Motion – NCERT Solved Questions
Question 1. An object has moved through a distance. Can it have zero displacement ? If yes, support your answer with an example.
Answer. Yes, it can have zero displacement.
Explanation : If we take a round trip and reach back at the starting point, then we have traveled some distance, but our displacement will be zero.
Question 2. 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 1 minutes 20 seconds ?
Answer.
If the farmer starts from point A, then at the end of 1 minutes and 20 seconds (= 80 seconds), he will reach the diagonally opposite corner C. The magnitude of displacement of the farmer is:
Question 3. Which of the following is true for displacement?
(a) It cannot be zero.
(b) Its magnitude is greater than the distance traveled by the object.
Answer.
Neither (a) nor (b) is true.
Question 4. Distinguish between speed and velocity.
Answer.
Difference between Speed and velocity:
| Speed | Velocity |
| 1. The distance traveled by a moving body per unit time is called its speed.
2. Speed is a scalar quantity. |
1. The distance traveled by a moving body in a particular direction per unit time is called its velocity.
2. velocity is a vector quantity. |
Question 5. Under what condition (s) is the magnitude of average velocity of an object equal to its average speed?
Answer.
When an object moves along a straight line in the same direction, its total path length is equal to the magnitude of displacement. Hence, its average speed is equal to the magnitude of average velocity.
Question 6. What does the odometer of an automobile measure?
Answer.
The odometer of an automobile measures the distance moved by it.
Question 7. What does the path of an object look like when it is in uniform motion?
Answer.
Straight line path.
Question 8. 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.
Answer.
Question 9. When will you say a body is in
(i) uniform acceleration,
(ii) non-uniform acceleration?
Answer. (i)
If a body travels in a straight line and its velocity changes by equal amounts in equal intervals of time, however small these time intervals may be, then the body is said to be in uniform acceleration.
Answer (ii).
If the velocity of a body changes by unequal amounts in equal intervals of time, then the body is said to be in non-uniform acceleration.
Question 10. A bus decreases its speed from 80 km/h to 60 km/h in 5s. Find the acceleration of the bus.
Answer.
Question 11. A train starting from the railway station and moving with a uniform acceleration attains a speed of 40 km/h in 10 minutes. Find its acceleration.
Answer.
Question 12. What is the nature of the distance time graphs for uniform and non-uniform motion of an object?
Answer.
(i) For uniform motion, the distance-time graph is a straight line inclined with the time-axis,
(ii) For non-uniform motion, the distance-time graph is a curve.
Question 13. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Answer.
The object is at rest.
Question 14. What can you about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
Answer.
The object is moving with a uniform speed.
Question 15. What is the quantity which is measured by the area occupied below velocity-time graph?
Answer.
Distance covered by the body in the given time interval.
Question 16. A bus starting from rest moves with a uniform acceleration of 0.1 ms-2 for 2 minutes. Find (a) the speed acquired,
(b) the distance traveled.
Answer.
Question 17. A train is traveling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of-0.5 ms-2. Find how far the train will go before it is brought to rest ?
Answer.
Question 18. A trolley while going down an inclined plane has an acceleration of 2 cms-2. What will be its velocity in three seconds after start ?
Answer.
Question 19. A racing car has a uniform acceleration of 4ms-2. What distance will it cover in 10 s after start ?
Answer.
Question 20. A stone is thrown in a vertically upward direction with a velocity of 5ms-1. If the acceleration of the stone during its motion is 10ms-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Answer.
Question 21. 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?
Answer.
Diameter of circular path (d) = 200 m
Radius of circular path, r = 100 m
Time period (T) = 40 s
Time (t) = 2 min, 20s
= (120 + 20)s = 140 s
Therefore, number of revolutions
= 3.5 revolution
(i) Distance traveled
= 2200 m
(ii) Now during 3 revolutions, the displacement is zero, since the athlete has reached the
starting point.
Therefore, actual displacement is due to half revolution
= Diameter of circular track
=200m
Question 22. 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) from A to C?
Answer.
(a) For motion from A to B:
Distance covered = 300 m
Displacement = 300 m
Time taken = 2 minutes 30 seconds
= 2(60) + 30=150 s
Average speed = Distance covered/ Time taken
Average velocity = Displacement/Time taken
(b) For motion from A to B To C:
Distance covered = 300 + 100 = 400 m
Displacement = AB – CB
= 300 – 100 = 200m
Time taken = 150 + 60 = 210 s
Average speed = Distance covered/Time taken
Average velocity = Displacement/Time taken
Question 23. 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 30 km/h. What is the average speed for Abdul’s trip?
Answer.
Let one way distance = x km
Time taken in forward trip at a speed of 20 km/h
Time taken in return trip at a speed of 30 km/h
Total time for the whole trip
Total distance covered = x + x km
