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ICSE Class IX Notes 2026 : Physics

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CHAPTER 8 MOTION INTEXT QUESTIONS PAGE NO. 100 1. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds? Ans. The farmer takes 40 s to cover 4 10 = 40 m. 40 In 2 min and 20 s (140 s), he will cover a distance = 140 140m 40 140 Therefore, the farmer completes 3.5 rounds (3 complete rounds and a half round) of 40 the field in 2 min and 20 s. That means, after 2 min 20 s, the farmer will be at the opposite end of the starting point. Now, there can be two extreme cases. Case I: Starting point is a corner point of the field. In this case, the farmer will be at the diagonally opposite corner of the field after 2 min 20 s. Therefore, the displacement will be equal to the diagonal of the field. Hence, the displacement will be 102 10 2 14.1m Case II: Starting point is the middle point of any side of the field. In this case the farmer will be at the middle point of the opposite side of the field after 2 min 20 s. Therefore, the displacement will be equal to the side of the field, i.e., 10 m. For any other starting point, the displacement will be between 14.1 m and 10 m. 2. Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object. Ans. (a) Not true. Displacement can become zero when the initial and final position of the object is the same. (b) Not true. Displacement is the shortest measurable distance between the initial and final positions of an object. It cannot be greater than the magnitude of the distance travelled by an object. However, sometimes, it may be equal to the distance travelled by the object. INTEXT QUESTIONS PAGE NO. 102 1. Distinguish between speed and velocity. Ans. Speed Velocity Speed is the distance travelled by an object Velocity is the displacement of an object in in a given interval of time. It does not have a given interval of time. It has a unique any direction. direction. The speed of an object can never be The velocity of an object can be negative, negative. At the most, it can become zero. positive, or equal to zero. This is because This is because distance travelled can never displacement can take any of these three be negative. values. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 1 - 2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed? Total distance covered Ans. Average speed and Total time taken Displacement Average velocity Total time taken If the total distance covered by an object is the same as its displacement, then its average speed would be equal to its average velocity. 3. What does the odometer of an automobile measure? Ans. The odometer of an automobile measures the distance covered by an automobile. 4. What does the path of an object look like when it is in uniform motion? Ans. An object having uniform motion has a straight line path. 5. During 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, 3 108 m s 1. Ans. Time taken by the signal to reach the ground station from the spaceship = 5 min = 5 60 = 300 s Speed of the signal = 3 108 m/s Distance travelled Speed Time taken Distance travelled = Speed Time taken = 3 108 300 = 9 1010 m Hence, the distance of the spaceship from the ground station is 9 1010 m. INTEXT QUESTIONS PAGE NO. 103 1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration? Ans. (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. (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. 2. A bus decreases its speed from 80 km h 1 to 60 km h 1 in 5 s. Find the acceleration of the bus. 5 Ans. Initial speed of the bus, u = 80 km/h = 80 22.22m / s 18 5 Final speed of the bus, v = 60 km/h = 60 16.66m / s 18 Time take to decrease the speed, t = 5 s v u 16.66 22.22 Acceleration, a 1.12m / s 2 t 5 Here, the negative sign of acceleration indicates that the velocity of the car is decreasing. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 2 - 3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h 1 in 10 minutes. Find its acceleration. Ans. Initial velocity of the train, u = 0 (since the train is initially at rest) 5 Final velocity of the train, v = 40 km/h = 40 11.11m / s 18 v u 11.11 0 Acceleration, a 0.0185m / s 2 t 5 Time taken, t = 10 min = 10 60 = 600 s Hence, the acceleration of the train is 0.0185 m/s2. INTEXT QUESTIONS PAGE NO. 107 1. What is the nature of the distance time graphs for uniform and non-uniform motion of an object? Ans. The distance time graph for uniform motion of an object is a straight line (as shown in the following figure). The distance time graph for non-uniform motion of an object is a curved line (as shown in the given figure). 2. What can you say about the motion of an object whose distance time graph is a straight line parallel to the time axis? Ans. When an object is at rest, its distance time graph is a straight line parallel to the time axis. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 3 - A 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. 3. What can you say about the motion of an object if its speed time graph is a straight line parallel to the time axis? Ans. Object is moving uniformly. A 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. 4. What is the quantity which is measured by the area occupied below the velocity time graph? Ans. Distance The graph shows the velocity time graph of a uniformly moving body. Let the velocity of the body at time (t) be v. Area of the shaded region = length breath Where, Length = t, Breath = v Area = vt = velocity time (i) Distance We know, Velocity Time Distance = Velocity Time (ii) From equations (i) and (ii), Area = Distance Hence, the area occupied below the velocity time graph measures the distance covered by the body. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 4 - EQUATIONS OF MOTION BY GRAPHICAL METHOD When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration during motion and the distance covered by it in a certain time interval by a set of equations known as the equations of motion. There are three such equations. These are: v = u + at --------------- (1) 1 s ut at 2 ------------- (2) 2 2a s = v2 u2 ----------- (3) where u is the initial velocity of the object which moves with uniform acceleration a for time t, v is the final velocity, and s is the distance travelled by the object in time t. Eq. (1) describes the velocity-time relation and Eq. (2) represents the position-time relation. Eq. (3), which represents the relation between the position and the velocity, can be obtained from Eqs. (1) and (2) by eliminating t. These three equations can be derived by graphical method. EQUATION FOR VELOCITY-TIME RELATION In the graph, AC gives the initial velocity (u). BE gives the final velocity (v). CE represents the time taken t. DF gives the change in velocity. Change in velocity Time DF OF OD a CE OE OC But OE OC = t v u a t Acceleration= v u = at (i) v = u + at (I) Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5 - EQUATION FOR POSITION-VELOCITY RELATION Let s be the displacement of the body in a time t. In the graph, Displacement = Area CABE s = Area of the rectangle CAGE + Area of the triangle ABG. 1 s = AC x CE + (AG x GB) 2 Here AC = u CE = t AG = t GB = v u = at [from(i)] 1 s = ut + x t x at 2 s = ut + 1 2 at (II) 2 EQUATION FOR POSITION-TIME RELATION In the graph, Displacement = Area of the trapezium CABE 1 s = (AC+EB) x CE 2 Here AC = u, EB = v, CE = t u v s t.............(ii ) 2 v u From (i), we have t a Substituting the value of t, u v v u 2 a 2 v u2 s 2a v 2 u 2 2as v2 = u2 + 2as .(III) (I), (II) and (III) are the equations of motion. s INTEXT QUESTIONS PAGE NO. 109 1. A bus starting from rest moves with a uniform acceleration of 0.1 m s 2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled. Ans. (a) Initial speed of the bus, u = 0 (since the bus is initially at rest) Acceleration, a = 0.1 m/s2 Time taken, t = 2 minutes = 120 s Let v be the final speed acquired by the bus. v u v 0 a 0.1 t 120 v = 12 m/s (b) According to the third equation of motion: v2 u2 = 2as Where, s is the distance covered by the bus Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6 - (12)2 (0)2 = 2(0.1) s s = 720 m Speed acquired by the bus is 12 m/s. Distance travelled by the bus is 720 m. 2. A train is travelling at a speed of 90 km h 1. Brakes are applied so as to produce a uniform acceleration of 0.5 m s 2. Find how far the train will go before it is brought to rest. Ans. Initial speed of the train, u = 90 km/h = 25 m/s Final speed of the train, v = 0 (finally the train comes to rest) Acceleration = 0.5 m s 2 According to third equation of motion: v2 = u2 + 2 as (0)2 = (25)2 + 2 ( 0.5) s Where, s is the distance covered by the train s 252 625m 2 0.5 The train will cover a distance of 625 m before it comes to rest. 3. A trolley, while going down an inclined plane, has an acceleration of 2 cm s 2. What will be its velocity 3 s after the start? Ans. Initial velocity of the trolley, u = 0 (since the trolley was initially at rest) Acceleration, a = 2 cm s 2 = 0.02 m/s2 Time, t = 3 s According to the first equation of motion: v = u + at Where, v is the velocity of the trolley after 3 s from start v = 0 + 0.02 3 = 0.06 m/s Hence, the velocity of the trolley after 3 s from start is 0.06 m/s. 4. A racing car has a uniform acceleration of 4 m s 2. What distance will it cover in 10 s after start? Ans. Initial velocity of the racing car, u = 0 (since the racing car is initially at rest) Acceleration, a = 4 m/s2 Time taken, t = 10 s 1 According to the second equation of motion: s ut at 2 2 Where, s is the distance covered by the racing car 1 400 s 0 4 102 200m 2 2 Hence, the distance covered by the racing car after 10 s from start is 200 m. 5. A stone is thrown in a vertically upward direction with a velocity of 5 m s 1. If the acceleration of the stone during its motion is 10 m s 2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there? Ans. Initially, velocity of the stone,u = 5 m/s Final velocity, v = 0 (since the stone comes to rest when it reaches its maximum height) Acceleration of the stone, a = acceleration due to gravity, g = 10 m/s2 (in downward direction) There will be a change in the sign of acceleration because the stone is being thrown upwards. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7 - Acceleration, a = 10 m/s2 Let s be the maximum height attained by the stone in time t. According to the first equation of motion: v = u + at 0 = 5 + ( 10) t 5 t 0.5s 10 According to the third equation of motion: v2 = u2 + 2 as (0)2 = (5)2 + 2( 10) s 52 s 1.25m 20 Hence, the stone attains a height of 1.25 m in 0.5 s. EXERCISE QUESTIONS PAGE NO. 112 1. An 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? Ans. Diameter of a circular track, d = 200 m d Radius of the track, r = 100m 2 Circumference = 2 r = 2 (100) = 200 m In 40 s, the given athlete covers a distance of 200 m. 200 In 1 s, the given athlete covers a distance = m 40 The athlete runs for 2 minutes 20 s = 140 s 200 22 Total distance covered in 140 s 140 2200m 40 7 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. He takes 3 rounds in 40 3 = 120 s. Thus, after 120 s his displacement is zero. Then, 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. Displacement of the athlete = 200 m Distance covered by the athlete in 2 min 20 s is 2200 m and his displacement is 200 m. 2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 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? Ans. (a) From end A to end B Distance covered by Joseph while jogging from A to B = 300 m Time taken to cover that distance = 2 min 50 seconds = 170 s Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 8 - Total distance covered Total time taken Total distance covered = 300 m Total time taken = 170 s Displacement = shortest distance between A and B = 300 m Time interval = 170 s 300 Average speed 1.765m / s 170 Displacement Average velocity Total time taken Displacement = shortest distance between A and B = 300 m Time interval = 170 s 300 Average velocity 1.765m / s 170 The average speed and average velocity of Joseph from A to B are the same and equal to 1.765 m/s. (b) From end A to end C Average speed Average speed Total distance covered Total time taken Total distance covered = Distance from A to B + Distance from B to C = 300 + 100 = 400 m Total time taken = Time taken to travel from A to B + Time taken to travel from B to C = 170 + 60 = 230 s 400 Average speed 1.739m / s 230 Displacement Average velocity Total time taken Displacement from A to C = AC = AB BC = 300 100 = 200 m Time interval = time taken to travel from A to B + time taken to travel from B to C = 170 + 60 = 230 s 200 Average velocity 0.87m / s 230 The average speed of Joseph from A to C is 1.739 m/s and his average velocity is 0.87 m/s. 3. Abdul, while driving to school, computes the average speed for his trip to be 20 km/h. On his return trip along the same route, there is less traffic and the average speed is 40 km/h. What is the average speed for Abdul s trip? Ans. Case I: While driving to school Average speed of Abdul s trip = 20 km/h Total distance covered Average speed Total time taken Total distance = Distance travelled to reach school = d Let total time taken = t1 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 9 - d d t1 ..........(i ) t1 20 Case II: While returning from school Total distance = Distance travelled while returning from school = d Now,total time taken = t2 d d 40 t1 ..........(ii) t2 40 Total distance covered in the trip Average Speed for Abdul's trip = Total time taken Where, Total distance covered in the trip = d + d = 2d Total time taken, t = Time taken to go to school + Time taken to return to school = t1 + t2 2d Average speed = t1 +t 2 From equations (i) and (ii), 2d 2 80 Average speed = 26.67m / s d d 2 1 3 + 20 40 40 20 Hence, the average speed for Abdul s trip is 26.67 m/s. 4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s2 for 8.0 s. How far does the boat travel during this time? Ans. Initial velocity, u = 0 (since the motor boat is initially at rest) Acceleration of the motorboat, a = 3 m/s2 Time taken, t = 8 s According to the second equation of motion: 1 s ut at 2 2 Distance covered by the motorboat, s 1 s 0 3 82 96m 2 Hence, the boat travels a distance of 96 m. 5. A driver of a car travelling at 52 km/h applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km/h 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? Ans. Case A: Initial speed of the car, u1 = 52 km/h = 14.4 m/s Time taken to stop the car, t1 = 5 s Final speed of the car becomes zero after 5 s of application of brakes. Case B: Initial speed of the car, u2 = 3 km/h = 0.833 m/s 0.83 m/s Time taken to stop the car, t2 = 10 s Final speed of the car becomes zero after 10 s of application of brakes. Plot of the two cars on a speed time graph is shown in the following figure: Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 10 - Distance covered by each car is equal to the area under the speed time graph. Distance covered in case A, 1 1 S1 OP OR 14.4 5 36m 2 2 Distance covered in case B, 1 1 S 2 OS OQ 0.83 10 4.15m 2 2 Area of OPR > Area of OSQ Thus, the distance covered in case A is greater than the distance covered in case B. Hence, the car travelling with a speed of 52 km/h travels farther after brakes were applied. 6. Fig 8.11 shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions: (a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C? Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 11 - Distance Time y-axis Distance Slope of the graph= x-axis Time Speed = slope of the graph Since slope of object B is greater than objects A and C, it is travelling the fastest. (b) All three objects A, B and C never meet at a single point. Thus, they were never at the same point on road. (c) On the distance axis: Ans. Speed 7 small boxes = 4 km 4 1 small box km 7 Initially, object C is 4 blocks away from the origin. 16 Initial distance of object C from origin km 7 Distance of object C from origin when B passes A = 8 km 16 56 16 40 Distance covered by C 8 5.714km 7 7 7 Hence, C has travelled a distance of 5.714 km when B passes A. (d) 4 36 Distance covered by B at the time it passes C for 9 boxes 9 5.143km 7 7 Hence, B has travelled a distance of 5.143 km when B passes A. 7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m/s2, with what velocity will it strike the ground? After what time will it strike the ground? Ans. Distance covered by the ball, s = 20 m Acceleration, a = 10 m/s2 Initially, velocity, u = 0 (since the ball was initially at rest) Final velocity of the ball with which it strikes the ground, v According to the third equation of motion: v2 = u2 + 2 as Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 12 - v2 = 0 + 2 (10) (20) v = 20 m/s According to the first equation of motion: v = u + at Where, Time, t taken by the ball to strike the ground is, 20 = 0 + 10 (t) t=2s Hence, the ball strikes the ground after 2 s with a velocity of 20 m/s. 8. The speed-time graph for a car is shown is Fig. 8.12. (a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period. (b) Which part of the graph represents uniform motion of the car? Ans: (a) The shaded area which is equal to 1 4 6 12m represents the distance travelled by 2 the car in the first 4 s. (b) The part of the graph in red colour between time 6 s to 10 s represents uniform motion of the car. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 13 - 9. State which of the following situations are possible and give an example for each of these: (a) an object with a constant acceleration but with zero velocity (b) an object moving in a certain direction with an acceleration in the perpendicular direction. Ans. (a) Possible When 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. (b) Possible When a car is moving in a circular track, its acceleration is perpendicular to its direction. 10. An 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? Ans. Radius of the circular orbit, r = 42250 km Time taken to revolve around the earth, t = 24 h Speed of a circular moving object, 2 r v t 2 3.14 42250 v 24 v 1.105 104 km / h 3.069km / s Hence, the speed of the artificial satellite is 3.069 km/s. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 14 - PRACTICE QUESTIONS MOTION Multiple Choice Questions 1. If the displacement of an object is proportional to square of time, then the object moves with (a) uniform velocity (b) uniform acceleration (c) increasing acceleration (d) decreasing acceleration 2. The distance time graph of a body coincides with its time axis. The body must be (a) in uniform motion (b) at rest (c) in uniformly accelerated motion (d) in zig-zag motion 3. From the given v t graph (see below Fig.), it can be inferred that the object is (a) in uniform motion (b) at rest (c) in non-uniform motion (d) moving with uniform acceleration 4. The velocity time graph of a body is parallel to the time axis. The body is (a) at rest (b) having uniform acceleration (c) having zero acceleration (d) having non-uniform acceleration 5. A particle is moving in a circular path of radius r. The displacement after half a circle would be: (a) Zero (b) r (c) 2 r (d) 2 r 6. A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is, (a) u/g (b) u2/2g (c) u2/g (d) u/2g Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 15 - 7. The numerical ratio of displacement to distance for a moving object is (a) always less than 1 (b) always equal to 1 (c) always more than 1 (d) equal or less than 1 8. Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of 10 m/s. It implies that the boy is (a) at rest (b) moving with no acceleration (c) in accelerated motion (d) moving with uniform velocity 9. Area under a v t graph represents a physical quantity which has the unit (a) m2 (b) m (c) m3 (d) m/s 10. Four cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown in below Fig.. Choose the correct statement (a) Car A is faster than car D. (b) Car B is the slowest. (c) Car D is faster than car C. (d) Car C is the slowest. 11. Slope of a velocity time graph gives (a) the distance (b) the displacement (c) the acceleration (d) the speed 12. In which of the following cases of motions, the distance moved and the magnitude of displacement are equal? (a) If the car is moving on straight road (b) If the car is moving in circular path (c) The pendulum is moving to and fro (d) The earth is revolving around the Sun Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 16 - 13. Which of the following figures (see below Figure) represents uniform motion of a moving object correctly? SHORT ANSWER QUESTIONS 14. The displacement of a moving object in a given interval of time is zero. Would the distance travelled by the object also be zero? Justify you answer. 15. How will the equations of motion for an object moving with a uniform velocity change? 16. A car starts from rest and moves along the x-axis with constant acceleration 5 m/s2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest? 17. A motorcyclist drives from A to B with a uniform speed of 30 km/h and returns back with a speed of 20 km/h. Find its average speed. 18. Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height. 19. The velocity-time graph (see below Figure) shows the motion of a cyclist. Find (i) its acceleration (ii) its velocity and (iii) the distance covered by the cyclist in 15 seconds. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 17 - 20. A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement time graph is shown in below figure. Plot a velocity time graph for the same. LONG ANSWER QUESTIONS 21. An object starting from rest travels 20 m in first 2 s and 160 m in next 4 s. What will be the velocity after 7 s from the start. 22. An electron moving with a velocity of 5 104 m/s enters into a uniform electric field and acquires a uniform acceleration of 104 m/s2 in the direction of its initial motion. (i) Calculate the time in which the electron would acquire a velocity double of its initial velocity. (ii) How much distance the electron would cover in this time? 23. Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds. 24. Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove that the heights reached by them would be in the ratio of u12 : u22 ( Assume upward acceleration is g and downward acceleration to be +g ). 25. An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2 s if both the objects drop with same accelerations? How does the difference in heights vary with time? Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 18 -

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