CBSE Class 9 Answered
Show that the area under velocity time graph for uniformly increasing velocity is equal to the distance covered by the body
Asked by snehanath217 | 09 Apr, 2023, 19:56: PM
Let the initial velocity of object be u . Let v be the velocity after time duration t.
If change in velocity with respect to time is uniform, then average velocity uavg in this time duration is
![begin mathsize 14px style u subscript a v g end subscript space equals space fraction numerator u plus v over denominator 2 end fraction end style](https://images.topperlearning.com/topper/tinymce/cache/2a4f19d6cb6beca9bc84e40ad25b6da6.png)
Distance S travelled is
S = uavg × t = (1/2) ( u + v ) t . . . . . . . . . . . . . . . . . . . . (1)
Graph below shows the typical variation of velocity that increases uniformly ( constant acceleration )
Line AB shows that velocity varies uniformly from u to v in time t . Time duration is shown as CD on time axis.
If we compare the expression given in eqn.(1) and area under the graph line AB , we see that area under the graph line AB is a trapezium as given by eqn.(1)
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/e266544b96b0234eb0eb5a00735016ac6432efac0894f8.45808681f4.png)
Hence , when velocity changes uniformly, then area under the velocity-time graph is the diplacement made in given time
Answered by Thiyagarajan K | 09 Apr, 2023, 22:40: PM
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