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 / t_{1}
t_{1} = d / 20…(i)
Case II: While returning from school
Total distance = d
Speed = 30 km/h
Now,total time taken = t_{2}
Speed = distance / time
30 = d / t_{2}
t_{2} = 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
= t_{1} + t_{2}
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/s^{2}
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, u_{1} = 52 km/h = 52 x (5 / 18) = 14.44 m/s
Time taken, t_{1} = 5 s
Final speed = 0 m/s
Case B:
Initial speed of the car, u_{2} = 3 km/h = 3 x (5 / 18) = 0.833 m/s ≅ 0.83 m/s
Time taken, t_{2} = 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 distancetime 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/s^{2}
Initially, velocity, u = 0
Final velocity, v = ?
Time, t = ?
According to the third equation of motion:
v^{2} = u^{2} + 2 as
v^{2} = 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 speedtime 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/s^{2}.
(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 
