{"id":3003,"date":"2020-08-26T08:39:31","date_gmt":"2020-08-26T03:09:31","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=3003"},"modified":"2021-07-02T15:04:28","modified_gmt":"2021-07-02T09:34:28","slug":"kseeb-solutions-class-9-science-chapter-8","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-science-chapter-8\/","title":{"rendered":"KSEEB Solutions for Class 9 Science Chapter 8 Motion"},"content":{"rendered":"

Karnataka Board Class 9 Science Chapter 8 Motion<\/h2>\n

KSEEB Solutions for Class 9 Science Chapter 8 Intext Questions<\/h3>\n

Question 1.
\nAn object has moved through a distance can it have zero displacement? If yes, support your answer with an example.
\nAnswer:
\n\"KSEEB
\nIn the figure distance of an object moved from 0 to A is 60 km and distance travelled is 60 km and displacement is 60 km. From 0 to A and back to B, distance travelled is 60 km + 25 km = 85 km. But displacement (35 km) is not equal to distance travelled (85 km). If we observe like this, displacement of a body is zero, but distance travelled is not zero, starting from point \u20180\u2019 and returning back to ‘O\u2019 final position is mixing with initial point. Hence displacement is zero.<\/p>\n

Question 2.
\nA farmer moves along the boundary of a square field of side 10 m in 40s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
\nAnswer:
\nA farmer moves along the boundary of a square field of side 10 m in 40s
\n2 minutes 20 seconds means 140 seconds.
\nDistance travelled by farmer
\n= \\(\\frac{40}{40}\\) \u00d7 140 m
\n= 140 m.
\n\u2234 Farmer moves for 2 minutes 20 seconds. It means \\(\\frac{140}{40}\\) = 3.5 rounds
\nFrom the original point, he moves for 2 minutes 20 seconds.
\nCase i)
\nOriginal point means any point in the corner of the field. In this case he moves for 2 min 20 seconds from the diagonal of the field.
\nDisplacement is equal to diagonal of the field displacement
\n= \\(\\sqrt{10^{2}+10^{2}}\\)
\n= 14.1 m.
\nCase ii)
\nOriginal point means any point in the middle of the side of the field.
\n\u2234 Displacement is equal to any side of the field = 10 m.
\nIt means displacement is between any original point ie in between 14.1 m and 10 m.<\/p>\n

Question 3.
\nWhich of the following is true for displacement?
\na) It cannot be zero.
\nb) Its magnitude is greater than the distance travelled by the object.
\nAnswer:
\nb) Its magnitude is greater than the distance travelled by the object.<\/p>\n

Question 4.
\nDistinguish between speed and velocity
\nAnswer:<\/p>\n\n\n\n\n\n\n
Speed<\/td>\nVelocity<\/td>\n<\/tr>\n
1. Scalar quantity<\/td>\nVector quantity<\/td>\n<\/tr>\n
2. Distance travelled in a given time.<\/td>\nDistance travelled along with a path with the given time.<\/td>\n<\/tr>\n
3. It is always is + ve.<\/td>\nIt is + ve or – ve<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Question 5.
\nUnder what condition (s) is the magnitude of the average velocity of an object equal to its average speed?
\nAnswer:
\nIf the speed of the object changing uniformly, the magnitude of the average velocity of an object equal to its average speed.<\/p>\n

Question 6.
\nWhat does the odometer of an automobile measure?
\nAnswer:
\nOdometer measures the speed of the vehicle.<\/p>\n

Question 7.
\nWhat does the path of an object look like when it is in uniform motion?
\nAnswer:
\nWhen the object is in uniform motion, it covers equal distance at equal intervals of time.<\/p>\n

Question 8.
\nDuring an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, which is 3 \u00d7 10 ms-1<\/sup>.
\nAnswer:
\nTime taken by the signal to reach the ground station from the spaceship = 5 min = 5 \u00d7 60 = 300 seconds
\nSpeed of the signal = 3 \u00d7 108<\/sup> m\/s.
\nSpeed = \\(\\frac{\\text { Total distance travelled }}{\\text { Total time taken }}\\)
\n\u2234 Distance travelled = speed \u00d7 time taken
\n= 3 \u00d7 108<\/sup> \u00d7 300
\n= 9 \u00d7 1010<\/sup>m
\n\u2234 Distance of change of velocity = 9 \u00d7 1010<\/sup> m<\/p>\n

Question 9.
\nWhen will you say a body is in<\/p>\n

    \n
  1. Uniform acceleration?<\/li>\n
  2. non-uniform acceleration?<\/li>\n<\/ol>\n

    Answer:<\/p>\n

      \n
    1. If the object covers equal distances in equal intervals of time, it is added to be in uniform motion.<\/li>\n
    2. If the object covers unequal distances in equal intervals of time, it is said to be in non-uniform motion.<\/li>\n<\/ol>\n

      Question 10.
      \nA bus decreases its speed from 80 km h-1<\/sup> to 60 km h-1<\/sup> in 5s. Find the acceleration of the bus.
      \nAnswer:
      \nInitial velocity of the bus u = 80 kmh-1<\/sup>
      \n\"KSEEB
      \nFinal velocity, V = 60 km\/h
      \ntime taken to decrease the velocity of bus, t = 5 seconds
      \nacceleration, a = 5 seconds
      \n\"KSEEB
      \n= -1.112 ms-2<\/sup>.<\/p>\n

      Question 11.
      \nA train starting from a railway station and moving with uniform acceleration attains a speed 40 km h-1<\/sup> in 10 minutes. Find its acceleration.
      \nAnswer:
      \nInitial speed of the train u = 0
      \n\"KSEEB
      \n\"KSEEB
      \n\u2234 acceleration of a train = 0.0185 ms-2<\/sup>.<\/p>\n

      Question 12.
      \nWhat is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
      \nAnswer:
      \n(i) The nature of the distance-time graphs for the uniform motion of an object is a straight line.
      \n\"KSEEB
      \n(ii) The nature of tire distance time graph for non-uniform motion is curved line.
      \n\"KSEEB
      \nDistance-time graph for a car moving with non-uniform speed<\/p>\n

      Question 13.
      \nWhat can you say about the motion of an object whose distance-time graph is straight line parallel to the time axis?
      \nAnswer:
      \nThis graph indicates the object is at rest.<\/p>\n

      Question 14.
      \nWhat can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
      \nAnswer:
      \nThis graph indicates the object is in uniform motion.<\/p>\n

      Question 15.
      \nWhat is the quantity which is measured by the area occupied below the velocity-time graph?
      \nAnswer:
      \nThe quantity which is measured by the area occupied below the velocity-time graph is length = l<\/p>\n

      Question 16.
      \nA bus starting from rest moves with a uniform acceleration of 0.1 ms-2<\/sup> for 2 minutes. Find
      \n(a) the speed acquired,
      \n(b) the distance travelled.
      \nAnswer:
      \n(a) Initial velocity of the bus u = 0 (Bus is at rest).
      \nacceleration, a = 0.1 ms-2<\/sup>
      \ntime t = 2 minutes = 120 seconds
      \nLet the speed of bus be ‘V’
      \n\"KSEEB
      \n\u2234 v = 12m\/s
      \n(b) As per third law of motion:
      \nv2<\/sup> – u2<\/sup> = 2as
      \n(12)2<\/sup> – (0)2<\/sup> = 2(0. 1)s
      \n\u2234 s = 720 m.
      \nthe speed of acquired by bus = 12 m\/s.
      \nthe distance travelled = 720 m.<\/p>\n

      Question 17.
      \nA train is travelling at a speed of 90 kmh-1<\/sup> Brakes are applied so as to produce a uniform acceleration of -0.5 ms-2<\/sup>. Find how far the train will go before it is brought to rest.
      \nAnswer:
      \nLet the initial speed of the train be u = 90 Km\/h
      \n= 25 m\/s.
      \nFinal speed of the train, v = 0 (train comes to rest)
      \nacceleration a = 0.5 ms-2<\/sup>
      \nAs per 3rd<\/sup> law of motion
      \nv2<\/sup> = u2<\/sup> + 2as
      \n(0)2<\/sup> = (25)2<\/sup> + 2(0.5)s
      \ns = train travelled distance
      \n\"KSEEB
      \nTrain will travel 625 km before it is brought to rest.<\/p>\n

      Question 18.
      \nA trolley, while going down an inclined plane has an acceleration of 2cms-2<\/sup> what will be its velocity 3s<\/sup> after the start?
      \nAnswer:
      \nInitial velocity of a trolley, u = 0 (at rest)
      \nacceleration, a = 2 cms-2<\/sup> = 0.02 m\/s2<\/sup> time t = 3S
      \nAs per 1 st law of motion
      \nv = u + at
      \nHere V means velocity of a trolley after 3 s
      \nV = 0 + 0.02 \u00d7 3
      \n= 0.06 m\/s.
      \n\u2234 = 0.06 m\/s.<\/p>\n

      Question 19.
      \nA racing car has a uniform acceleration of 4ms-2<\/sup>, what distance will it cover in 10s after start?
      \nAnswer:
      \nInitial velocity of a racing car, u = 0 (at rest)
      \nacceleration, a = 4 m\/s2<\/sup>
      \ntime t = 10 s
      \nAs per second law of motion
      \n\"KSEEB
      \nHere S means distance travelled by car after 10S
      \n\"KSEEB
      \n\u2234 Distance covered by car after 10 S = 200 m.<\/p>\n

      Question 20.
      \nA stone is thrown in a vertically upward direction with a velocity of 5ms-1<\/sup> If the acceleration of the stone during its motion is 10ms-2<\/sup> in the downward direction, what will he the height attain by the stone and how much time will it take to reach there?
      \nAnswer:
      \nThe initial velocity of a stone, u = 5 ms-1<\/sup>.
      \nFinal velocity of a stone v = 0 (at rest)
      \nIf the acceleration of the stone during its motion is = -10ms-2<\/sup>.
      \nLet the maximum height be ‘s’
      \nv = u + at
      \n0 = 5 + (-10)t
      \n\"KSEEB
      \ntime taken by stone = 0.5s
      \nAs per 3rd Law of motion
      \nv2<\/sup> = u2<\/sup> + 2as
      \n(0)2<\/sup> = (5)2<\/sup> + 2(-10)s
      \n\"KSEEB
      \nThe height attained by the store = 1.25 m.<\/p>\n

      KSEEB Solutions for Class 9 Science Chapter 8 Textbook Exercises<\/h3>\n

      Question 1.
      \nAn athlete completes one round of a circular track of diameter 200 m. in the 40s. What will be the distance covered and the displace\u00acment at the end of 2 minutes 20s?
      \nAnswer:
      \nDiameter of circular track, d = 200 m.
      \nradius of the circular track,
      \nr = \\(\\frac{\\mathrm{d}}{2}\\) = 100m
      \nCircumference = 2\u03c0r = 2p(100) = 200\u03c0cm.
      \nin 40s, athlete covers distance 200\u03c0m
      \nin 1s, distance covered by athlete = \\(\\frac{200 \\pi}{40}\\) m
      \nAthlete runs for 2 minutes 20s means 140s
      \n\u2234 Displacement of all athlete is equal to the diameter of circle
      \n= \\(\\frac{200 \\times 22}{40 \\times 7}\\) \u00d7 140
      \n= 2200 m
      \n\u2234 Displacement = 200 m.<\/p>\n

      Question 2.
      \nJoseph jogs from one end A to the other end B of a straight 300 M road in 2 minutes 30 seconds and then turns around and jogs 100 M back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
      \nAnswer:
      \na) from A to B distance covered by Joseph = 300 M.
      \nTime taken to cover this distance
      \n= 2 min 30 sec. = 150 seconds
      \n\"KSEEB
      \nDisplacement = Nearest distance between A and B = 300 m.
      \nTime taken = 150s
      \nAverage velocity = ygy = 2 m\/s.<\/p>\n

      b) From A to C :
      \n\"KSEEB
      \nTotal distance covered = Distance between A to B + distance between B to C
      \n= 300 + 100 = 400 m.
      \nTotal time taken to cover from A to B + Time taken to cover from B to C
      \n= 150 + 60= 210s.
      \n\"KSEEB
      \n\"KSEEB
      \nDisplacement from A to C
      \n= AC + AB – BC
      \n= 300 – 100
      \n= 200m
      \nTotal time taken = Time taken to travel from A to B + Time taken to travel from B to C
      \n= 210 s
      \n\u2234 Average velocity = \\(\\frac{200}{210}\\) =0.952 ms-1<\/sup> .<\/p>\n

      Question 3.
      \nAbdul, while during to school, computes the average speed for his trip to be 20Kmh-1<\/sup> on his return trip along the some route, there is less traffic and the average speed is 30 Kmh-1<\/sup>. What is the average speed for Abdul’s trip?
      \nAnswer:
      \ni) Average speed of Abdul’s trip while driving to school = 20 Kmh-1<\/sup>
      \nAverage speed = \\(\\frac{\\text { Total distance }}{\\text { Total time taken }}\\)
      \nTotal distance = Distance travelled to reach school = t1<\/sub>
      \n\u2234 20 = \\(\\frac{\\mathrm{d}}{\\mathrm{t}_{1}}\\)
      \nt1<\/sub> = \\(\\frac{d}{20}\\) …………… (i)
      \nWhile returning from school
      \nTotal distance = Distance travelled while returning from school = d Now total time taken = t2<\/sub>.
      \n\"KSEEB<\/p>\n

      Question 4.
      \nA motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms-2<\/sup> for 8.0 s. How far does the boat travel during this time?
      \nAnswer:
      \nInitial velocity of motor boat, u =0
      \n(at rest) ,
      \nAccieratlon of a motor boat a = 3m\/s2<\/sup>
      \nTime taken, t = 85
      \nAccording to second equation of motion, s
      \nS = 0 +\\(\\frac{1}{2}\\) \u00d7 (8)2 <\/sup>= 96m.
      \n\u2234 Motorboat travels = 96m.<\/p>\n

      Question 5.
      \nA driver of a car travelling at 52kmh-1<\/sup> applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5s. Another driver going at 3kmh-1<\/sup> in another car applies his brakes slowly and stop In 10s. on the same graph paper. Plot the speed versus time graphs for the two cars, which of the two cars travelled farther after the brakes were applied?
      \nAnswer:
      \ninitial velocity of A car u1<\/sub> = 52 kmh-1<\/sup>
      \n= 14.4 m\/s.
      \nTime taken to stop car t1<\/sub> = 5 s.
      \nAfter 5s car comes to rest Initial velocity of B car, u2<\/sub> = 3Kmh-1<\/sup>
      \n= 0.833 m\/s. = 0.83 m\/s.
      \nTime taken to stop car t2<\/sub> = 10s
      \nAfter 10s car comes to rest.
      \nGraph:
      \n\"KSEEB
      \n\"KSEEB
      \nArea of \u2206OPR = Area of \u2206OSQ
      \nWhen compared car A to car B, A car travelled father. After brake it travels 52 Kmh-1<\/sup><\/p>\n

      Question 6.
      \nFig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions.
      \n\"KSEEB
      \na) Which of the three is travelled the fastest?
      \nb) Are all three ever at the same point on the road?
      \nc) How far has C travelled when B passes A?
      \nd) How far has B travelled by the time passes C?
      \nAnswer:
      \n\"KSEEB
      \nSlope of B object is greater than slopes A and C. This is travelling fastest.<\/p>\n

      b) All three objects never meet at a point, hence these are not at the same point.<\/p>\n

      c) There are 7 boxes in X-axis = 4 Km
      \n1 box = \\(\\frac{4}{7}\\) Km
      \nIn the beginning, C object is 4 boxes away from 0
      \n= \\(\\frac{16}{7}\\) km
      \nDistance to C from origin = Distance of C when it moves from B to A =8 Km
      \n\"KSEEB
      \n\u2234 Distance travelled by C when B passes A = 5.714 km.<\/p>\n

      d) Distance B travelled by the time it passes C = 9 boxes
      \n\"KSEEB
      \n\u2234 B has travelled 5.143 k.m. by the time when it passes C.<\/p>\n

      Question 7.
      \nA ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10ms-1<\/sup> with what velocity will it strike the ground? After what time will it strike the ground?
      \nAnswer:
      \nDistance ball covers, S = 20 m
      \nacceleration, a = 10Kmh-1<\/sup> = 10 m\/s
      \ninitial velocity, u = 0 (at rest)
      \nWhen ball strikes, its final velocity,
      \nAccording to 3rd law of motion
      \nv2<\/sup> = u2<\/sup> + 2as
      \nv2<\/sup> = 0 + 2(10)(20)
      \nv = 20 m\/s.
      \nAccording to 1 st Law of motion v = u + at
      \nTime taken by ball to strike the ground, t,
      \n20 = 0 + 10 (t)
      \n\u2234 t = 2s
      \n\u2234 2s is required to ball for striking the ground and velocity is 20 m\/s<\/p>\n

      Question 8.
      \nThe speed-time graph for a car is shown in figure
      \n\"KSEEB
      \na) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by car during the period.
      \nb) Which part of the graph represents uniform motion of the car?
      \nAnswer:
      \na) Shade lightly in the graph about distance travelled by car in 4s.
      \nb) Shade with red colour about 6 cm to 10 cm. It represents the uniform motion of the car.<\/p>\n

      Question 9.
      \nState which of the following situations are possible and give an example for each of these;
      \na) an object with a constant acceleration but with zero velocity.
      \nb) an object moving in a certain direction with an acceleration in the perpendicular direction.
      \nAnswer:
      \na) This is possible
      \nEg: when a ball is thrown to a height its velocity is zero.
      \n(For a ball G = 9.8 m\/s2<\/sup>)<\/p>\n

      b) This is possible
      \nEg: when a car moves in a curved line, its acceleration is perpendicular to the given direction)<\/p>\n

      Question 10.
      \nAn artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
      \nAnswer:
      \nSatellite is moving in circular orbit radium of satellite r = 42250 km
      \ntime t = 24 Hrs = 3600 \u00d7 24 S
      \n\"KSEEB
      \n\u2234 Speed of the satellite when it moves in circular orbit = 3.07 kms-1<\/sup>.<\/p>\n

      KSEEB Solutions for Class 9 Science Chapter 8 Additional Questions<\/h3>\n

      Fill in the blanks with suitable words.<\/p>\n

      Question 1.
      \nThe two physical quantities required to know final position of an object are _________ and _________
      \nAnswer:
      \nDistance travelled, displacement.<\/p>\n

      Question 2.
      \nS.I. unit of speed is ________
      \nAnswer:
      \nm\/s<\/p>\n

      Question 3.
      \nDistance travelled by the object in unit time is ________
      \nAnswer:
      \nSpeed.<\/p>\n

      Question 4.
      \nIf acceleration is in the direction of velocity it is _________ acceleration.
      \nAnswer:
      \npositive.<\/p>\n

      Question 5.
      \nWhen an object moves in a circular path with uniform speed, its motion is called ________
      \nAnswer:
      \nuniform circular motion.<\/p>\n

      Answer the following questions:<\/p>\n

      Question 1.
      \nGive an express10n for centripetal force?
      \nAnswer :
      \n\\(f=\\frac{m v^{2}}{r}\\)<\/p>\n

      Question 2.
      \nWhat is uniform circular mot10n?
      \nAnswer :
      \nBody moving in a circular path arbitrary any instant along a tangential to the posit10n of the body on the circular path at that instant or time.<\/p>\n

      Question 3.
      \nA body falls freely. What is constant?
      \nAnswer :
      \nWhen the body falls freely. It has a constant acceleration.<\/p>\n

      Question 4.
      \nDefine uniform speed.
      \nAnswer :
      \nThe speed of an object is said to be uniform speed if it travels equal distances in equal intervals of time.<\/p>\n

      KSEEB Solutions for Class 9 Science<\/a><\/h5>\n","protected":false},"excerpt":{"rendered":"

      Karnataka Board Class 9 Science Chapter 8 Motion KSEEB Solutions for Class 9 Science Chapter 8 Intext Questions Question 1. An object has moved through a distance can it have zero displacement? If yes, support your answer with an example. Answer: In the figure distance of an object moved from 0 to A is 60 …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/3003"}],"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=3003"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/3003\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=3003"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=3003"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=3003"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}