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 agoA 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?