Question 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?
Solution :
Displacement after 3.5 rounds = diameter of the track = 200 m
Question 2. Joseph 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)and from A to C?
Solution:
(a) From A to B.
(b) From A to C.
Time taken = A to B + B to C, 150 + 60 = 210 s Total distance = 300 + 100 = 400 m
Question 3. 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 30 km h-1 What is the average speed for Abdul’s trip?
Solution: Let the school be at a distance of x km. If t is time taken to reach the school, then
Question 4. A motorboat starting from rest on a lake accelerates in a straight tine at a constant rate of 3.0 m s-2 for 8.0 s. How far does the boat travel during this time?
Solution:
Question 5. 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?
Solution: The data given in this numerical problem are in different units. So, we should first convert km h-1 unit into m s-1 unit.
For first car:
The distance traveled by a moving body is given by the area under its speed-time graph. so, Distance traveled by the first car=Area of the triangle AOB
Similarly, Distance travelled by the second car = Area of triangle COD
Thus, the second car travels 4.1 m and the first car travels 36 m before coming to rest. So, the second car traveled farther after the brakes were applied.
Question 6.
Figure given below 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?
Solution:
(a) B is travelling fastest.
(b) As three lines do not meet at any point, the three objects never meet on the road.
(c) B passes A at D. At this time, C is at E, which corresponds to 7 km. Hence when B crosses A, then C is at 7 km from the origin.
(d) By the time B passes C, it has travelled 4.5 km.
Question 7.
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?
Solution:
The ball strike the ground after 2 s with the velocity of 20 m s–
Question 8. The speed-time graph for a car is shown in the figure.
Solution
(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?
Solution: The motion during first 4 seconds is not uniformly accelerated. So, distance travelled by car in first 4 seconds is calculated by graphical method
(a) Number of squares in shaded part of the graph = 320.5 one small square in x axis represent t = 
(b) The limiting flat portion of the curve describes the constant speed of the car, i.e., a speed of 6.0 m s-1. At this stage, the acceleration of the car is zero.
Therefore, portion of the graph between t = 6 s to 10 s, describes the uniform motion of the car.
Question 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.
Solution: (a) Free fall of an object.
(b) Object moving in a circular path
Question 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.
Solution:
Radius of the orbit = 42250 km
= 42250 × 1000 m Time taken for one revolution = 24 hours
= 24 × 60 × 60 s