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# Chapter 8 Motion

We have seen things around us moving but we don't understand the physics behind this. Motion is due to speed, velocity, acceleration. In this chapter we will deal with these term and try to understand the concept and its physics.

Download pdf of NCERT Examplar with Solutions for Class Science Chapter 8 Motion

### Exercise 1

•  Q1 An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example. Ans: Yes. An object that has moved through a distance can have zero displacement.  Example: If an object moves from point A and reaches to the same point A. In this case, the total distance covered by the man is 20 m + 20 m + 20 m + 20 m = 80 m. Hence, his displacement is zero because the shortest distance between his initial and final position is zero. Q2 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: In fig. ABCD is a square field of side 10 m.   Time for one round = 40 s Total time = 2 min 20 s                 = (2 x 60 + 20)s = 140s Number of round completed = If farmer starts from A, it will complete 3 rounds (A → B → C → D → A) at A. In the last 0.5 round starting from A, he will finish at C.  Displacement of farmer Q3 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 ; If an object moving from point A and reaches to the same point A, then displacement can be zero. (b) Not true ; Its magnitude can not be greater than the distance travelled by an object. However, sometimes, it can be either equal or less the distance travelled by the object.

### Exercise 3

•  Q1 When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration? Ans: (i) Uniform acceleration : A body is said to be in uniform acceleration when its velocity is changing at the same rate i.e when its acceleration in unit time is the same.  example : An object in freefall is in uniform acceleration. (ii) Non-uniform acceleration : When a body moves with unequal velocity in the equal interval of time, the body is said to be moving with non-uniform acceleration. example : If a car covers 10 meters in the first two seconds and 15 meters in the next two seconds. Q2 A bus decreases its speed from 80 km h-1 to 60 km h-1 in 5 s. Find the acceleration of the bus. Ans: Initial speed of the bus, u = 80 km/h = Final speed of the bus, v = 60 km/h = Time take to decrease the speed, t = 5 s Here, the acceleration of bus is -1.112 m/s2 Q3 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) Final velocity of the train, v = 40 km/h = 40 x 1000 / 60 x 60 = 11.1 m/s Time taken, t = 10 min = 10 × 60 = 600 s Now acceleration is given by the relation : Hence, the acceleration of the train is 0.0185 m/s2.

### Exercise 4

•  Q1 What is the nature of the distance-time graphs for uniform and non-uniform motion of an object? Ans: For uniform motion, the distance−time graph of an object is a straight line (as shown in the following figure). For non-uniform motion, the distance−time graph of an object is a curved line (as shown in the given figure). Q2 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 the slope of the distance-time graph is a straight line parallel to the time axis, the object is at the same position as the time passes. Thus, the object is at rest. Q3 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: The speed of the object is not changing with the time it means the object is in uniform motion. So , the object is moving with uniform motion. Q4 What is the quantity which is measured by the area occupied below the velocity-time graph? Ans: The area occupied under the speed time graph gives you distance traveled by object.

### Exercise 5

•  Q1 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  Acceleration, a = 0.1 m/s2 Time taken, t = 2 minutes = 60 x 2 = 120 s Let v be the final speed acquired by the bus.   v = u + at = 0 + 0.1 x 120 = 12 m/s ∴ The speed acquired by the bus is 12 m/s (b) According to the equation of motion:  v2 = u2 + 2as v2 − u2 = 2as (12)2 - (0)2 =2(0.1)s 144 = 0.2 s s = 144 / 0.2 s = 720 m  Distance travelled by the bus is 720 m. Q2 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 = 90 x 1000 / 60 x 60 = 25 m/s Final speed of the train, v = 0  Acceleration = −0.5 m s-2 Using equation of motion: v2 = u2 + 2 as (0)2 = (25)2 + 2 (−0.5) s 0 = 625 - s  s = 625 m Where, s is the distance covered by the train The train will cover a distance of 625 m before it comes to rest. Q3 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 Acceleration, a = 2 cm s−2 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 + 2 × 3 = 6 cm/s Hence, the velocity of the trolley after 3 s from start is 6 cm/s. Q4 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 Acceleration, a = 4 m/s2 Time taken, t = 10 s Distance, s = ? According to the  equation of motion: where s is the distance covered by the racing car, Hence, the distance covered by the racing car after 10 s from start is 200 m. Q5 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 Acceleration of the stone, a = acceleration due to gravity, g = 10 m/s2 Since, u is upward & a is downward, it is a retarded motion. 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 According to the third equation of motion: v2 = u2 + 2 as (0)2 = (5)2 + 2(−10) s However , the stone attains a height of 1.25 m in 0.5 s.

### Exercise 6

•  Q1 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 Radius of the track, r = d/2 = 100 m Circumference = 2πr = 2π (100) = 200π m In 40 s, the given athlete covers a distance of 200π m. In 1 s, the given athlete covers a distance = 200π / 40 m The athlete runs for 2 minutes 20 s = 140 s ∴Total distance covered in 140 s = 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. Q2 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  A to B = 300 m Time taken to cover that distance = 2 min 30 seconds = 2 x 60 + 30 = 150 s Average speed =  300 / 150 = 2 m/s Average velocity = 300 / 150 = 2 m/s The average speed and average velocity of Joseph from A to B are the same and equal to 2 m/s. (b) From end A to end C  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  = 150 + 60 = 210 s Average speed = 400 / 210 = 1.90 m/s Average velocity = 200 / 210 = 0.95 m/s The average speed of Joseph from A to C is 1.90 m/s and his average velocity is 0.95 m/s. Q3 Abdul, while driving to school, computes the average speed for his trip to be 20 km h-1. On his return trip along the same route, there is less traffic and the average speed is 40 km h-1. 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 =  d Let total time taken = t1 Speed = Distance / time 20 = d / t1 t1 = d / 20…(i) Case II: While returning from school Total distance = d Speed = 30 km/h Now,total time taken = t2 Speed = distance / time  30 = d / t2 t2 = d / 30  ..... (ii) 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 Total time taken = d / 20 + d / 30 = 3d + 2d / 60 = 5d / 60 = d / 12 From equations (i) and (ii), Average Speed = 120/5                            = 24  km/h Hence, the average speed for Abdul’s trip is 24 km/h. Q4 A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s-2 for 8.0 s. How far does the boat travel during this time? Ans: Initial velocity, u = 0 Acceleration of the motorboat, a = 3 m/s2 Time taken, t = 8 s According to equation of motion: Distance covered by the motorboat, s Hence, the boat travels a distance of 96 m. Q5 A driver of a car travelling at 52 km h-1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h-1 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 = 52 x (5 / 18) = 14.44 m/s Time taken, t1 = 5 s Final speed = 0 m/s Case B: Initial speed of the car, u2 = 3 km/h = 3 x (5 / 18) = 0.833 m/s ≅ 0.83 m/s Time taken, t2 = 10 s Final speed = 0 m/s Plot of the two cars on a speed−time graph is shown in the following figure: Distance covered by each car is equal to the area under the speed−time graph. Distance covered in case A, Distance covered in case B, Then, the car1 travelling with a speed of 52 km/h travels farther after brakes were applied. Q6 Fig 8.11 shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions: Fig. 8.11 (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? Ans: (a) Object B (b) No (c) 5.714 km (d) 5.143 km (a) From the graph, it is clear B covers more distance in less time. Therefore, B is the fastest  (b) All three objects A, B and C never come at the same point . (c) On the distance axis: 7 small boxes = 4 km ∴1 small box = 4 / 7 km Initially, object C is 4 blocks away from the origin. ∴Initial distance of object C from origin= 16 / 7  Distance of object C from origin when B passes A = 8 km Distance covered by C Hence, C has travelled a distance of 5.714 km when B passes A. (d)  Distance covered by B at the time it passes C = 9 boxes Hence, B has travelled a distance of 5.143 km when B passes C. Q7 A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, 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  Final velocity, v = ? Time, t = ? According to the third equation of motion: v2 = u2 + 2 as v2 = 0 + 2 (10) (20) v = 20 m/s According to the first equation of motion: v = u + at 20 = 0 + 10 (t) t = 2 s Hence, time taken to reach at the ground is 2 seconds. Q8 The speed-time graph for a car is shown is Fig. 8.12. 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) Shaded area = The distance covered by the car in first 4 seconds is 12 m. (b) The red colour between time 6 s to 10 s represents uniform motion of the car. Q9 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 an object 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 an object is moving in a circular track, its acceleration is perpendicular to its direction. Q10 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: Distance covered by satellite in 24 hours, s = 2πr = 2 × 3.14 × 42250 = 265330 km  Time = 24 h Speed = 265330 / 24  = 11055.4 km/h  = 11055.4 / 3600 = 3.07 km/s

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