1<\/sup>. Verify that the graph shown in Fig. 3.27 correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take
                        \ng = 10 ms-2<\/sup>. Give the equations for the linear and curved parts of the plot.
                        \n\"1st
                        \nAnswer:
                        \nInitial velocity of stone 1 is u1<\/sub> = 15 ms-1<\/sup>
                        \nInitial velocity of stone 2 = u2<\/sub> = 30 ms-1<\/sup>
                        \nAcceleration = g = -10ms-2<\/sup> (vertically upwards direction is chosen as positive)
                        \nx1<\/sub> = 200 + u1<\/sub>t + 1\/2 at\u00b2
                        \n= 200 + 15 t – 5 t\u00b2
                        \nx2<\/sub> = 200 + u2<\/sub>t + 1\/2 at\u00b2
                        \n= 200 + 30 t – 5 t\u00b2.
                        \nAlso x1<\/sub> = 0 for t = 8 s and
                        \nx2<\/sub> = 0 for t = 10s.
                        \n\u2234 For 0 \u2264 t \u2264 8 s
                        \nx2<\/sub> – x1<\/sub> = 15 t
                        \nAt t = 8 s,
                        \nx2<\/sub> – x1<\/sub> = 120 m
                        \nFor 8 < t; t \u2264 10 s
                        \nx2<\/sub> – x1<\/sub> = 200 + 30 t – 5 t\u00b2 – 0
                        \n= 200 + 30 t – 5 t\u00b2.
                        \nFor t > 10 s,
                        \nx2<\/sub> – x1<\/sub> = 0
                        \n\u2234 The given graph is correct and because it matches with the equations obtained.<\/p>\n

                        \"KSEEB<\/p>\n

                        Question 27.
                        \nThe speed-time graph of a particle moving along a fixed direction is shown In Fig. 3.28. Obtain the distance traversed by the particle between
                        \n(a) t = 0 s to 10 s.
                        \n(b) t = 2 s to 6 s.
                        \n\"1st
                        \nWhat is the average speed of the particle over the intervals in (a) and (b)?.
                        \nAnswer:
                        \n(a) Distance covered = Area under the speed – time graph
                        \n= 1\/2 \u00d7 10 \u00d7 12 = 60 m
                        \nAverage speed = \\(\\frac{\\text { Distance covered }}{\\text { Time taken }}\\)
                        \n= \\(\\frac{60 m}{10 s}\\) = 6 ms-1
                        \n<\/sup>
                        \n(b) For 0 \u2264 t \u2264 5 s,
                        \nSpeed v = \\(\\frac{12}{5}\\) t ms-1<\/sup>
                        \nFor 5 < t \u2264 10 s,
                        \nSpeed v = 12 – \\(\\frac{12}{5}\\) (t – 5) ms-1<\/sup>
                        \n\u2234 Speed at t = 2 s is 4.8 ms-1<\/sup> and speed at t = 6 s is 9.6 ms-1<\/sup>
                        \nDistance covered = Area of trapezium ABEF + Area of trapezium BCDE
                        \n\"1st
                        \n= 1\/2 (4.8 + 12) \u00d7 3 + 1\/2 (12 + 9.6) \u00d7 1
                        \n= 36 m
                        \nAverage speed = \\(\\frac{\\text { Distance covered }}{\\text { Time taken }}\\)
                        \n= \\(\\frac{36}{6-2}\\) = 9 ms-1<\/sup><\/p>\n

                        Question 28.
                        \nThe velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29:
                        \n\"1st
                        \nWhich of the following formulae are correct for describing the motion of the particle over the time-interval t1<\/sub> to t2<\/sub><\/p>\n

                          \n
                        1. x(t2<\/sub>) = x(t1<\/sub> + v (t1<\/sub>) (t2<\/sub> – t1<\/sub> + (1\/2) a (t2<\/sub>– t1<\/sub>)\u00b2<\/li>\n
                        2. v(t2<\/sub>) = v(t1<\/sub>) + a(t2<\/sub> – t1<\/sub>)<\/li>\n
                        3. vaverage<\/sub> = (x(t2<\/sub>) – x(t1<\/sub>)\/(t2<\/sub>-t1<\/sub>)<\/li>\n
                        4. average = (v(t2<\/sub> – v(t1<\/sub>))\/(t2<\/sub> – t1<\/sub>)<\/li>\n
                        5. x(t2<\/sub>) = x(t1<\/sub>) + vaverage<\/sub> (t2<\/sub> – t1<\/sub>) + (1\/2) aaverage<\/sub> (t2<\/sub> – t1<\/sub>)\u00b2<\/li>\n
                        6. x(t2<\/sub>) – x(t1<\/sub>) = area under the v-t curve bounded by the t-axis and the dotted line shown.<\/li>\n<\/ol>\n

                          Answer:<\/p>\n

                            \n
                          1. Wrong, because it is not known whether the acceleration \u2018a\u2019 is constant.<\/li>\n
                          2. Wrong, because it is not known whether the acceleration \u2018a\u2019 is constant.<\/li>\n
                          3. Correct. By definition.<\/li>\n
                          4. Correct. By definition<\/li>\n
                          5. Wrong. The formula should contain v1<\/sub> (t) instead of vaverage<\/sub>.<\/li>\n
                          6. Correct By definition.<\/li>\n<\/ol>\n

                            1st PUC Physics Motion in a Straight Line One Mark Questions and Answers<\/h3>\n

                            Question 1.
                            \nWhen do you say that a body is in motion?
                            \nAnswer:
                            \nA body is said to be in motion when it changes its position with respect to time and surrounding.<\/p>\n

                            Question 2.
                            \nWhat is a particle?
                            \nAnswer:
                            \nA particle is a geometrical mass point.<\/p>\n

                            Question 3.
                            \nDefine displacement.
                            \nAnswer:
                            \nThe change of position of a body in a particular direction is called displacement.<\/p>\n

                            Question 4.
                            \nCan the displacement of a moving body be zero?
                            \nAnswer:
                            \nYes<\/p>\n

                            Question 5.
                            \nDefine speed.
                            \nAnswer:
                            \nThe rate of change of position of a body in any direction is known as speed.<\/p>\n

                            Question 6.
                            \nDefine velocity.
                            \nAnswer:
                            \nThe rate of change of position of a body in a particular direction is called velocity or Rate of displacement of a body is called velocity.<\/p>\n

                            \"KSEEB<\/p>\n

                            Question 7.
                            \nDefine uniform velocity.
                            \nAnswer:
                            \nIf a body covers equal distances in equal intervals of time in a given direction, however small the intervals maybe then its velocity is said to be uniform.<\/p>\n

                            Question 8.
                            \nDefine variable velocity.
                            \nAnswer:
                            \nIf the velocity of a body either changes in magnitude or in direction or both, then its velocity is said to be variable.<\/p>\n

                            Question 9.
                            \nDefine average velocity.
                            \nAnswer:
                            \nThe average velocity of a body is defined as the ratio of its total displacement to the total time.<\/p>\n

                            Question 10.
                            \nDefine acceleration.
                            \nAnswer:
                            \nRate of change of velocity of a body is called acceleration.<\/p>\n

                            Question 11.
                            \nDefine uniform acceleration.
                            \nAnswer:
                            \nIf the velocity of a body changes by an equal amount in equal interval of time, however, small the interval maybe then its acceleration is said to be uniform.<\/p>\n

                            Question 12.
                            \nWhat is a position-time graph?
                            \nAnswer:
                            \nA graph drawn by taking time along the x-axis and displacement along the y-axis is called position-time graph.<\/p>\n

                            Question 13.
                            \nWhat does the slope of the position-time graph indicate?
                            \nAnswer:
                            \nThe slope of the position-time graph indicates velocity.<\/p>\n

                            Question 14.
                            \nDraw the position-time graph of a particle which is at rest.
                            \nAnswer:
                            \n\"1st<\/p>\n

                            Question 15.
                            \nDraw the position-time graph of an object starting from rest and moving with uniform velocity.
                            \nAnswer:
                            \n\"1st<\/p>\n

                            Question 16.
                            \nWhat is a velocity-time graph?
                            \nAnswer:
                            \nA graph drawn by taking time along x-axis and velocity along the y-axis is called velocity-time graph.<\/p>\n

                            Question 17.
                            \nWhat does the slope of the velocity-time graph indicate?
                            \nAnswer:
                            \nThe slope of the velocity-time graph indicates the acceleration.<\/p>\n

                            Question 18.
                            \nWhat does the area under the velocity-time graph indicate?
                            \nAnswer:
                            \nArea under velocity-time graph indicates the displacement.<\/p>\n

                            Question 19.
                            \nAn object is moving with uniform velocity. What Is its acceleration?
                            \nAnswer:
                            \nZero<\/p>\n

                            Question 20.
                            \nHow do you determine the instantaneous velocity of a particle from the position-time graph?
                            \nAnswer:
                            \nInstantaneous velocity at any point is given by the slope of the tangent drawn to the position-time graph at that point.<\/p>\n

                            Question 21.
                            \nDraw the velocity-time graph of a particle moving with<\/p>\n

                              \n
                            1. uniform velocity<\/li>\n
                            2. variable velocity.<\/li>\n<\/ol>\n

                              Answer:
                              \n1)
                              \n\"1st
                              \n2)
                              \n\"1st
                              \nQuestion 22.
                              \nDraw the velocity-time graph of an object starting from rest and moving with uniform acceleration.
                              \nAnswer:
                              \n\"1st<\/p>\n

                              Question 23.
                              \nDraw the velocity-time graph of an object moving with constant retardation.
                              \nAnswer:
                              \n\"1st<\/p>\n

                              Question 24.
                              \nDraw the position time of particle which is initially at rest and moving with uniform acceleration.
                              \nAnswer:
                              \n\"1st<\/p>\n

                              Question 25.
                              \nWhat is the acceleration time graph?
                              \nAnswer:
                              \nA graph drawn by taking time along x-axis and acceleration along the y-axis is called velocity-time graph.<\/p>\n

                              Question 26.
                              \nWrite the expression for the distance travelled by a body during the nth second of its motion.
                              \nAnswer:
                              \nDistance travelled by a body during the nth<\/sup> second of motion is given by
                              \nsn<\/sub> = u + a(n – 1\/2)
                              \nwhere u is the initial velocity and a is the uniform acceleration.<\/p>\n

                              \"KSEEB<\/p>\n

                              Question 27.
                              \nWrite the expression for the acceleration in terms of distance travelled in two consecutive intervals of time.
                              \nAnswer:
                              \nAcceleration is given by,
                              \na = \\(\\frac{s_{2}-s_{1}}{t^{2}}\\)
                              \nwhere s1<\/sub> and s2<\/sub> are the distances travelled in two consecutive intervals of time \u2018t\u2019 seconds each.<\/p>\n

                              Question 28.
                              \nWhat is relative velocity?
                              \nAnswer:
                              \nThe velocity of a particle in motion relative to another particle is called relative velocity.<\/p>\n

                              Question 29.
                              \nWhen is the relative velocity of two bodies maximum?
                              \nAnswer:
                              \nRelative velocity of two bodies is maximum when they are moving in opposite directions.<\/p>\n

                              Question 30.
                              \nWhat does the area of a-t graph indicate?
                              \nAnswer:
                              \nArea of a-t graph represents the change in velocity of the body in a given time interval.<\/p>\n

                              Question 31.
                              \nDraw the a-t graph of a body moving with uniform acceleration.
                              \nAnswer:
                              \n\"1st<\/p>\n

                              Question 32.
                              \nWhy is it not necessary for a body following another, to stop, to avoid collision?
                              \nAnswer:
                              \nIf the relative velocity is zero, no collision will occur.<\/p>\n

                              Question 33.
                              \nIf in case df a motion, displacement is directly proportional to the square of the tune elapsed, what do you think about its acceleration, le, constant or variable? Explain why?
                              \nAnswer:
                              \nx \u221d t\u00b2
                              \n\u21d2 x = kt\u00b2 Where k is a constant
                              \na = \\(\\frac{d}{d t}\\left(\\frac{d x}{d t}\\right)\\) = \\(\\frac{d}{d t}\\) (2 k t)
                              \n= 2k = constant
                              \n\u2234 Acceleration is constant.<\/p>\n

                              Question 34.
                              \nWhy does the earth Impact the same acceleration to all bodies?
                              \nAnswer:
                              \nSince acceleration is the force on unit mass, the acceleration due to gravity is constant.<\/p>\n

                              Question 35.
                              \nWhat will be the nature of the velocity-time graph for a uniform motion?
                              \nAnswer:
                              \nThe v -t graph will be a line parallel to the time axis.<\/p>\n

                              \"KSEEB<\/p>\n

                              Question 36.
                              \nThe position coordinate of a moving particle is given by x = 6 + 18t + 9t2(x is in metres and t is in seconds). What is its velocity at t s 2s?
                              \nAnswer:
                              \nx = 6 + 18 t + 9t\u00b2
                              \nv= \\(\\frac{\\mathrm{d} x}{\\mathrm{dt}}\\) = 18 + 18 t
                              \nv at t = 2 s = 18 + 18 (2)
                              \n= 54 ms-1<\/sup><\/p>\n

                              Question 37.
                              \nA ball is thrown straight up. What is its velocity and acceleration at the top?
                              \nAnswer:
                              \nAt the top, v = 0, a = g.<\/p>\n

                              1st PUC Physics Motion in a Straight Line Two Mark Questions and Answers<\/h3>\n

                              Question 1.
                              \nDistinguish between speed and velocity.
                              \nAnswer:<\/p>\n

                                \n
                              1. Speed is the rate of change of position of a body in any direction while velocity is the rate of change of position of a body in a particular direction.<\/li>\n
                              2. Speed is a scalar quantity whereas velocity is a vector quantity.<\/li>\n<\/ol>\n

                                Question 2.
                                \nWhat is a velocity-time graph? What Is Its importance?
                                \nAnswer:
                                \nA graph drawn by taking time along x-axis and velocity along the y-axis is called velocity-time graph. The velocity-time graph can be used<\/p>\n

                                  \n
                                1. To find the nature of the motion of a body.<\/li>\n
                                2. To determine the velocity of a body at any instant.<\/li>\n
                                3. To calculate the displacement of a body in a given time.<\/li>\n<\/ol>\n

                                  Question 3.
                                  \nIf V1<\/sub> and v2<\/sub> are the velocities of two bodies then what is their relative velocity when they move in the<\/p>\n

                                    \n
                                  1. same and<\/li>\n
                                  2. opposite direction?<\/li>\n<\/ol>\n

                                    Answer:<\/p>\n

                                      \n
                                    1. When the bodies move in the same direction, the relative velocity of 1st<\/sup> body w.r.t 2nd<\/sup> is v12<\/sub> = v1<\/sub> – v2<\/sub><\/li>\n
                                    2. When the bodies move in the opposite directions, the relative velocity of 1st<\/sup> body w.r.t 2nd<\/sup> is v12<\/sub> = v1<\/sub> + v2<\/sub>.<\/li>\n<\/ol>\n

                                      Question 4.
                                      \nA car moving with a uniform velocity 54kmph is brought to rest in travelling a distance of 5m What is the retardation produced by brake?
                                      \nAnswer:
                                      \nInitial velocity of the car.,
                                      \nu = 54km\/hr
                                      \n= \\(\\frac{54 \\times 1000}{3600}\\) = 15 ms-1<\/sup>
                                      \nFinal velocity of the car, v =0.
                                      \nLet a be the retardation of the car. Then using the relation.
                                      \nv\u00b2 = u\u00b2 + 2as
                                      \n0\u00b2 = 15\u00b2 + 2a \u00d7 5
                                      \na = \\( -\\frac{15 \\times 15}{2 \\times 5}\\) = \\(\\frac{225}{10}\\) = – 22.5 ms-2<\/sup>.<\/p>\n

                                      Question 5.
                                      \nAn automobile moving with a uniform velocity of 500ms-1<\/sup> is brought to rest in travelling a distance of 5m. What is the acceleration produced by the brakes?
                                      \nAnswer:
                                      \nInitial velocity of the car, u = 500ms-1<\/sup>
                                      \nFinal velocity of the car v = 0
                                      \nLet a be the acceleration of the car.
                                      \nThen using the relation.
                                      \nv\u00b2 = u\u00b2 + 2aS
                                      \n0\u00b2 = 500\u00b2 + 2a \u00d7 5
                                      \na = \\( -\\frac{500^{2}}{10}\\) = – 25 \u00d7 103<\/sup>mS-2<\/sup>.<\/p>\n

                                      \"KSEEB<\/p>\n

                                      Question 6.
                                      \nThe equation of motion of a body is given by S = 2t + 2t\u00b2, where S is in metre and time in second. What is the acceleration of the body?
                                      \nAnswer:
                                      \nS = 2t + 2t\u00b2. This is of the standard form S = ut + 1\/2 at\u00b2
                                      \nBy comparison, 1\/2 a = 2 or a = 4m\/s\u00b2.<\/p>\n

                                      Question 7.
                                      \nDraw a velocity-time graph for a particle in the following situations:<\/p>\n

                                        \n
                                      1. Starts from rest and moving with uniform acceleration<\/li>\n
                                      2. Moving with uniform retardation.<\/li>\n<\/ol>\n

                                        Answer:
                                        \n1. Body moving the uniform acceleration.
                                        \n\"1st
                                        \n2. Body moving with uniform retardation.
                                        \n\"1st<\/p>\n

                                        Question 8.
                                        \nDefine velocity and acceleration.
                                        \nAnswer:
                                        \nVelocity is defined as the rate of displacement. Acceleration is known as the rate of change of velocity.<\/p>\n

                                        Question 9.
                                        \nTwo parallel nail tracks run North-South. Train A moves due north with a speed of 54 km h-1<\/sup> and train B moves due south with a speed of 90 km h-1<\/sup>. What is the relative speed of B with respect to A in ms-1<\/sup>?
                                        \nAnswer:
                                        \nVA<\/sub> = 54 km h-1<\/sup>
                                        \nVB<\/sub> = – 90 km h-1<\/sup>
                                        \n(north direction is chosen as positive)
                                        \nVBA<\/sub> = VB<\/sub> – VA<\/sub>
                                        \n= – 90 – 54 = -144 km h-1<\/sup>
                                        \n= – 40 ms-1<\/sup><\/p>\n

                                        Question 10.
                                        \nDifferentiate between average speed and instantaneous speed of an object.
                                        \nAnswer:
                                        \nThe distance covered in unit time is called average speed.
                                        \nVavg<\/sub> = \\(\\frac{x\\left(t_{2}\\right)-x\\left(t_{1}\\right)}{t_{2}-t_{1}}\\)
                                        \nThe speed at any instant of time is called instantaneous speed.
                                        \nVinst<\/sub>= \\(\\Delta \\mathrm{t} \\rightarrow 0 \\frac{\\Delta \\mathrm{x}}{\\Delta \\mathrm{t}}\\)<\/p>\n

                                        Question 11.
                                        \nVelocity time graph of a moving object is shown below. What is the acceleration of the object? Also, draw displacement time graph for the motion of the object.
                                        \n\"1st
                                        \nAnswer:
                                        \nAcceleration = Zero (because velocity is constant) Displacement time graph is shown below
                                        \n\"1st<\/p>\n

                                        Question 12.
                                        \nTwo straight lines are drawn on the same displacement time graph make angles 30\u00b0 and 60\u00b0 with the time axis respectively in the figure. Which line represents greater velocity? What is the ratio of the velocity of line A to that of line B?
                                        \n\"1st
                                        \nAnswer:
                                        \nVelocity = Slope of x -t graph
                                        \nVA<\/sub> = tan 30\u00b0 = \\(1_{\\sqrt{3}}\\)
                                        \nVB<\/sub> = tan 60\u00b0 = Clearly, VB<\/sub> > VA<\/sub>.
                                        \n\\(\\frac{V_{A}}{V_{B}}\\) = \\(\\frac{1_{\\sqrt{3}}}{\\sqrt{3}}\\) = \\(\\frac{1}{3}\\)<\/p>\n

                                        1st PUC Physics Motion in a Straight Line Three Mark Questions and Answers<\/h3>\n

                                        Question 1.
                                        \nThe distance x travelled by a body In a straight line is directly proportional to t\u00b2. Decide on the type of motion associated. If x \u221d t3<\/sup> what change will you observe?
                                        \nAnswer:
                                        \nx \u221d t\u00b2 = kt\u00b2 where k is a constant
                                        \na = \\(\\frac{d}{d t}\\left(\\frac{d x}{d t}\\right)\\) = \\(\\frac{\\mathrm{d}}{\\mathrm{dt}}\\) (2 k t )
                                        \n= 2 k
                                        \n\u2234 Acceleration is constant \/ uniform.
                                        \nIf x \u221d t3<\/sup> where k is a constant
                                        \na = \\(\\frac{d}{d t}\\left(\\frac{d x}{d t}\\right)\\) = \\(\\frac{\\mathrm{d}}{\\mathrm{dt}}\\) (3 k t\u00b2)
                                        \n= 6 k t
                                        \n\u2234 Acceleration is non-uniform<\/p>\n

                                        Question 2.
                                        \nDraw the following graph for an object under free-fall:<\/p>\n

                                          \n
                                        1. Variation of acceleration with respect to time.<\/li>\n
                                        2. Variation of velocity with respect to time<\/li>\n
                                        3. Variation of distance with respect to time.<\/li>\n<\/ol>\n

                                          Answer:
                                          \n1.
                                          \n\"1st
                                          \n2.
                                          \n\"1st
                                          \nLine 2 appears only when the body bounces.
                                          \n3.
                                          \n\"1st<\/p>\n

                                          Question 3.
                                          \nA body starts from rest accelerates uniformly along a straight line at the rate of 10 ms-2<\/sup> for 5 seconds. It moves for 2 seconds with a uniform velocity of 50 ms-1<\/sup>. Then it retards uniformly and comes to rest in 3s. Draw the velocity-time graph of the body and find the total distance travelled by the body.
                                          \nAnswer:
                                          \n\"1st
                                          \nTotal distance travelled = Area under the v – t graph
                                          \n= 1\/2 \u00d7 50 \u00d7 (10 + 2)
                                          \n= 300 m<\/p>\n

                                          Question 4.
                                          \nThe displacement (in metre) of a particle moving along x-axis given by x = 18 t + 5t\u00b2. Calculate:-<\/p>\n

                                            \n
                                          1. the instantaneous velocity at t = 2s<\/li>\n
                                          2. average velocity between ta 2 s and 3 s<\/li>\n
                                          3. instantaneous acceleration<\/li>\n<\/ol>\n

                                            Answer:
                                            \nx = 18 t + 5 t\u00b2
                                            \nv(t) = \\(\\frac{d x}{d t}\\) = 18 + 10 t,
                                            \na(t) = \\(\\frac{d v}{d t}\\) = 10 ms-2<\/sup>
                                            \n1. v (2) = 18 + 10 (2) = 38 ms-1<\/sup>
                                            \n2. v (2) = 38 ms-1
                                            \n<\/sup>
                                            \nv(3) = 18 + 10 (3) = 48 ms-1<\/sup>
                                            \nSince acceleration is constant,
                                            \nVavg<\/sub> \\(\\frac{v(3)+v(2)}{2}\\) = \\(\\frac{48+38}{2}\\)
                                            \n= 43 ms-1<\/sup>.<\/p>\n

                                            3. Instantaneous acceleration a = 10 ms-2<\/sup>.<\/p>\n

                                            1st PUC Physics Motion in a Straight Line FourFive Mark Questions and Answers<\/h3>\n

                                            Question 1.
                                            \nDerive an expression for velocity of the particle after time \u2018t\u2019 or Derive v = u + at using v-t graph.
                                            \nAnswer:
                                            \nConsider a particle moving along a straight line with a constant acceleration ‘a’. Let u be its initial velocity and v be its velocity after a time t. The velocity-time graph AB of this particle is shown in the figure.
                                            \n\"1st
                                            \nAcceleration of the particle is given by the slope of the v-t graph.
                                            \nFrom the graph,
                                            \nslope of the line AB = \\(\\frac{B C}{A C}=\\frac{B C}{O D}=\\frac{B D-C D}{O D}\\)
                                            \nBut, BD = v, the final velocity,
                                            \nOA = u, the initial velocity,
                                            \nOD = t, the time.
                                            \n\u2234 Slope of the graph AB is,
                                            \na = \\(\\frac{B D-C D}{O D}\\)
                                            \n= \\(\\frac{v-u}{t}\\)
                                            \nThus a = \\(\\frac{v-u}{t}\\) or v = u + at.<\/p>\n

                                            Question 2.
                                            \nWhat is the v-t graph? Derive an expression for distance covered by the particle in time \u2018t\u2019 or Derive the equation s = ut + 1\/2 at\u00b2 using v-t graph.
                                            \nAnswer:
                                            \nA graph drawn by taking time along x-axis and velocity along the y-axis is called velocity-time graph. Consider a particle moving along a straight line with a constant acceleration ‘a’. Let u be its velocity at time t = 0 and v be the velocity after t seconds. The velocity-time graph AC of this particle is shown in the figure.
                                            \n\"1st
                                            \nDistance travelled by the particle in t seconds is given by,
                                            \ns = area under v-t graph
                                            \n= area of the triangle ABC + area of the rectangle ABDO
                                            \n= 1\/2 BC \u00d7 AB + OA \u00d7 OD From the graph,
                                            \nBC = CD – BD = v – u,
                                            \nAB = OD = t and OA = u
                                            \n\u2234 s=1\/2 ( v-u) t + ut …….. (1)
                                            \nBut, a = Slope of the graph = \\(\\frac{v-u}{t}\\)
                                            \n\u2234 v – u = at
                                            \nOn substituting in (1) we have,
                                            \ns = 1\/2 at.t + ut
                                            \ns = ut + 1\/2 at\u00b2.<\/p>\n

                                            \"KSEEB<\/p>\n

                                            Question 3.
                                            \nDerive an expression for the velocity of the particle after covering distance \u2018s\u2019 OR Derive the equation v\u00b2 = u\u00b2 + 2as using the velocity-time graph.
                                            \nAnswer:
                                            \nparticle moving along a straight line with a constant acceleration ‘a’. Let u be its velocity at time t = 0 and ‘v’ be the velocity after ‘t’ seconds. The velocity-time graph AC of this particle is shown in the figure.
                                            \n\"1st
                                            \nDistance travelled by the particle is given by,
                                            \ns = area under v-t graph
                                            \n= area of the trapezium OACD
                                            \n= 1\/2 (OA+CD) \u00d7 OD
                                            \nBut OA = u, CD = v and OD = t
                                            \n\u2234 s = 1\/2 (u + v) t or
                                            \n2s = (u+v) t ………(1)
                                            \nBut, a = slope of the graph = \\(\\frac{v-u}{t}\\)
                                            \n\u2234 t = \\(\\frac{v-u}{a}\\)
                                            \nOn substituting in (1),
                                            \n2s = \\((u+v)\\left(\\frac{v-u}{a}\\right)\\)
                                            \ni.e., v\u00b2 – u\u00b2 = 2as
                                            \nor v\u00b2 = u\u00b2 + 2as.<\/p>\n

                                            Question 4.
                                            \nDerive an expression for distance travelled during the nth<\/sup> second of motion.
                                            \nOR
                                            \nDerive sn<\/sub>= u + a(n – 1\/2)
                                            \nAnswer:
                                            \nConsider a particle moving with uniform acceleration \u2018a\u2019.Let u be the initial velocity of the particle.
                                            \nDistance travelled during nth<\/sup> second = Distance travelled in \u2018n\u2019 seconds – Distance travelled in (n-1) seconds
                                            \nsn<\/sub> = un + 1\/2 an\u00b2 – [ u(n-1) + 1\/2 a(n-1)\u00b2]
                                            \n= un + 1\/2 an\u00b2 – [ un-u + 1\/2 a(n\u00b2-2n +1)]
                                            \n= un + 1\/2 an\u00b2 – [ un-u + 1\/2 an\u00b2-an + 1\/2 a]
                                            \n= un + 1\/2 an\u00b2 – un+u – 1\/2 an\u00b2+an – 1\/2 a
                                            \n= u + an – 1\/2 a
                                            \nsn<\/sub> =u + a(n – 1\/2)<\/p>\n

                                            Question 5.
                                            \nDerive an expression for acceleration in terms of distance travelled in two successive equal interval of time.
                                            \nOR
                                            \nDerive the equation a = \\(\\frac{s_{2}-s_{1}}{t^{2}}\\)
                                            \nAnswer:
                                            \nConsider a particle moving with uniform acceleration a. Let it start from a point with initial velocity \u2018u\u2019 and covers the distances s1<\/sub> and s2<\/sub> in two successive equal interval of time ‘t’ each. Distance covered in the first interval
                                            \ns1<\/sub> = ut + 1\/2 at\u00b2 ………. (1)
                                            \nDistance covered at the end of time 2t,
                                            \ns1<\/sub> + s2<\/sub> = u(2t) + 1\/2 a (2t)\u00b2
                                            \ns1<\/sub> + s2<\/sub> = 2ut + 2at\u00b2……………. (2)
                                            \n(s1<\/sub> + s2<\/sub> ) – 2s1<\/sub> = (2ut + 2at\u00b2) – (2ut + at\u00b2)
                                            \ns2<\/sub> – s1<\/sub> = at\u00b2
                                            \na = \\(\\frac{s_{2}-s_{1}}{t^{2}}\\)<\/p>\n

                                            Question 6.
                                            \nDefine uniform velocity obtain v\u00b2 – u\u00b2 = 2as from the v – t graph, where the symbols have the usual meaning.
                                            \nAnswer:
                                            \nIf a body undergoes equal displacements in equal intervals of time, however small interval of time may be then velocity is said to be uniform velocity.
                                            \nparticle moving along a straight line with a constant acceleration ‘a’. Let u be its velocity at time t= 0 and ‘v’ be the velocity after ‘t’ seconds. The velocity-time graph AC of this particle is shown in the figure.
                                            \n\"1st
                                            \nDistance travelles by the particle is given by,
                                            \ns = area under v-t graph
                                            \n= area of the trapezium OACD
                                            \n= 1\/2 (OA+CD) \u00d7 OD
                                            \nBut OA = u, CD = v and OD = t
                                            \n\u2234 s = 1\/2 (u + v) t or 2s = (u+v) t ………(1)
                                            \nBut, a = slope of the graph = \\(\\frac{v-u}{t}\\)
                                            \n\u2234 t = \\(\\frac{v-u}{a}\\)
                                            \nOn substituting in (1), 2s = \\((u+v)\\left(\\frac{v-u}{a}\\right)\\)
                                            \ni.e., v\u00b2 – u\u00b2 = 2as
                                            \nor v\u00b2 = u\u00b2 + 2as.<\/p>\n

                                            \"KSEEB<\/p>\n

                                            Question 7.
                                            \nA particle is moving with uniform acceleration covers a distance of 45 m in 6th<\/sup> second and 75 m in 12th<\/sup> second during its motion. Calculate the displacement of the particle after 20s?
                                            \nAnswer:
                                            \nS6<\/sub> = 45m, S12<\/sub>=75m
                                            \nusing Sn<\/sub> = u + a (n – 1\/2 )
                                            \ni.e., S6<\/sub> \u00a0= u + a or 45 = u + 5.5a ……… (1)
                                            \nS12<\/sub>= u + a(12 – 1\/2) or 75 = u + a(11.5) …….. (2)
                                            \nEq(2) – (1) gives 6 a = 30
                                            \nOr a = 30\/6 = 5m\/s\u00b2
                                            \nFrom (1), 45 = u + 5 \u00d7 5.5
                                            \nu = 17.5m\/s.
                                            \nDistance travelled in t = 20s,
                                            \nS = ut+ 1\/2at2
                                            \n= 17.5 \u00d7 20 + 1\/2 \u00d7 5 \u00d7 20\u00b2
                                            \n=1350m.<\/p>\n

                                            Question 8.
                                            \nDraw a velocity-time graph of uniformly accelerated motion in one dimension. From the velocity-time graph of uniformly accelerated motion deduce the equations of motion in distance and time.
                                            \nAnswer:
                                            \nA graph drawn by taking time along x-axis and velocity along y-axis is called velocity-time graph. Consider a particle moving along a straight line with a constant acceleration ‘a’. Let u be its velocity at time t = 0 and v be the velocity after t seconds. The velocity-time graph AC of this particle is shown in the figure.
                                            \n\"1st
                                            \nDistance travelled by the particle in t seconds is given by,
                                            \ns = area under v-t graph
                                            \n= area of the triangle ABC + area of the rectangle ABDO
                                            \n= 1\/2 BC \u00d7 AB + OA \u00d7 OD From the graph,
                                            \nBC = CD – BD = v – u,
                                            \nAB = OD = t and OA = u
                                            \n\u2234 s=1\/2 ( v-u) t + ut …….. (1)
                                            \nBut, a = Slope of the graph = \\(\\frac{v-u}{t}\\)
                                            \n\u2234 v – u = at
                                            \nOn substituting in (1) we have,
                                            \ns = 1\/2 at.t + ut
                                            \ns = ut + 1\/2 at\u00b2.<\/p>\n

                                            Question 9.
                                            \nDerive an equation for the distance covered by a uniformly accelerated body in the nth<\/sup> second of its motion. A body travels half its total path in the last second of its fall from rest. Calculate the time of its fall.
                                            \nAnswer:
                                            \nConsider a particle moving with uniform acceleration \u2018a\u2019.Let u be the initial velocity of the particle.
                                            \nDistance travelled during nth second = Distance travelled in \u2018n\u2019 seconds – Distance travelled in (n-1) seconds
                                            \nsn<\/sub> = un + 1\/2 an\u00b2 – [ u(n – 1) + 1\/2 a(n – 1)\u00b2]
                                            \n= un + 1\/2 an\u00b2 – [ un – u + 1\/2 a(n\u00b2 – 2n + 1)]
                                            \n= un + 1\/2 an\u00b2 – [ un – u + 1\/2 an\u00b2 – an + 1\/2 a]
                                            \n= un + 1\/2 an\u00b2 – un + u – 1\/2 an\u00b2 + an – 1\/2 a
                                            \n= u + an – 1\/2 a
                                            \nsn<\/sub> =u + a(n – 1\/2)
                                            \nsn<\/sub> = 4 + a (n – 1\/2)
                                            \nGiven u = 0, a = g
                                            \ns = 1\/2 g t\u00b2 = 1\/2 g n\u00b2
                                            \nBy the given condition 1\/2 s = sn<\/sub>
                                            \n\u21d2 \\(\\frac{\\mathrm{g} n^{2}}{4}\\) = g(n – 1\/2)
                                            \n\u21d2 n\u00b2 – 4 n + 2 = 0
                                            \n\u21d2 n = \\(\\frac{4 \\pm \\sqrt{16-8}}{2}\\)
                                            \n\u21d2 n = 2 \u00b1 \\(\\sqrt{2}\\)
                                            \n\u2234 Time of the body\u2019s fall = (2 \u00b1 \\(\\sqrt{2}\\)) s<\/p>\n

                                            1st PUC Physics Motion in a Straight Line Numerical Problems Questions and Answers<\/strong><\/p>\n

                                            Question 1.
                                            \nA man runs from his home to office at a speed of 2 ms-1<\/sup> on a straight road and returns back to home at a speed of 4 ms-1<\/sup>. Find<\/p>\n

                                              \n
                                            1. Average speed and<\/li>\n
                                            2. average velocity.<\/li>\n<\/ol>\n

                                              Solution:
                                              \n1. To find the average speed. Let ‘s’ is the distance between the home and the office.
                                              \n\u2234 Time taken to reach the office
                                              \nt1<\/sub> \\(\\frac{\\text { distance }}{\\text { velocity }}\\) = \\(\\frac{s}{2}\\)
                                              \nSimilarly, time taken to walk back to the home from the office is t2<\/sub> = \\(\\frac{s}{4}\\)
                                              \n\u2234 Total time taken = t1<\/sub> + 2<\/sub> = \\(\\frac{s}{2}\\) + \\(\\frac{s}{4}\\) = \\(\\frac{3 s}{4}\\)
                                              \nTotal distance moved = s + s = 2s
                                              \nHence. the average speed \\(=\\frac{\\text { toal distance moved }}{\\text { total time taken }}\\)
                                              \n\"1st = 2.67 ms-1<\/sup>
                                              \n2. To find the average velocity Since the man comes back to home, the final and initial position is same.
                                              \n\u2234 Total displacement is zero.
                                              \nHence average velocity
                                              \n\\(=\\frac{\\text { toal displacement }}{\\text { total time }}\\) = 0<\/p>\n

                                              Question 2.
                                              \nA car moving along a straight high-way with a speed of 35 m\/s is brought to stop within a distance of 200 m. What Is the retardation, and how long does it take for the car to stop? Solution:
                                              \nInitial speed of the car, u = 35 ms-1<\/sup>
                                              \nFinal speed of the car, v = 0
                                              \nLet a be the retardation of the car.
                                              \nThen using the relation,
                                              \nv\u00b2 = u\u00b2 + 2as, we have 0 = (35)\u00b2 + 2a \u00d7 200
                                              \na = – \\(\\frac{35 \\times 35}{2 \\times 200}\\) = – 3.06 ms-2<\/sup>
                                              \nLet t be the time taken by the car to come to stop. Then,
                                              \nv = u + at ;
                                              \n0 = 35 – 3.06 \u00d7 t or
                                              \nt = \\(\\frac{35}{3.06}\\) = 11.43 s.<\/p>\n

                                              Question 3.
                                              \nA car starts from rest and accelerates from rest uniformly for 10 s to a velocity of 36 kmhr-1<\/sup>. It then runs at a constant velocity and is finally brought to rest in 50 m with uniform retardation. If the total distance covered by the car is 500 m, find<\/p>\n

                                                \n
                                              1. acceleration and<\/li>\n
                                              2. retardation.<\/li>\n<\/ol>\n

                                                Solution:
                                                \n1. Initial velocity of the car u = 0;
                                                \nFinal velocity v = 36 kmhr-1<\/sup>
                                                \n= \\(\\frac{36 \\times 1000}{3600}\\) ms-1<\/sup>
                                                \n= 10 ms-1<\/sup>
                                                \ntime taken t=10 s
                                                \n\u2234 acceleration a = \\(\\frac{\\mathrm{v}-\\mathrm{u}}{\\mathrm{t}}\\) = \\(\\frac{\\mathrm{10}-\\mathrm{0}}{\\mathrm{10}}\\) 1ms-2
                                                \n<\/sup>
                                                \n2. To find the retardation
                                                \nDuring the motion with retardation, Initial velocity of the car u = 10 ms-1<\/sup>
                                                \nFinal velocity v = 0, distance travelled, s = 50 m
                                                \nUsing the relation v\u00b2 = u\u00b2 + 2as,
                                                \n0 = 10\u00b2 – 2a \u00d7 50
                                                \na = \\(-\\frac{100}{2 \\times 50}\\) = -1 ms-2<\/sup><\/p>\n

                                                Question 4.
                                                \nA body travelling with a uniform acceleration travels a distance of 100m in the first 5 second and 200m in the next 5 second, calculate the initial velocity and acceleration of the body.
                                                \nSolution:
                                                \nIf s1<\/sub> and s2<\/sub> are the distances travelled by a body in two successive intervals of t seconds, then the acceleration is given
                                                \nby, a = \\(\\frac{s_{2}-s_{1}}{t^{2}}=\\frac{200-100}{5^{2}}=\\frac{100}{25}\\) = 4ms-2<\/sup>
                                                \nwe have, s = ut + 1\/2 at\u00b2
                                                \nHere, s = s1<\/sub> = 100m, a = 4ms-2<\/sup>, t = 5 s
                                                \n100 = u \u00d7 5 + 1\/2 \u00d7 4 \u00d7 25 OR
                                                \n5u = 50 \u2234u= 10ms-1<\/sup>.<\/p>\n

                                                \"KSEEB<\/p>\n

                                                Question 5.
                                                \nA body moving with uniform acceleration along a straight line covers 60m in the 5th<\/sup> second and 80m in the 8th<\/sup> second of its motion. Calculate the initial velocity and uniform acceleration.
                                                \nSolution:
                                                \nThe distance travelled during the nth<\/sup> second is given by,
                                                \nsn<\/sub> = u + a\/2(2n – 1), n = 5 and sn<\/sub> = 60
                                                \n\u2234 60 = u+ a\/2 (2 \u00d7 5 – 1)
                                                \n60 = u + 9a\/2 ………… (1)
                                                \nSimilarly, for n = 8, sn<\/sub> =80
                                                \n\u2234 80 = u+ a\/2 (2 \u00d7 8 – 1)
                                                \n80 = u + 15a\/2 …………….. (2)
                                                \n(2) – (1), gives
                                                \n20 = 6a\/2 = 3a
                                                \n\u2234 a = 20\/3 ms-2<\/sup>
                                                \nUsing this in equation (1), we get
                                                \n60 = u + \\(\\frac{9 \\times 20}{2 \\times 3}\\)
                                                \nu = 60 – 30 = 30 ms-1<\/sup>.<\/p>\n

                                                Question 6.
                                                \nA stone is thrown vertically upwards with an initial velocity of 19.6 m-1<\/sup>. After how long will the stone strike the ground? Take g = 9.8 ms-2<\/sup>.
                                                \nSolution:
                                                \nInitial velocity u = +19.6 ms-1<\/sup>. The stone after reaching the highest point comes back to the initial point
                                                \n\u2234 The displacement = 0.
                                                \na = g = – 9.8 ms-2<\/sup>
                                                \nLet t be the time taken to reach the ground
                                                \nFrom s = ut + 1\/2at\u00b2, we have
                                                \n0=19.6t – 1\/2 \u00d7 9.8 \u00d7 t\u00b2
                                                \n4.9t\u00b2 – 19.6t = 0
                                                \nDividing throughout by 4.9t, we have t = 4s. Hence, the stone reaches the ground after 4s.<\/p>\n

                                                \"KSEEB<\/p>\n

                                                Question 7.
                                                \nA stone is thrown vertically upwards with a velocity of 10ms-1<\/sup> from the top of a tower 40m tall and it finally falls to the ground,<\/p>\n

                                                  \n
                                                1. Find the time taken by the stone to reach the ground<\/li>\n
                                                2. After how long will it pass through the point of projection<\/li>\n
                                                3. Calculate the velocity when it strikes the ground. Take g = 10 ms-2<\/sup><\/li>\n<\/ol>\n

                                                  Solution:
                                                  \nLet the stone be thrown upward with a velocity u = 10ms-1<\/sup> from the tower AB of height 40 m. Let C be the highest point reached.
                                                  \n\"1st
                                                  \n1. Let t be the total time taken by the stone to move from B to C and back to the ground.
                                                  \n\u2234 The displacement = BA = – 40 m (directed downward since displacement is measured from the initial to the final position)
                                                  \na = -10 ms-2<\/sup>.
                                                  \nFrom s = ut + 1\/2at\u00b2, we have
                                                  \n– 40= 10 \u00d7 t – \\(\\frac{10 t^{2}}{2}\\)
                                                  \n– 40 = 10t – 5t\u00b2
                                                  \n– 8 = 2t -t\u00b2
                                                  \nt\u00b2 – 2t – 8 = 0
                                                  \n(t – 4) (t + 2) = 0.
                                                  \nOR
                                                  \nt = + 4s or – 2s
                                                  \nAs time cannot be negative total time taken to reach the ground = 4s.<\/p>\n

                                                  2. For the motion from B to C and back to B,
                                                  \ndisplacement = 0
                                                  \nu = 10ms-1<\/sup>, a = g = -10ms-2<\/sup>
                                                  \nLet t1<\/sub> be the time taken.
                                                  \nFrom s = ut + 1\/2at\u00b2, we have
                                                  \n0= 10t1<\/sub> – \\(\\frac{10 t_{1}^{2}}{2}\\) or 101<\/sub>\u00b2 – 5t1<\/sub>\u00b2 = 0
                                                  \n\u2234 t1<\/sub> = 2s.
                                                  \nHence, the stone reaches the point of projection after 2 seconds.<\/p>\n

                                                  3. Let ‘v’ be the velocity of the stone when it strikes the ground. For the motion from B to C and back to A, we have,
                                                  \nu = 10 ms-1<\/sup>, a = g = -10 ms-2<\/sup>, t = 4s
                                                  \nFrom, v = u + at, we have
                                                  \nv = 10 – 10 \u00d7 4 = – 30 ms-1<\/sup>.
                                                  \nThe -ve sign indicates that it is directed downward.<\/p>\n

                                                  Question 8.
                                                  \nA stone is projected vertically upwards from the ground with a velocity of 49 ms-1<\/sup>. At the same time, another stone is dropped from a height 98m to fall freely along the same path as the first. Find where and when the two stones meet each other.
                                                  \nSolution:
                                                  \n\"1st
                                                  \nLet P be the stone dropped from B at a height 98m from the ground A. Q is the other stone thrown up vertically upwards with a velocity 49ms-1<\/sup> at the same instant. Let the two stones meet at C after a time t seconds.
                                                  \nFor the stone P, we have
                                                  \nu = 0, a = + g, s = + h (=BC), t = t
                                                  \nFrom s = ut + 1\/2 at\u00b2, we have,
                                                  \n+h = 0 \u00d7 t + 1\/2 gt\u00b2 or h = 1\/2 gt\u00b2 ………. (1)
                                                  \nFor the stone Q,
                                                  \nu = 49ms-1<\/sup>,
                                                  \na = -g,
                                                  \ns = +AC = +(98 – h),
                                                  \n(98 – h) = 49t – 1\/2 gt\u00b2
                                                  \nUsing eqn. (1) in (2), we have,
                                                  \n(98 – h) = 49t – h or 49t =98
                                                  \n\u2234 t = 2s.
                                                  \nUsing this in (1), we have
                                                  \nh = BC = 1\/2 \u00d7 9.8 \u00d7 4 = 19.6 m
                                                  \n\u2234 AC =96 – 19.6 = 76.4 m
                                                  \n\u2234 Two stones meet at a height of 76.4 m from the ground, after 2 seconds.<\/p>\n

                                                  Question 9.
                                                  \nA train is moving southwards with a speed of 30ms-1<\/sup>. A man is running on the roof of the train with speed of 5ms-1<\/sup> with respect to the train. Find the velocity of the man as observed by the person on the ground if the man is running<\/p>\n

                                                    \n
                                                  1. southwards<\/li>\n
                                                  2. northwards.<\/li>\n<\/ol>\n

                                                    Solution:
                                                    \nIf vA<\/sub> and vB<\/sub> are the velocity of the two bodies moving along the same direction with respect to the ground, relative velocity of A with respect to B is given by,
                                                    \nVAB<\/sub> = VA<\/sub> – VB<\/sub> …………. (1)
                                                    \nHere, the velocity of the man with respect to the train,
                                                    \nvAB<\/sub> = 5ms-1<\/sup>.
                                                    \nvelocity of the train with respect to the ground is,
                                                    \nvB<\/sub> = 30 ms-1<\/sup>
                                                    \n\u2234 velocity of the man with respect to the ground,
                                                    \nvA<\/sub> = vB<\/sub> + vAB<\/sub> = 30 + 5 = 35 ms-1<\/sup> if the man moves southwards.
                                                    \nOn the other hand, if the man moves northwards,
                                                    \nvAB<\/sub> = – 5 ms-1<\/sup>
                                                    \n\u2234 VA<\/sub> = VB<\/sub> – VAB<\/sub>
                                                    \n= 25 ms-1<\/sup>.<\/p>\n

                                                    \"KSEEB<\/p>\n

                                                    Question 10.
                                                    \nA police van moving on a highway with a speed of 10ms-1<\/sup> fires a bullet at a smuggler’s car speeding away in the same direction with a speed of 30ms-1<\/sup>. If the muzzle velocity of the bullet is 140ms-1<\/sup>, with what speed the bullet will hit the smuggler’s car?
                                                    \nSolution:
                                                    \nSpeed of the police car vp<\/sub> = 10 ms-1<\/sup>.
                                                    \nMuzzle velocity of the bullet vM<\/sub>=140 ms-1<\/sup>
                                                    \nNet velocity of the bullet,
                                                    \nvA<\/sub> = vp<\/sub> + vM<\/sub> = 10 + 140 = 150 ms-1<\/sup>
                                                    \n(\u2235 gun is mounted on the van)
                                                    \nspeed of the smuggler’s car vB<\/sub> = 30 ms-1<\/sup>
                                                    \n\u2234 Velocity of the bullet relative to smuggler’s car,
                                                    \nvAB<\/sub> = VA<\/sub> – VB<\/sub>
                                                    \n= 150 – 30 = 120 ms-1<\/sup>.<\/p>\n

                                                    Question 11.
                                                    \nTwo trains A and B are moving on two parallel tracks with uniform speed of 20ms-1<\/sup> in the same direction with A ahead of B. The driver of B decides to overtake A and accelerates by 1 ms-2<\/sup>. After 50 seconds, they are moving together, find the original distance of separation.
                                                    \nSolution:
                                                    \nInitially, both A and B are moving with the same velocity in the same direction,
                                                    \n\u2234 The initial relative velocity .
                                                    \nuAB<\/sub> = uA<\/sub> – uB<\/sub> = 0
                                                    \nwhen t = 50 s,
                                                    \nrelative acceleration = aAB<\/sub> = 1 ms-2<\/sup>
                                                    \nIf s is the distance moved, then
                                                    \nFrom s = ut + 1\/2at\u00b2
                                                    \nS = uAB<\/sub> \u00d7 t + 1\/2 aAB<\/sub>t\u00b2
                                                    \nS = 0 \u00d7 t + 1\/2 \u00d7 1 \u00d7 50\u00b2
                                                    \n= 2500\/2 = 1250 m.
                                                    \nHence, the Initial distance of separation is 1250m.<\/p>\n

                                                    Question 12.
                                                    \nA body is starting from rest and is subjected to a uniform acceleration of 10ms-2<\/sup> determine.<\/p>\n

                                                      \n
                                                    1. Velocity of the body at the end of 5s<\/li>\n
                                                    2. Displacement of the body In the first 3s<\/li>\n
                                                    3. velocity of the body after a displacement of 20m.<\/li>\n
                                                    4. Displacement of the body In the 4th<\/sup> second of its motion (Tumkur 05)<\/li>\n<\/ol>\n

                                                      Solutions:
                                                      \n1. Initial velocity, u = 0
                                                      \nacceleration, a = 10ms-2<\/sup>
                                                      \ntime, t = 5s,
                                                      \nv=?
                                                      \nFrom, the Equation, v = u + at we have,
                                                      \nv= 0 + 10 \u00d7 5
                                                      \nv= 50 ms-1<\/sup><\/p>\n

                                                      2. Initial velocity, u = 0
                                                      \nacceleration a = 10 ms-2<\/sup>
                                                      \ntime, t = 36s
                                                      \nDistance travelled, s =?
                                                      \nFrom the equation s = ut+ 1\/2 at\u00b2 we have
                                                      \n= 0 + 1\/2 \u00d7 10 \u00d7 9
                                                      \ns = 45 m<\/p>\n

                                                      3. Initial velocity u= 0,
                                                      \nacceleration a = 10ms-2<\/sup>
                                                      \nDisplacement s = 20m
                                                      \nvelocity v=?
                                                      \nFrom the equation v\u00b2= u\u00b2 + 2as ,
                                                      \n= 0\u00b2 + 2 \u00d7 10 \u00d7 20
                                                      \n= 400
                                                      \nv = \\(\\sqrt{400}\\) = 20ms-1<\/sup><\/p>\n

                                                      4. Initial velocity u=0
                                                      \nAcceleration a = 10ms-1<\/sup>
                                                      \nn = 4 sec
                                                      \nDisplacement S4<\/sub> = ?
                                                      \nFrom the Equation, Sn<\/sub> = u+ a\/2 (2n-1)
                                                      \nS4<\/sub>=0 + 10\/2 (2 \u00d7 4 – 1) = 0 + 5(7)
                                                      \nS4<\/sub> = 35m.<\/p>\n

                                                      \"KSEEB<\/p>\n

                                                      Question 13.
                                                      \nA stone dropped from the top of a building travels 24.5 m in the last second of its fall. Find the height of the tower, (g = 9.8 m\/s2)
                                                      \nAnswer:
                                                      \nDistance travelled in the last second of its fall is Sn<\/sub> = 24.5m
                                                      \nAlso, Sn<\/sub> = ut + a (n – 1\/2)
                                                      \nBut u = 0, a = g
                                                      \n\u2234 Sn<\/sub> = 0 + g (n – 1\/2)
                                                      \n\"1st
                                                      \n24.5 = 9.8n – 4.9
                                                      \nn =\\(\\frac{24.5+4.9}{9.8}\\) = 3 seconds.
                                                      \nwe know that S = ut + 1\/2 at\u00b2
                                                      \n\u2234 Height of the tower;
                                                      \nh = 0 \u00d7 3+ 1\/2 \u00d7 9.8(3)\u00b2
                                                      \n(\u2235 s = h, u = 0, a = g = 9.8m\/s)
                                                      \n\u2234 h = 0 + 4.9(3)\u00b2
                                                      \n= 44.1m.<\/p>\n

                                                      Question 14.
                                                      \nA body moving along a straight line with uniform acceleration covers 23m and 35m respectively In the 5th<\/sup> and 8th<\/sup> second of its motion calculate the distance travelled by the body in<\/p>\n

                                                        \n
                                                      1. 10th<\/sup> second<\/li>\n
                                                      2. complete 10 second and what is its velocity at the end of 6 seconds.<\/li>\n<\/ol>\n

                                                        Answer:
                                                        \nS5<\/sub> = 23m, S8<\/sub> = 35m
                                                        \nW.k.that sn<\/sub> = u + a(n – 1\/2)
                                                        \ni.e., S5<\/sub> = u + a(5 – 1\/2) or 23 = u + 4.5 a….(1)
                                                        \nS8<\/sub> = u + a(8 – 1\/2) or 35 = u + a(7.5)……….(2)
                                                        \nEq(2) – (1) gives 3a = 12
                                                        \nor a = 12\/3 = 4 m\/s\u00b2
                                                        \nFrom (1), 23 = u + 4.5 \u00d7 4
                                                        \nu = 5 m\/s
                                                        \n1. Distance traveled in 10th<\/sup> second
                                                        \nS10<\/sub>=u + a(n – 1\/2) = 5 + 4(10 – 1\/2) = 43m<\/p>\n

                                                        2. Distance traveled in 10s is
                                                        \nS= ut + 1\/2at\u00b2 = 5 \u00d7 10 + 1\/2 \u00d7 4 \u00d7 10\u00b2 = 250m
                                                        \nVelocity at the end of 6 seconds is v = u + at = 5 + 4 \u00d7 6 =29 m\/s.<\/p>\n

                                                        Question 15.
                                                        \nA point object is thrown vertically upwards at such a speed that it returns the thrown after 6 seconds. With what speed was it thrown up and how high did it rise? Plot speed-time graph for the object and use it to find the distance travelled by it in the last second of its journey.
                                                        \nAnswer:
                                                        \nTime of rise t1<\/sub> = time of fall = t2<\/sub>
                                                        \nt = 6 s
                                                        \nv = u – g t (upward direction is chosen as positive)
                                                        \nv = – u
                                                        \n– u=u-gt
                                                        \n\u21d2 – 2u = – 9.8 \u00d7 6
                                                        \n\u21d2 u = 29.4 ms-1<\/sup>
                                                        \nv\u00b2 = u\u00b2 – 2 gh
                                                        \nAt the top, v = 0
                                                        \n\u21d2 h = \\(\\frac{u^{2}}{2 g}\\)
                                                        \n= \\(\\frac{(29.4)^{2}}{2 \\times 9.8}\\)
                                                        \nh = 44.1 m
                                                        \nInitial speed = 29.4 ms-1<\/sup>
                                                        \nHeight the object attained = 44.1 m
                                                        \n\"1st
                                                        \nDistance covered in last second = area under speed – time graph.
                                                        \n= 1\/2 \u00d7 (6 – 5) \u00d7\u00a0(19.6 + 29.4)
                                                        \n= 24.5 m<\/p>\n

                                                        Question 16.
                                                        \nA race car is moving on a straight road with a speed of 180 km h-1<\/sup>. If the driver stops the car in 25 s by applying the brakes, calculate the distance covered by the car during the time brakes are applied. Assume acceleration of the car is uniform throughout the retarding motion.
                                                        \nAnswer:
                                                        \nInitial speed \u03bc = 180 km h-1<\/sup>
                                                        \n\u03bc = 50 ms-1<\/sup>
                                                        \nt = 25 s
                                                        \nv = u – at
                                                        \nv = 0
                                                        \n\u2234 0 = 50 – a \u00d7 25
                                                        \n\u21d2 a = 2 ms-2<\/sup>
                                                        \nv\u00b2 = u\u00b2 – 2as
                                                        \n\u21d2 s = \u03bc\u00b2\/2 a
                                                        \n\\(\\frac{50^{2}}{2 \\times 2}\\)
                                                        \n\u2234 Stopping distance = 625 m<\/p>\n

                                                        Question 17.
                                                        \nA body covers 12 m in the 2nd<\/sup> second and 20 m in the 4th<\/sup> second. Find what distance the body will cover in the 4 seconds after the 5th<\/sup> second.
                                                        \nAnswer:
                                                        \nsn<\/sub>= u + a\/2 (2n -1)
                                                        \ns2<\/sub> = 12
                                                        \n\u21d2 12 = u+ a\/2 (2 \u00d7 2 – 1)
                                                        \n\u21d2 12 = u + 3 a\/2 ………. (1)
                                                        \ns4<\/sub> = 20
                                                        \n\u21d2 20 = u+ a\/2 (2 \u00d7 4 – 1)
                                                        \n\u21d2 20 = u + 7a\/2 ………. (2)
                                                        \nSubtract equation (1) from (2)
                                                        \ns = 4a\/2
                                                        \n\u21d2 a = 4 ms-2<\/sup>
                                                        \n12 = u + 3\/2 \u00d7 4
                                                        \n\u21d2 u = 6 ms-1<\/sup>
                                                        \nDistance covered in the 4 seconds after 5th<\/sup> second,
                                                        \n= S9<\/sub> – S5<\/sub>
                                                        \n= u (9) + 1\/2 a (9)\u00b2 – [u(5) + 1\/2 a (5)\u00b2 ]
                                                        \n= 4u + a\/2 (81 – 25)
                                                        \n= 4 \u00d7 6 + 4\/2 \u00d7 56 = 136 m.<\/p>\n

                                                        \"KSEEB<\/p>\n

                                                        Question 18.
                                                        \nA ball is thrown vertically upwards with a velocity of 20 ms-1<\/sup> from the top of a building of height 25 m from the ground.<\/p>\n

                                                          \n
                                                        1. How high will the ball reach?<\/li>\n
                                                        2. How long will the ball tale to reach the ground ? (g = 10 ms-2<\/sup>)<\/li>\n<\/ol>\n

                                                          Answer:
                                                          \nu = 20 ms-1<\/sup>
                                                          \nh = 25 m
                                                          \ng = 10ms-2<\/sup>
                                                          \n1. v\u00b2 = u\u00b2 – 2gh1<\/sub>
                                                          \nAt the top, v = 0
                                                          \n\u21d2 h1<\/sub> = \\(\\frac{\\mathrm{u}^{2}}{2 \\mathrm{g}}\\)
                                                          \n= \\(\\frac{20^{2}}{2 \\times 10}\\)
                                                          \n= 20 m
                                                          \n\u2234 Height the ball reaches = 20 + 25 = 45 m from the ground.
                                                          \n2. – h = u t – 1\/2 g t\u00b2
                                                          \n– 25 = 20 t – 1\/2 \u00d7 10 t\u00b2
                                                          \n\u21d2 5 t\u00b2 – 20 t – 25 = 0
                                                          \n\u21d2 t\u00b2 – 4 t – 5 = 0
                                                          \n\u21d2 (t – 5 ) (t + 1 ) = 0
                                                          \n= t = 5, -1
                                                          \n\u2234 t = 5 s (Since t cannot be negative) Total time to reach the ground = 5 s<\/p>\n

                                                          1st PUC Physics Motion in a Straight Line Hard Questions and Answers<\/strong><\/p>\n

                                                          Question 1.
                                                          \nThe speed of a train increases at a constant rate a from zero to v and then remains constant for an interval, and finally decreases to zero at a constant rate p. If L be the total distance covered prove that the total time taken
                                                          \nis \\(\\frac{L}{v}+\\frac{v}{2}\\left(\\frac{1}{\\infty}+\\frac{1}{\\beta}\\right)\\)
                                                          \nAnswer:
                                                          \nDuring time t1<\/sub>, the train accelerates uniformly from 0 to V.
                                                          \nv = \u03b1 t1<\/sub>
                                                          \nDistance covered d1<\/sub> = 1\/2 \u03b1 t1<\/sub>\u00b2
                                                          \n= 1\/2 (\u221d t1<\/sub>) t1<\/sub>
                                                          \nd1<\/sub> = 1\/2 v t1<\/sub>
                                                          \n\u21d2 \\(t_{1}=\\frac{2 d_{1}}{v} \\alpha\\)
                                                          \nDuring time t2<\/sub>, the train travels at a uniform speed of v
                                                          \nDistance covered d2<\/sub> = vt2<\/sub>
                                                          \n\u21d2 t2<\/sub> = \\(\\frac{\\mathrm{d}_{2}}{\\mathrm{v}}\\)
                                                          \nDuring time t3<\/sub>, the train decelerates uniformly from v to 0.
                                                          \n0 = v – \u03b2 t3<\/sub>
                                                          \n\u21d2 v = \u03b2 t3<\/sub>
                                                          \nDistance covered d3<\/sub> = vt3<\/sub> – 1\/2 \u03b2 t3<\/sub>\u00b2
                                                          \n= vt3<\/sub> – 1\/2 (\u03b2 t3<\/sub>) t3<\/sub>
                                                          \nd3<\/sub> = vt3<\/sub> – 1\/2 v t3<\/sub>
                                                          \n\u21d2 d3<\/sub> = 1\/2 v t3<\/sub>
                                                          \n\u21d2 t3<\/sub> = \\(\\frac{2 \\mathrm{d}_{3}}{\\mathrm{v}}\\)
                                                          \nTotal time T = t1<\/sub> + t2<\/sub> + t3<\/sub>
                                                          \n\"1st
                                                          \nUsing v\u00b2 = u\u00b2 + 2as
                                                          \nd1<\/sub> = \\(\\frac{v^{2}}{2 \\alpha}\\) and d3<\/sub> = \\(\\frac{v^{2}}{2 \\beta}\\)
                                                          \n\u2234 T = \\(\\frac{L}{v}+\\frac{v}{2 \\alpha}+\\frac{v}{2 \\beta}\\)
                                                          \nT = \\(\\frac{L}{v}+\\frac{v}{2}\\left(\\frac{1}{\\alpha}+\\frac{1}{\\beta}\\right)\\)
                                                          \nHence proved.<\/p>\n

                                                          Question 2.
                                                          \nIn a car race, car A takes ‘t’ seconds less than car B and passes the finish line with a velocity ‘V’ more than that of the car B. Of the cars start from rest and travel with constant acceleration a1<\/sub> and a2<\/sub> respectively, show that
                                                          \nV = \\(t \\sqrt{a_{1} a_{2}}\\).
                                                          \nAnswer:
                                                          \nLet the car A take tA<\/sub> seconds to finish the race.
                                                          \nCar B takes tB<\/sub> seconds to finish the race. Let the final velocities of the cars be vA<\/sub> and vB<\/sub>
                                                          \nBy the given conditions,
                                                          \ntB<\/sub> – tA<\/sub> = t
                                                          \nvA<\/sub> = a1<\/sub> tA<\/sub>
                                                          \nvB<\/sub> = a2<\/sub> tB<\/sub>
                                                          \nvA<\/sub> – vB<\/sub> = v
                                                          \nSince the distance d covered is the same,
                                                          \n\"1st
                                                          \nNow, vA<\/sub> = a1<\/sub> tA<\/sub>
                                                          \n\u21d2 vB<\/sub> + v = a1<\/sub> (tB<\/sub> – t)
                                                          \n\u21d2 a2<\/sub> tB<\/sub> + v = a1<\/sub> (tB<\/sub> – t)
                                                          \n\u21d2 v = (a1<\/sub> – a2<\/sub>) tB<\/sub> – a1<\/sub> t
                                                          \nSubstitute for t from (1),
                                                          \n\"1st
                                                          \nSubstitute for tB<\/sub> from (2),
                                                          \n\"1st
                                                          \nHence proved.<\/p>\n

                                                          Question 3.
                                                          \nAn object moves with a deceleration \\(\\alpha \\sqrt{\\mathbf{v}}\\) in a straight line (a is a constant) At time t = 0, the velocity is v0<\/sub>. What is the distance it traverses before coming to rest? What will be the total time taken?
                                                          \nAnswer:
                                                          \nDeceleration = \\(\\alpha \\sqrt{\\mathbf{v}}\\)
                                                          \n\u21d2 Acceleration a = – \\(\\alpha \\sqrt{\\mathbf{v}}\\)
                                                          \n\\(\\frac{\\mathrm{d} \\mathrm{v}}{\\mathrm{dt}}\\) = – \u03b1v1\/2<\/sup>
                                                          \n\u21d2 \\(\\frac{\\mathrm{d} \\mathrm{v}}{\\mathrm{v}^{1 \/ 2}}\\) = – \u03b1dt
                                                          \nIntegrating,
                                                          \n\"1st
                                                          \n2v1\/2<\/sup> = – \u03b1t + c
                                                          \nAt t = 0, v = v0<\/sub>
                                                          \n\u21d2 2v0<\/sub>1\/2<\/sup> = c
                                                          \n\u2234 2v1\/2<\/sup> = – \u03b1t + 2v0<\/sub>1\/2<\/sup> …………… (1)
                                                          \nWhen t = T, v = 0
                                                          \n\u2234 2(0)1\/2<\/sup> = – \u03b1T + 2v0<\/sub>1\/2<\/sup>
                                                          \n\u21d2 T = \\(\\frac{2 \\sqrt{v_{0}}}{\\alpha}\\)
                                                          \n\u2234 Total time taken = \\(\\frac{2 \\sqrt{v_{0}}}{\\alpha}\\)
                                                          \nFrom (1), we get
                                                          \n\"1st
                                                          \n\"1st<\/p>\n","protected":false},"excerpt":{"rendered":"

                                                          You can Download Chapter 3 Motion in a Straight Line Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 1st PUC Physics Question Bank Chapter 3 Motion in a Straight Line 1st PUC Physics Motion …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/9260"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=9260"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/9260\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=9260"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=9260"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=9260"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}