# CBSE 9th Science | Motion – NCERT Solved Questions

## 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 ?

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?

#### (b) Its magnitude is greater than the distance traveled by the object.

Neither (a) nor (b) is true.

### Question 4. Distinguish between speed and velocity.

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?

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?

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?

Straight line path.

### (ii) non-uniform acceleration?

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.

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 12. What is the nature of the distance time graphs for uniform and non-uniform motion of an object?

(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?

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?

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?

Distance covered by the body in the given time interval.

### 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?

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?

##### (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

##### (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