Average Velocity Calculator<\/a> to find the average velocity. It will generate the accurate average velocity by taking initial, final velocities.<\/p>\nQuestion 5. \nUnder what condition(s) is the magnitude of average velocity of an object equal to its average speed? \nAnswer: \nIf the total distance covered by an object is the same as its displacement; then its average speed would be equal to its average velocity.<\/p>\n
Question 6. \nWhat does the odometer of an automobile measure? \nAnswer: \nThe odometer of an automobile measures the distance covered by an automobile.<\/p>\n
Question 7. \nWhat does the path of an object look like when it is in uniform motion? \nAnswer: \nAn object having uniform motion has a straight line path.<\/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, that is, 3 \u00d7 108<\/sup> m s-1<\/sup>. \nAnswer: \nTime taken by the signal to reach the ground station from the spaceship \n= 5 min = 5 \u00d7 60 = 300 s \nSpeed of the signal = 3 \u00d7 108<\/sup> m\/s \nSpeed = Speed = \\(\\frac{\\text { Distance travelled }}{\\text { Time taken }}\\) \n\u2234 Distance travelled = Speed \u00d7 Time taken = 3 \u00d7 108<\/sup> \u00d7 300 = 9 \u00d7 1010<\/sup> m \nHence, the distance of the spaceship from the ground station is 9 \u00d7 1010<\/sup> m.<\/p>\nQuestion 9. \nWhen will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration? \nAnswer: \n(i) A body is said to have uniform acceleration if it travels in a straight path in such a way that its velocity changes at a uniform rate, i.e., the velocity of a body increases or decreases by equal amounts in an equal interval of time.<\/p>\n
(ii) A body is said to have non-uniform acceleration if it travels in a straight path in such a way that its velocity changes at a non-uniform rate, i.e., the velocity of a body increases or decreases in unequal amounts in an equal interval of time.<\/p>\n
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Question 10. \nA bus decreases its speed from 80 km h-1<\/sup> to 60 km h-1<\/sup> in 5 s. Find the acceleration of the bus. \nAnswer: \nInitial speed of the bus, \nu = 80 km\/h = 80\\(\\frac{5}{18}\\) = 22.22 m\/s \nFinal speed of the bus, \nv = 60km\/h = 60 \u00d7 \\(\\frac{5}{18}\\) = 16.66 m\/s \nTime take to decrease the speed, t = 5 s \nAcceleration a = \\(\\frac{v-u}{t}=\\frac{16.66-22.22}{5}\\) \n= -1.112 m\/s2<\/sup> \nHere, the negative sign- of acceleration indicates that the velocity of the car is decreasing.<\/p>\nQuestion 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 velocity of the train, u = 0 (since the train is initially at rest) \nFinal velocity of the train, \nv = 40 km\/h = 40 \u00d7 \\(\\frac{5}{18}\\) = 11.11 m\/s \nTime taken, t = 10 min = 10 \u00d7 60 = 600 s \nAcceleration, \na = \\(\\frac{v-u}{t}=\\frac{11.11-0}{600}\\) = 0.0185 m\/s2<\/sup> \nHence, the acceleration of the train is 0.0185 m\/s2<\/sup>.<\/p>\nQuestion 12. \nWhat is the nature of the distance-time graphs for uniform and non-uniform motion of an object? \nAnswer: \nThe distance-time graph for uniform motion of an object is a straight line (as shown in the following figure). \n \nThe distance-time graph for non-uniform motion of an object is a curved line (as shown in the given figure). \n <\/p>\n
Question 13. \nWhat can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis? \nAnswer: \nWhen an object is at rest, its distance-time graph is a straight line parallel to the time axis. \n \nA straight line parallel to the x-axis in a distance-time graph indicates that with a change in time, there is no change in the position of the object. Thus, 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: \nObject is moving uniformly. \n \nA straight line parallel to the time axis in a speed-time graph indicates that with a change in time, there is no change in the speed of the object. This indicates the uniform motion of the object.<\/p>\n
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Question 15. \nWhat is the quantity which is measured by the area occupied below the velocity-time graph? \nAnswer: \nDistance \n \nThe graph shows the velocity time graph of a uniformly moving body. \nLet the velocity of the body at time (t) be v. \nArea of the shaded region = length \u00d7 breath \nWhere, \nLength = t \nBreath = v \nArea = vt = velocity \u00d7 time …….. (i) \nWe know, \nVelocity = \\(\\frac{\\text { Displacement }}{\\text { Time }}\\) \n\u2234 Displacement = Velocity \u00d7 Time, ……… (ii) \nFrom equations (i) and (ii), \nArea = Displacement \nHence, the area occupied below the velocity-time graph measures the displacement covered by the body.<\/p>\n
Class 9 Science Chapter 8 Motion Textbook Questions and Answers<\/span><\/h3>\nQuestion 1. \nA bus starting from rest moves with a uniform acceleration of 0.1 m s-2<\/sup> for 2 minutes. Find (a) the speed acquired, (b) the distance travelled. \nAnswer: \n(a) 12 m\/s (b) 720 m \n(a) Initial speed of the bus, u = 0 (since the bus is initially at rest) \nAcceleration, a = 0.1 m\/s2<\/sup> \nTime taken, t = 2 minutes = 120 s \nLet v be the final speed acquired by the bus. \n\u2234 a = \\(\\frac{v-u}{t}\\) \n0.1 = \\(\\frac{v-0}{120}\\)<\/p>\n(b) According to the third equation of motion: v2<\/sup> – u2<\/sup> = 2as \nWhere, s is the distance covered by the bus (12)2<\/sup> – (0)2<\/sup> = 2(0.1)s \ns = 720 m \nSpeed acquired by the bus is 12 m\/s. \nDistance travelled by the bus is 720 m.<\/p>\nQuestion 2. \nA train is travelling at a speed of 90 km h-1<\/sup>. Brakes are applied so as to produce a uniform acceleration of -0.5 m s-2<\/sup>. Find how far the train will go before it is brought to rest. \nAnswer: \nInitial speed of the train, K = 90 km\/h = 25 m\/s \nFinal speed of the train, v = 0 (finally the train comes to rest) \nAcceleration = -0.5 m s-2<\/sup> \nAccording to third equation of motion: \nv2<\/sup> = u2<\/sup> + 2 as \n(0)2<\/sup> = (25)2<\/sup> + 2 (-0.5) s \nWhere, s is the distance covered by the train \ns = \\(\\frac{(25)^{2}}{2(0.5)}\\) = 625 m \nThe train will cover a distance of 625 m before it comes to rest.<\/p>\n <\/p>\n
Question 3. \nA trolley, while going down an inclined plane, has an acceleration of 2 cm s-2<\/sup>. What will be its velocity 3 s after the start? \nAnswer: \nInitial velocity of the trolley, u = 0 (since the trolley was initially at rest) \nAcceleration, a = 2 cm s-2<\/sup> = 0.02 m\/s2<\/sup> \nTime, t = 3 s \nAccording to the first equation of motion: \u03bd = u + at \nWhere, \u03bd is the velocity of the trolley after 3 s from start \n\u03bd = 0 + 0.02 \u00d7 3 = 0.06 m\/s \nHence, the velocity of the trolley after 3 s from start is 0.06 m\/s.<\/p>\nQuestion 4. \nA racing car has a uniform acceleration of 4 m s-2<\/sup>. What distance will it cover in 10 s after start? \nAnswer: \nInitial velocity of the racing car, u = 0 (since the racing car is initially at rest) Acceleration, a = 4 m\/s2<\/sup> \nTime taken, t = 10 s \nAccording to the second equation of motion: S= ut + \\(\\frac {1}{2}\\)at2<\/sup> \nWhere, s is the distance covered by the racing car \nHence, the distance covered by the racing car after 10 s from start is 200 m.<\/p>\nQuestion 5. \nA stone is thrown in a vertically upward direction with a velocity of 5 m s-1<\/sup>. If the acceleration of the stone during its motion is 10 m s-2<\/sup> in the downward direction, what will be the height attained by the stone and how much time will it take to reach there? \nAnswer: \nInitially, velocity of the stone, u = 5 m\/s \nFinal velocity, \u03bd = 0 (since the stone comes to rest when it reaches its maximum height) \nAcceleration of the stone, a = acceleration due to gravity, g = 10 m\/s2<\/sup> \n(in downward direction) \nThere will be a change in the sign of acceleration because the stone is being thrown upwards. \nAcceleration, a = -10 m\/s2<\/sup> \nLet s be the maximum height attained by the stone in time t. \nAccording to the first equation of motion: \n\u03bd = u + at \n0 = 5 + (-10) t \n\u2234 t = \\(\\frac {-5}{10}\\) = 0.5 S \nAccording to the third equation of motion: \u03bd2<\/sup> = u2<\/sup> + 2 as \n(0)2<\/sup> = (5)2<\/sup> + 2 (-10)s \n= S = \\(\\frac {52}{20}\\) = 1.25 \nHence, the stone attains a height of 1.25 m in 0.5 s.<\/p>\nQuestion 6. \nAn athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s? \nAnswer: \nDiameter of a circular track, d = 200 m \nRadius of the track, r = r = \\(\\frac{d}{2}\\) = 100 m \nCircumference = 2\u03c0r = 2\u03c0 (100) = 200\u03c0 m \nIn 40 s, the given athlete covers a distance of 200p m. \nIn 1 s, the given athlete covers a distance = \\(\\frac{200 \\pi}{40} m\\) \nThe athlete runs for 2 minutes 20 s = 140 s \n\u2234 Total distance covered in \ns = \\(\\frac{200 \\times 22}{40 \\times 7}\\) \u00d7 140 = 2200m<\/p>\n
The athlete covers one round of the circular track in 40 s. This means that after every 40 s, the athlete comes back to his original position. Hence, in 140 s he had completed 3 rounds of the circular track and is taking the fourth round.<\/p>\n
He takes 3 rounds in 40 \u00d7 3 = 120 s. Thus, after 120 s his displacement is zero. \nThen, the net displacement of the athlete is in 20 s only. In this interval of time, he moves at the opposite end of the initial position. Since displacement is equal to the shortest distance between the initial and final position of the athlete, displacement of the athlete will be equal to the diameter of the circular track.<\/p>\n
\u2234 Displacement of the athlete = 200 m \nDistance covered by the athlete in 2 min 20 s is 2200 m and his displacement is 200 m.<\/p>\n
Question 7. \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: \n(a) 2m\/s, 2 m\/s (b) 1.90 m\/s, 0.95 m\/s \n(a) From end A to end B \n \nDistance covered by Joseph while jogging from A to B = 300 m \nTime taken to cover that distance = 2 min 30 seconds = 150 s \nAverage speed = \\(\\frac{\\text { Total distance covered }}{\\text { Total time taken }}\\) \nTotal distance covered = 300 m \nTotal time taken = 150 s \nAverage velocity = \\(\\frac {300}{150}\\) = 2m\/s \nAverage velocity = \\(\\frac{\\text { Displacement }}{\\text { Time interval }}\\) \nDisplacement = shortest distance between A and B = 300 m \nTime interval = 150 s \nAverage velocity = \\(\\frac {300}{150}\\) = 2m\/s \nThe average speed and average velocity of Joseph horn A to B are the same and equal to 2 m \/ s.<\/p>\n
(b) From end A to end C \n \nAverage speed = \\(\\frac{\\text { Total distance covered }}{\\text { total time taken }}\\) \nTotal distance covered = Distance from A to B + Distance from B to C \n= 300 + 100 = 400 m \nTotal time taken = Time taken to travel from A to B + Time taken to travel from B to C = 150 + 60 = 210 s \nAverage speed = \\(\\frac {400}{210}\\) = 1.90 m\/s \nAverage velocity = \\(\\frac{\\text { Displacement }}{\\text { Time interval }}\\) \nDisplacement from A to C = AC = AB – BC = 300 – 100 = 200 m \nTime interval = time taken to travel from A to B + time taken to travel from B to C \n= 150 + 60 = 210 s \nAverage velocity = \\(\\frac {200}{210}\\) = 0.95 tn \/ s \nThe average speed of Joseph from A to C is 1.90 m\/s and his average velocity is 0.95 m\/s.<\/p>\n
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Question 8. \nAbdul, while driving to school, computes the average speed for his trip to be 20 km h-1<\/sup>. On his return trip along the same route, there is less traffic and the average speed is 30 km h-1<\/sup>. What is the average speed for Abdul’s trip? \nAnswer: \nCase I: While driving to school \nAverage speed of Abdul’s trip = 20 km\/h \nAverage speed = \\(\\frac{\\text { Total distance }}{\\text { Total time taken }}\\) \nTotal distance = Distance travelled to reach school = d \nLet total time taken = t1<\/sub> \n\u2234 20\\(\\frac{d}{t_{1}}\\) \nt1<\/sub> = \\(\\frac{d}{20}\\) ………(i)<\/p>\nCase II: While returning from school \nTotal distance = Distance travelled while returning front school = d \nNow, total time taken = t2<\/sub> \n30 = \\(\\frac{d}{t_{2}}\\) ……….(ii) \nAverage speed for Abdul’s trip = \\(\\frac{\\text { Total distance covered in the trip }}{\\text { Total timetaken }}\\)<\/p>\nWhere, \nTotal distance covered in the trip = d + d = 2d \nTotal time taken, t = Time taken to go to school + Time taken to return to school \n= t1<\/sub> + t2<\/sub> \n\u2234 Average speed = \\(\\frac{2 d}{t_{1}+t_{2}}\\) \nFrom equations (i) and (ii), \nAverage Speed = \\(\\frac{2 d}{\\frac{d}{20}+\\frac{d}{30}}=\\frac{2}{\\frac{3+2}{60}}\\) \nAverage Speed = \\(\\frac {120}{5}\\) =24 m\/s \nHence, the average speed for Abdul’s trip is 24 m\/s.<\/p>\nQuestion 9. \nA motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s-2<\/sup> for 8.0 s. How far does the boat travel during this time? \nAnswer: \nInitial velocity, u = 0 (since the motor boat is initially at rest) \nAcceleration of the motorboat, a = 3 m\/s2<\/sup> \nTime taken, t = 8 s \nAccording to the second equation of motion: \ns = ut + \\(\\frac {1}{2}\\)at2<\/sup> \nDistance covered by the motorboat, s \ns = 0 + \\(\\frac {1}{2}\\)3(8)2<\/sup> = 96 m \nHence, the boat travels a distance of 96 m.<\/p>\nQuestion 10. \nA driver of a car travelling at 52 km h-1<\/sup> applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h-1<\/sup> in another car applies his brakes slowly and stops in 10 s. 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: \nCase A: \nInitial speed of the car, u1<\/sub> = 52 km\/h = 14.4 m\/s \nTime taken to stop the car, t1<\/sub> = 5 s \nFinal speed of the car becomes zero after 5 s of application of brakes.<\/p>\nCase B: \nInitial speed of the car, u2<\/sub> = 3 km\/h = 0.833 m\/ss \u2245 0.83 m\/s \nTime taken to stop the car, t2<\/sub> = 10 s \nFinal speed of the car becomes zero after 10 s of application of brakes. \nPlot of the two cars on a speed-time graph is shown in the following figure: \n \nDistance covered by each car is equal to the area under the speed-time graph. \nDistance covered in case A, \ns1 = \\(\\frac {1}{2}\\) \u00d7 OP \u00d7 OR = \\(\\frac {1}{2}\\)14.4 \u00d7 5 = 36 m \nDistance covered in case B, \ns2 = \\(\\frac {1}{2}\\) \u00d7 OS \u00d7 OQ = \\(\\frac {1}{2}\\) \u00d7 0.83 \u00d7 10 = 4.15m \nArea of \u0394OPR > Area of \u0394OSQ \nThus, the distance covered in case A is greater than the distance covered in case B. \nHence, the car travelling with a speed of 52 km\/h travels farther after brakes were applied.<\/p>\nQuestion 11. \nFig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions: \n \n(a) Which of the three is travelling the fastest? \n(b) Are all three ever at the same point on the road? \n(c) How far has C travelled when B passes A? \n(d) How far has B travelled by the time it passes C? \nAnswer: \n(a) Object B \n(b) No \n(c) 5.714 km \n(d) 5.143 km \n \n(a) Speed = \\(\\frac{\\text { Distance }}{\\text { Time }}\\) \nslope of the graph = \\(\\frac{y-\\text { axis }}{x- \\text { axis }}=\\frac{\\text { Distance }}{\\text { Time }}\\) \n\u2234 Speed = slope of the graph \nSince slope of object B is greater than objects A and C, it is travelling the fastest.<\/p>\n
(b) AD three objects A, B and C never meet at a single point. Thus, they were never at the same point on road.<\/p>\n
(c) \n \nOn the distance axis: \n7 small boxes = 4 km \n\u2234 1 small box = \\(\\frac {4}{7}\\) km \nInitially, object C is 4 blocks away from the origin. \n\u2234 Initial distance of object C from origin = \\(\\frac {16}{7}\\) km \nDistance of object C from origin when B passes A = 8 km \nDistance covered by C \n= \\(8-\\frac{16}{7}=\\frac{56-16}{7}=\\frac{40}{7}\\) = 5.714 km \nHence, C has travelled a distance of 5.714 km when B passes A.<\/p>\n
(d) \n \nDistance covered by B at the time it passes C = 9 boxes \n= \\(\\frac {4}{7}\\) \u00d7 9 = \\(\\frac {36}{7}\\) = 5.143 km \nHence, B has travelled a distance of 5.143 km when B passes A.<\/p>\n
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Question 12. \nA ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2<\/sup>, with what velocity will it strike the ground? After what time will it strike the ground? \nAnswer: \nDistance covered by the ball, s = 20 m \nAcceleration, a= 10 m\/s \nInitially, velocity, u = 0 (since the ball was initially at rest) \nFinal velocity of the ball with which it strikes the ground, v \nAccording to the third equation of motion: \n\u03bd2<\/sub> = u2<\/sub> + 2 as \n\u03bd2<\/sub> = 0 + 2 (10) (20) \nv = 20 m\/s \nAccording to the first equation of motion: \n\u03bd = u + at \nWhere, \nTime, f taken by the baU to strike the ground is, \n20 = 0 + 10(t) \nt = 2s \nHence, the ball strikes the ground after 2 s with a velocity of 20 m\/ s.<\/p>\nQuestion 13. \nThe speed-time graph for a car is shown is Fig. 8.12. \n \n(a) Find out how far the car travels in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period. \n(b) Which part of the graph represents uniform motion of the car? \nAnswer: \n(a) \n \nThe shaded area which is equal to represents the distance travelled by the car in the first 4 s.<\/p>\n
(b) \n \nThe part of the graph in red colour between time 6 s to 10 s represents uniform motion of the car.<\/p>\n
Question 14. \nState which of the following situations are possible and give an example for each of these: \n(a) an object with a constant acceleration but with zero velocity. \n(b) an object moving in a certain direction with an acceleration in the perpendicular direction. \nAnswer: \n(a) Possible: \nWhen a ball is thrown up at maximum height, it has zero velocity, although it will have constant acceleration due to gravity, which is equal to 9.8 m\/s2<\/sup>.<\/p>\n(b) Possible: \nWhen a car is moving in a circular track, its acceleration is perpendicular to its direction.<\/p>\n
Question 15. \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: \nSpeed = \\(\\frac{\\text { Distance }}{\\text { Time }}\\) \nr Time \nDistance = 2\u03c0r = 2 \u00d7 3.14 \u00d7 42 \u00d7 42250 = 265330 km \nTime = 24 h \nSpeed = \\(\\frac{265330}{24}\\) = 11055.4 km\/h<\/p>\n
Class 9 Science Chapter 8 Motion Additional Important Questions and Answers<\/span><\/h3>\nMultiple choice Questions<\/span><\/p>\nQuestion 1. \nThe change in velocity of an object per unit time is called \n(a) Average velocity \n(b) Acceleration \n(c) Force \n(d) Momentum \nAnswer: \n(b) Acceleration<\/p>\n
Question 2. \nIf the distance-time graph of a body \u00a1s parabola, then its motion is \n(a) Uniform \n(b) Non-uniform \n(d) Body’s in rest \n(d) Body is in vibratory motion \nAnswer: \n(a) Uniform<\/p>\n
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Question 3. \nRetardation is \n(a) Positive acceleration \n(b) Zero acceleration \n(c) Non-uniform acceleration \n(d) Zero acceleration \nAnswer: \n(b) Zero acceleration<\/p>\n
Question 4. \nIf the displacement of an object is proportional to square of time, then the object moves with \n(a) uniform velocity \n(b) uniform acceleration \n(c) increasing acceleration \n(d) decreasing acceleration, \nAnswer: \n(b) uniform acceleration<\/p>\n
Question 5. \nSuppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of 10 ms-1<\/sup>. It implies that the boy is \n(a) at rest \n(b) moving with no acceleration \n(c) in accelerated motion \n(d) moving with uniform velocity \nAnswer: \n(c) in accelerated motion<\/p>\nQuestion 6. \nIn which of the following cases of motions, the distance moved and the magnitude of displacement are equal? \n(a) If the car is moving on straight road \n(b) If the car is moving in circular path \n(c) The pendulum is moving to and fro \n(d) The earth is revolving around the Sun \nAnswer: \n(a) If the car is moving on straight road<\/p>\n
Question 7. \nWhen a body completes one round in a circular path of radius ‘r’, its displacement is equal to \n(a) r \n(b) 2r \n(c) 2nr \n(d) Zero \nAnswer: \n(d) Zero<\/p>\n
Question 8. \nThe S.I. unit used to express acceleration is \n(a) ms \n(b) ms-1<\/sup> \n(c) Ms-2<\/sup> \n(d) m2<\/sup>m-1<\/sup> \nAnswer: \n(c) Ms-2<\/sup><\/p>\nQuestion 9. \nA body falls freely from a point and reaches the ground in 1 second, then the height of the point from the earth is equal to \n(a) 9.8 m \n(b) 4.9 m \n(c) 10m \n(d) 19.6 m \nAnswer: \n(b) 4.9 m<\/p>\n
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Question 10. \nA body starts moving from rest with a uniform acceleration of 2ms-2<\/sup>. \n(a) 6 m \n(b) 9m \n(c) 12 m \n(d) 18 m \nAnswer: \n(b) 9m<\/p>\nQuestion 11. \nWhich of the following figures (Fig.) represents uniform motion of a moving object correctly? \n \nAnswer: \n <\/p>\n
Question 12. \nWhich of the following is a vector quantity? \n(a) Speed \n(b) Distance \n(c) Velocity \n(d) energy \nAnswer: \n(a) Speed<\/p>\n
Very Short Answer Type Questions<\/span><\/p>\nQuestion 1. \nHow will you describe the position of an object? \nAnswer: \nThe position of an object can only be described with respect to a fixed point called reference point or origin.<\/p>\n
Question 2. \nWhen is the motion of a body said to be rectilinear motion? \nAnswer: \nWhen a body moves in a straight line, its motion is said to be rectilinear motion.<\/p>\n
Question 3. \nDefine velocity. Give its S.I. unit. \nAnswer: \nVelocity is defined as the displacement produced per unit time. It’s S.I. unit is ms-1<\/sup>.<\/p>\n <\/p>\n
Question 4. \nDefine acceleration. Give its S.I. unit. \nAnswer: \nThe change in velocity of an object per unit time is called its acceleration. \nAcceleration = \\(\\frac{\\text { Final velocity – Intitial velocity }}{\\text { Time }}\\) \nIt’s S.I. unit is ms-2<\/sup>.<\/p>\nQuestion 5. \nWhat does the slope of the following graphs indicate? \n(a) Distance-time graph passing through the origin and making an angle with the time axis. \n(b) Speed-time graph making an angle with lime axis and a straight line passing through the origin. \nAnswer: \n(a) Speed of the object. \n(b) Acceleration of the object.<\/p>\n
Question 6. \nDraw a velocity-time graph showing negative acceleration of the body. \nAnswer: \n <\/p>\n
Question 7. \nThe graph of two friends A and B are shown in the figure 8.13 below: \n \n(a) Which of the two friends is moving faster? \n(b) Where and when will they meet? \nAnswer: \n(a) A is moving faster than B because of its greater slope. \n(b) The two will meet after 4 minutes at distance of 80 meters.<\/p>\n
Question 8. \nWhat does the following graph indicate? \n(a) Distance-time graph is a straight line making an angle with time-axis \n(b) Speed-time graph is a straight line making an angle with the time-axis and passing through the origin. \nAnswer: \n(a) Uniform motion of body. \n(b) Uniform positive acceleration.<\/p>\n
Question 9. \nDraw a graph describing the osillatory motion of a body. \nAnswer: \n <\/p>\n
Question 10. \nWhat is the nature of uniform circular motion? \nAnswer: \nThe uniform circular motions are always accelerated motion.<\/p>\n
Short Answer Type Questions<\/span><\/p>\nQuestion 1. \nWhen is a body said to be in motion? Explain with an example that the motion of a body is relative to the position of other objects. \nAnswer: \nA body is said to be in motion if its position continues to change with the surroundings. Motion is relative i.e. an object\/ body may appeard to be moving to some but not others. Such as two students travelling in a bus while sitting on same seat. For people outside they are in motion but with respect to each other, the two friends are in state of rest.<\/p>\n
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Question 2. \nWhat do you mean by distance and displacement? How are they different? \nAnswer: \nThe length of the actual path travelled or covered between initial and final position of an object is called distance. \nThe change in position of an object is a given direction from its initial position is called displacement.<\/p>\n
The differences between distance and displacements are:<\/p>\n
\n\n\nDistance<\/td>\n Displacement<\/td>\n<\/tr>\n \n1. It refers to the actual length covered by a body between its initial and final position.<\/td>\n It refers to the minimum distance between initial and final position of the body.<\/td>\n<\/tr>\n \n2. It’s a scalar quantity.<\/td>\n It’s a vector quantity.<\/td>\n<\/tr>\n \n3. It is always positive.<\/td>\n It can be positive or negative.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nQuestion 3. \nA bus starts from point A and goes to point B after travelling 60 km on a straight road and then returns to point A. \n(i) What is the total distance covered by the bus? \n(ii) What is its displacement? \nAnswer: \n(i) Total distance travelled by bus from A to B and back to point A, then \nAB + BA = 60m + 60m = 120m \n(ii) The net displacement is zero because the final position coincides with initial position.<\/p>\n
Question 4. \nWhat do you mean by uniform and non-uniform acceleration? \nAnswer: \nUniform motion: When a body covers equal distance in equal interval of time, how small time interval may be, the motion is called uniform motion.<\/p>\n
If a body covers 60 km in an hour moving with a uniform motion, then it would cover 30 km in half hour and 15 km in 15 minutes.<\/p>\n
Non-uniform motion: A body is said to be in non-uniform motion which it covers unequal distances in equal interval of time or regular changes its direction. Most of the motion we see in daily life like that of vehicles on city roads are non-uniform motion.<\/p>\n
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Question 5. \nWhat is the speed and average speed of a body in motion? \nAnswer: \nThe distance covered by a body in motion in unit time is called its speed. \nSpeed = \\(\\frac{\\text { Distance travelled }}{\\text { Time }}\\) \nIf a car covers 40 km in 1st hours, 60 km in 2nd hour and 50 km in 3rd hour, then its, average speed is \n\\(\\frac{40+60+50}{3}\\) = 50 km\/h<\/p>\n
Question 6. \nOn a stormy night, the ligtning is seen first and thunder is heard later, though they are produced simultaneously at same time. Why? \nAnswer: \nThe lightning is seen first and thunder is heard later because of the difference in speed of light and sound. The light travelling with speed of 3 \u00d7 108<\/sup> ms-1<\/sup> reaches the earth earlier than sound travelling with speed of 340 ms-1<\/sup>.<\/p>\nQuestion 7. \nOn a rainy day, the sound of thunder is heard after 5 seconds of the flash of lightning. : \nCalculate the distance of the nearest point of ligtning. \nAnswer: \nIn atmosphere, lightning and thunder occur at same time. \nTime taken by sound to reach to earth = 5s \nSpeed of sound = 346 ms-1<\/sup> \n\u2234 Distance of nearest point of lightning = 346 \u00d7 5 = 1730 m<\/p>\nQuestion 8. \nWhat do you mean by positive and negative acceleration? Give examples. \nAnswer: \nWhen the acceleration of a body increases with time, it is called positive aceleration such as the free falling body experiences an increase in its velocity by 9.8 m\/s, every second.<\/p>\n
When the acceleration of a body decreases with time, it is called negative acceleration such as the stopping of car after application of breakes in time t, covering a distance of s metres.<\/p>\n
Question 9. \nGive examples of the following motion: \n(a) Uniform motion in a straight line in which the velocity is increasing \n(b) A body moving with constant speed in which velocity is changing at uniform rate. \n(c) Non-uniform motion with constant acceleration in the direction of motion. \n(d) Non-uniform motion in which acceleration is not constant. \nAnswer: \n(a) The motion of a train between two stations on a straight track between certain interval of time when its velocity becomes constant is an example of uniform motion in which velocity remains constant.<\/p>\n
(b) When a body is moving in a circular path with constant speed, its direction changes at every point hence, due to change in its direction, its velocity also changes simultaneously.<\/p>\n
(c) The motion of a free falling body is an example of non-uniform motion with the constant acceleration in tire direction of motion as it fall downward with constant acceleration of 9.8 ms-2<\/sup> due to gravity of the earth.<\/p>\n(d) When a train starts moving from a railway station, it increases its speed by unequal amount in equal interval of time till it attains the maximum speed.<\/p>\n
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Question 10. \nWhat is the importance of graphs in daily life. \nAnswer: \nA graph provides a convenient method to represent pictorially the basic information about an event. A graph can be bar graph, line graph or a pie chart. These graphs are used in:<\/p>\n
\nComparing and solving the linear motion and equations with two variable.<\/li>\n Finding the motion of a body to be uniform or non-uniformotion.<\/li>\n Calculating the distance travelled by a body from speed-time graph.<\/li>\n<\/ul>\nQuestion 11. \nDraw the distance-time graph of a uniform motion. How will you calculate the speed of the object from this graph. \n \nAnswer: \nTire given graph is the distance time graph. To determine the speed of object between A and B, draw,a line from A, parallel to X-axis and an another line parallel to speed-axis from point B. These two lines meet at point C and form a triangle ABC. In \u0394DBC, AC denotes the time interval and BC denotes the distance travelled. Hence, \nSpeed (V) = \\(\\frac{\\mathrm{BC}}{\\mathrm{AC}}=\\frac{S_{2}-S_{1}}{t_{2}-t_{1}}\\) Sloped line A<\/p>\n
Question 12. \nThe time of arrival and departure of a train at three station A, B and C and their distances from station A are given the table below. Plot the distance for graph for the motion of the tain assuming that its motion between the onation is uniform.<\/p>\n
\n\n\nStation<\/td>\n Distance from Station<\/td>\n Time of arrival<\/td>\n Time of departure<\/td>\n<\/tr>\n \nA<\/td>\n 0 km<\/td>\n 8 : 00<\/td>\n 8 : 15<\/td>\n<\/tr>\n \nB<\/td>\n 120 km<\/td>\n 11 : 15<\/td>\n 11 : 30<\/td>\n<\/tr>\n \nC<\/td>\n 180 km<\/td>\n 13 : 00<\/td>\n 13 : 15<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nAnswer: \nThe distance timetable for the motion of the train \n \n <\/p>\n
Question 13. \nMegha and his sister Anu goes to the school on bicycle. Both of them start at the same time but take different time to reach the school although they follow the same route. Table below shows the distance moved by both at different instant of time, plot the distance-time graph for their motion on the same graph and interpret the graph. \n \nAnswer: \n \nThe distance-time gaph shows both are moving with non-uniform motion. The average speed of Megha is greater than that of Anu.<\/p>\n
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Question 14. \nThe average walking speed of a boy while going to school is 4 km\/h. It take half an hour to reach the school. Calculate the school’s distance from home. \nAnswer: \nAverage speed of the boy = 4 km\/h \nTime taken to reach school = \\(\\frac {1}{2}\\) hour \nDistance of school from house = Speed \u00d7 Time \n= 4 km\/h \u00d7 \\(\\frac {1}{2}\\) hour = 2 km<\/p>\n
Question 15. \nThe odometer of a car reads 2000 km from the start of a trip by John. The trip takes 8 hours and shows reading of 2400 km. What is the average speed of John’s car in km\/h and m\/ s? \nAnswer: \nDistance covered by car = Final reading – Initial read \n= 2400 – 2000 = 400 km \nTime taken by John to cover 400 km = 8 hr \n\u2234 Average speed = \\(\\frac{400 \\mathrm{~km}}{8 \\mathrm{hr}}\\) = 50km\/hr \nAverage speed (in ms-1<\/sup>1) \\(\\frac{50 \\times 1000}{60 \\times 60}\\) = 13.9 ms-1<\/sup><\/p>\nQuestion 16. \nUsha swims in a 90 m long swimming pool. She covers 180 m in one minute swimming from one end of pool and back to the pool along a straight path. Find the average speed and average velocity? \nAnswer: \nTotal distance covered by Usha in one minium = 180 m. \n <\/p>\n
Question 17. \nNarang is moving his car with a velocity of 90 km\/h. How much distance will he cover in (a) one minute (b) one second. \nAnswer: \nVelocity of car = 90 km\/h \n(a) Distance covered in one minute \n= \\(\\frac{90}{60}=\\frac{3}{2}=\\frac{3 \\times 1000}{2}\\) = 1500 m \n(b) Distance covered in one second \n\\(\\frac{90}{60 \\times 60} k m=\\frac{90 \\times 1000}{60 \\times 60}\\) = 25 m<\/p>\n
Question 18. \nAn electric train is moving with an aver age velocity of 120 km\/h. How much distance will it cover in 30s ? \nAnswer: \nVelocity of train (V) = 120 km\/h. \n= \\(\\frac{120 \\times 1000}{60 \\times 60}=\\frac{100}{3} \\mathrm{~m} \/ \\mathrm{s}\\) \nTime taken = 30s \n\u2234 Distance covered in 30s \n= Average \u00d7 Time velocity = \\(\\frac {100}{3}\\) \u00d7 30 = 1000 m<\/p>\n
Question 19. \nIn a long distance race, the athletes were expected to take four rounds of the track such that the line of finish was same as the line of start. The length of track is 200 m. \n(a) Calculate the total distance covered by the athletes. \n(b) Displacement of athletes when they touch the finishing line. \n(c) Is the motion of athletes uniform or non-uniform? \nAnswer: \n(a) In four rounds, total distance covered \n= 4 \u00d7 length of track = 4 \u00d7 200 = 800 m \n(b) Displacement upon touching the finishing line would be zero as track is a circular track. \n(c) Motion of athletes would be non-uniform as they are not completing the race at the same time i.e. distance covered by them in equal interval of time is not equal.<\/p>\n
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Question 20. \nA circular track has a circumference of 314 m with AB as one of its diameter. A cyclist traveels from A to B along the circular path in 30s. Find \n(a) Speed of cyclist. \n(b) Displacement of cyclist. \n \nAnswer: \n(a) Circumference of circular track = 314 \n\u2234 Distance from A to B = \\(\\frac {314}{2}\\) = 157 m. \nTime taken by cyclist = 30 s. \nTime = \\(\\frac{\\text { Distance }}{\\text { Time }}=\\frac{157}{30}\\) = 5.23 m\/s<\/p>\n
(b) Diplsacement of cyclist from A to B which represents diameter of circular track \n2\u03c0r = 314 \n2 \u00d7 3.14 \u00d7 r =314 \nr = \\(\\frac{314}{3.14 \\times 2}\\) = 50m \nWhen r = 50 m, d = 2r = 50 \u00d7 2 = 100 m<\/p>\n
Question 21. \nA car travels a certain distance at an average speed of 30 km\/h and returns back with an average speed of 50 km\/h. Calculate the average speed of car. \nAnswer: \nLet distance covered by car travelling at 30 km\/h = x \n\u2234 Time taken = \\(\\frac{x}{30}\\)h \nSimilarly when coming back time taken = \\(\\frac{x}{50}\\)h \nTotal time taken = \\(\\frac{x}{30}+\\frac{x}{50}=\\frac{5 x+3 x}{150}=\\frac{8 x}{150} h\\) \nTotal distance covered = x + x = 2x \nAverage speed of car = \\(\\frac{2 x}{8 x \/ 50}=\\frac{2 x \\times 150}{8 x}=\\frac{75}{2}\\) = 37.5 km\/h<\/p>\n
Question 22. \nRahul starting from rest paddles his bicycle to attain a velocity of 6 m s-1<\/sup> in 30s. Then he applies brakes so that the velocity of bicycle comes to 4ms-1<\/sup> in next 5 seconds. Calculate acceleration of bicycle in both cases. \nAnswer: \nIn first case, initial velocity (u) = 0 \nFinal velocity (\u03bd) =6 ms-1<\/sup> \nTime taken = 30s \nAcceleration (a) = ? \nWe know that, \na = \\(\\frac{v-u}{t}\\) \nd = \\(\\frac{6-0}{30}=\\frac{1}{5}\\) = 0.2 ms-1<\/sup> \nIn second case, u = 6ms-1<\/sup>, \u03bd = 0.2 ms-1<\/sup>, \nTime = 68 \n\u2234 a = \\(\\frac{4-6}{5}=\\frac{-2}{5}\\) = -0.4 ms-2<\/sup><\/p>\nQuestion 23. \nA train starting from rest attains a velocity of 72 km\/h in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and distance travelled by the train while attaining this velocity. \nAnswer: \nInitial velocity, u = 0 final velocity (\u03bd) = 72 km\/h \nTime (t) = 5 minutes = 5 \u00d7 60 = 300 s \nAcceleration (a) = ? \na = \\(\\frac{v-u}{t}=\\frac{20-0}{300}=\\frac{1}{15} m s^{-2}\\) \nLet distance travelled by train = S \n2as = \u03bd2<\/sup> – u2<\/sup> \n2 \u00d7 \\(\\frac {1}{5}\\) \u00d7 5 = (20)2<\/sup> (72 km\/h = 20 mis) \nS = 2 \u00d7 \\(\\frac {1}{5}\\) \u00d7 5 = 3000m = 3 km.<\/p>\nQuestion 24. \nA can accelerates uniformly from 18 km\/h to 36 km\/h in 5 seconds. Calculate (i) The acceleration, (ii) Distance covered by car is that time. \nAnswer: \nInitial velocity of car (h) = 18 km\/h \n= \\(\\frac{18 \\times 100}{60 \\times 60} 5 \\mathrm{~m} \/ \\mathrm{s}\\) \nFinal velocity of car (\u03bd) =36 km\/h \n= \\(\\frac{36 \\times 100}{60 \\times 60}=10 \\mathrm{~m} \/ \\mathrm{s}\\) \nTime (t) = 5s. \n(i) \u2234 Acceleration, a = \\(\\frac{v-u}{t}=\\frac{10-5}{5} 1 \\mathrm{~ms}^{-2}\\) \n(ii) Let distance covered by car = S \n2as = \u03bd2<\/sup> – u2<\/sup> \n2 \u00d7 1 \u00d7 s = (10)2<\/sup> – (5)2<\/sup> \n2 \u00d7 1 \u00d7 s = 100 – 25 \ns = \\(\\frac {75}{2}\\) = 37.5m<\/p>\n <\/p>\n
Question 25. \nThe brakes applied to a car produce a negative acceleration of – 6 m\/s2<\/sup>. The car takes 2 seconds to stop after applying the brakes. Calculate the distance it would cover before coming to a stop. \nAnswer: \nAcceleration of car, a = – 6 ms-2<\/sup> \nTime taken by car to stop = 2s \nFinal velocity \u03bd = 0 \nWe know that \n\u03bd = u + at \n= 4 – 6 \u00d7 2 \nor u = 12m\/s<\/p>\nLet distance covered = s \nu = ut + \\(\\frac {1}{2}\\)at–<\/sup> \ns = 12 \u00d7 2 + \\(\\frac {1}{2}\\) \u00d7 (6) \u00d7 (2)–<\/sup> \n= 24 + 12 = 36 m.<\/p>\nQuestion 26. \nFind the initial velocity of car which is stopped in 10 seconds after appplication of brakes. Also calculate the distance travelled by the car if its negative acceleation is – 2.5 ms-2<\/sup>. \nAnswer: \nFinal velocity (\u03bd) = 0 \nLet initial velocity =u \nAcceleration (a) = -2.5 ms-2<\/sup> \nTime taken (t) = 10 seconds \nWe know that, \n\u03bd = u + at \n0 = u – 2.5 \u00d7 10 \nu = 25 ms-1<\/sup><\/p>\nLet distance covered = s \ns = ut + \\(\\frac {1}{2}\\)at2<\/sup> \n= 25 x 10 –\\(\\frac {1}{2}\\) \u00d7 2.5(10)2<\/sup> \n= 250 + 125 = 375 m<\/p>\nQuestion 27. \nA ball is thrown upward with initial velocity of 90 km\/h in upward direction. Calculate \n(i) Maximum height reached. \n(ii) Time taken to reach the maximum height. \nAnswer: \nInitial velocity of ball, u = 90 km\/h \n= \\(\\frac{90 \\times 1000}{3600}=25 \\mathrm{~m} \/ \\mathrm{s}\\) \nFinal velocity of ball, \u03bd = 0 (at max height) \nAcceleration, a = -10ms-2<\/sup> \nLet the height gained by ball = s \n2as = \u03bd2<\/sup> – u2<\/sup> \n2 \u00d7 (-10) \u00d7 s = 0 – (25)2<\/sup> \n– 20S = – 625 \ns = \\(\\frac {625}{20}\\) = 31.25 m \nLet time taken to achieve the height = t \n\u03bd = u + at \n0 = 25 – 10 \u00d7 t \nt = \\(\\frac {25}{10}\\) = 2.5<\/p>\nQuestion 28. \nCan the average speed of a moving object be zero? Can the average velocity of a moving object be zero? Give example. \nAnswer: \nThe average speed of a moving object cannot be zero while the average velocity of a moving object can be zero. \nA car start from point A and travels 40 km in an hour and then returns to point A again in 40 minutes.<\/p>\n
Then Average speed = \\(\\frac{(40+40) \\mathrm{km}}{1.40 \\mathrm{hr}}=\\frac{80}{1.40} \\mathrm{hr}=57.14 \\mathrm{~km} \/ \\mathrm{hr}\\) \nThe displacement produced after the comple-tion of return to point A = 0. \nAverage velocity = \\(\\frac{\\text { Total displacement }}{\\text { Total time }}\\) = 0.<\/p>\n
Question 29. \nA radar spots an enemy plane. A radio pulse emitted by radar and reflected by the plane reaches back in 0.5 x 10-3<\/sup>\u00a0s. Find the distance between plane and radar. \n(Speed of radar pulse = 3 \u00d7 108<\/sup> m\/s) \nAnswer: \nTotal time taken for pulse to reach and get reflected = 0.5 \u00d7 10-3<\/sup> s \n= \\(\\frac{0.5 \\times 10^{-3}}{2}\\) \nDistance = Speed \u00d7 Time \n= 3 \u00d7 108<\/sup> \u00d7 \\(\\frac{0.5 \\times 10^{-3}}{2}\\) = 0.75 \u00d7 105<\/sup>m = 7.5 km<\/p>\n <\/p>\n
Question 30. \nDraw a velocity time graph for a uniformly accelerated motion. How will you determine the distance covered by a body in a given interval of time using the graph. \nAnswer: \n \nThe velocity time graph for uniformly accelerated motion is a straight line curve passing through the origin and makes an angle with time axis. Using the graph, the distance travelled by a body between any given time interval can eas easily be determined.<\/p>\n
Let the time interval be t1<\/sub> and t2<\/sub> The distance covered would be equal to area of ABCDE \nArea of ABCDE = Area of \u0394ADE + Area of rectangle ABCD \n= \\(\\frac {1}{2}\\) (AD \u00d7 DE) + (AB \u00d7 BC)<\/p>\nLong Answer Type Questions<\/span><\/p>\nQuestion 1. \nDraw a velocity time graph of a car moving with uniform velocity of 40 km\/h. Using the graph determine the distance travelled by car in a given time interval of t1<\/sub> and t2<\/sub>. \nAnswer: \nThe velocity-time graph of a car moving with the uniform velocity shown in the figure the graph is parallel to time-axis.<\/p>\nTo determine the distance travelled between the time interval t1 and t2, perpendicular are drawn from point t1<\/sub> and t2<\/sub> on the graph. \nDistance travelled = Velocity \u00d7 time \n= AC \u00d7 CD = 40 \u00d7 (t2<\/sub> – t1<\/sub>) km \n= Area of rectangle ABCD<\/p>\nQuestion 2. \nDerive the relation V = u + at from the velocity-time graph for uniform motion. \nAnswer: \nThe velocity-time graph for a uniformly accelerated motion is given in figure. A represents \n \ninitial velocity and B represent final velocity in time t shown by OC. \nDraw AD || OC, so that \nBD + DC = BC or \nBA + CA = BC \nIf BC = u, OA = u \nThem \u03bd = BD + u \n\u03bd – u = BD ….. (i) \nBut accleration (a) = \\(\\frac{\\text { Change in velocity }}{\\text { Time }}\\) \n= \\(\\frac{\\mathrm{BD}}{\\mathrm{AD}}=\\frac{\\mathrm{BD}}{\\mathrm{BC}}\\) \nor a = \\(\\frac{\\mathrm{BD}}{\\mathrm{t}}\\) \nBD = at \nor \u03bd = u + at<\/p>\n
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Question 3. \nDerive the relation s = ut + 1at2 <\/sup>from velocity-time graph of a uniform accelerated motion. \n \nAnswer: \nThe velocity-time graph for a uniform accelerated motion is shown in figure. Initial velocity = u, at point A, \nFinal velocity = \u03bd, at point B \nTime = t \nFrom point, perpendicular BL is drawn on time-axis and BE on velocity-axis respectively. OA represents initial velocity OE or BC represents final velocity in time interval, t or OC.<\/p>\nThe distance travelled by object is given by the area enclosed with in OBC under the velocity time graph AB. \nArea of OBC = Area of rectangle OADC + Area of \u0394ADB \n= (OA \u00d7 OC) + \\(\\frac {1}{2}\\)(AD \u00d7 BD) \nAcceleration (a) \n <\/p>\n
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Question 4. \nDerive the relation v2<\/sup> – u2<\/sup> = 2a from velocity-time graph of a uniformly accelerated motion. \nAnswer: \nThe velocity time graph of an acclerated motion is shown in the figure. \nThe distance covered by object moving under uniform accleration ‘a’ is given by area of trapezium OABC \n\u2234 Distance s = \\(\\frac {1}{2}\\)(OA + BC) \u00d7 OC \nOA = u, BC = \u03bd and OC = t \ns = \\(\\left(\\frac{u+v}{a}\\right) \\times t\\) \nBut we know \u03bd = u + at \n \n <\/p>\n","protected":false},"excerpt":{"rendered":"These NCERT Solutions for Class 9 Science Chapter 8 Motion Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams. Motion NCERT Solutions for Class 9 Science Chapter 8 Class 9 Science Chapter 8 Motion InText Questions and Answers Question 1. An object has moved through …<\/p>\n
NCERT Solutions for Class 9 Science Chapter 8 Motion<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Science Chapter 8 Motion - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n