**Considering an object in uniform motion, its velocity-time graph can be represented as follows.**

Now, the area below the velocity-time graph is the area of the rectangle OABC, which is given by

OA*OC. But OA is the velocity of the object and OC represents time. Therefore, the shaded area can be represented as:

Area under the velocity-time graph = velocity*time.

Substituting the value of velocity as \( \frac{displacement}{time}\) in the previous equation, it is found that the area under the velocity-time graph represents the total displacement of the object.

Answered by Shivani Kumari | 1 year ago**A bus starting from rest moves with a uniform acceleration of 0.1 m s ^{-2} for 2 minutes. Find**

**(a)** the speed acquired,

**(b) **the distance traveled.

A farmer moves along the boundary of a square field of side 10m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

**A bus starting from rest moves with a uniform acceleration of 0.1 m s ^{-2} for 2 minutes. Find**

**(a)** the speed acquired,

**(b)** the distance travelled.

What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?