Figure given below is the distance-time graph of the straight line motion of an object.

(a) What will be the position of the object at 20 s?
(b) What will be the distance travelled by the object in 12 s?
(c) What is the average speed of the object?

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Solution

(a) At 20 s, the object will be 8 m away from the starting point.
(b) In 12 s, distance travelled by the object will be 6 m.
(c) Average speed of the object,
$A v e r a g e S p e e d = \frac{T o t a l d i s tan c e c o v e r e d}{T o t a l t i m e t a k e n} = \frac{8 m}{20 s} = 0.4 m / s$

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Question 18

Distance-time graph of an object is given below.

(i) What will be the position of the object at 20s?
(ii) What will be the distance traveled by the object in 12s?
(iii) What is the average speed of the object?

Q. The object starts from

O

having a straight line motion along the path

O P Q R S

is represented by the distance-time graph shown in figure. Match the given statements to the different parts of the graph correctly.
(i) Object is moving with maximum speed.
(ii) Object starts moving in reversed direction.
(iii) Object is at rest.
(iv) Average velocity of object is zero.

Q. An object is moving along a straight line whose velocity-time graph is given below:
What will be the ratio of the distance travelled to the displacement of object from

t = 0

and

t = 5

sec?

Q. What is the nature of the distance time graphs for uniform and non-uniform motion of an object moving in a straight line?

Q. The acceleration-time graph of motion of an object is given. What will be the corresponding most probable velocity-time graph of the object?

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Figure given below is the distance-time graph of the straight line motion of an object. (a) What will be the position of the object at 20 s? (b) What will be the distance travelled by the object in 12 s? (c) What is the average speed of the object?

(a) At 20 s, the object will be 8 m away from the starting point. (b) In 12 s, distance travelled by the object will be 6 m. (c) Average speed of the object, Average Speed = Total distance coveredTotal time taken= 8 m20 s= 0.4 m/s

Figure given below is the distance-time graph of the straight line motion of an object.

(a) What will be the position of the object at 20 s?
(b) What will be the distance travelled by the object in 12 s?
(c) What is the average speed of the object?

(a) At 20 s, the object will be 8 m away from the starting point.
(b) In 12 s, distance travelled by the object will be 6 m.
(c) Average speed of the object,
$A v e r a g e S p e e d = \frac{T o t a l d i s tan c e c o v e r e d}{T o t a l t i m e t a k e n} = \frac{8 m}{20 s} = 0.4 m / s$