CBSE Class 9 Answered
Show that the area under velocity time graphs for uniform acceleration gives the distance travelled s=ut+1/2atsquare
Asked by Shauryapratapsingh310 | 28 May, 2018, 13:43: PM
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/9277f6d4867a1452d0cfb7137dc325065b0bca1f1f68a2.29063898may2802.png)
By Geomtry :-
Let a body starts with initial velocity u at time t1, moves with uniform acceleration and its final velocity at time t2 is v.
The velocity-time graph shows the plot for this movement.
The area under the line BC that represents the above movement is a trapezium ABCD.
Area of trapezeium = (1/2)[AB+CD]×(t2-t1)
let (t2-t1) = t , then Area of trapezium = (1/2)[u+v]×t ..............(1)
if we substitute the definition of acceleration i.e., v = u+at in (1) we get area = u×t +(1/2)a×t2
Since area under the curve in a velocity-time graph is the distance travelled S, we write the equation as
S = u×t +(1/2)a×t2
By Calculus:
velocity = dS/dt = v(t) ...........(1)
where S is displacement, v(t) is velocity which is function of t.
hence we write, dS = v(t)dt ..............(2)
By integrating eqn.(2) , we get the total displacement S which is written as
![begin mathsize 12px style integral d s space space equals space S space equals space integral subscript 0 superscript t v left parenthesis t apostrophe right parenthesis space d t apostrophe space space equals space integral subscript 0 superscript t left parenthesis u plus a t apostrophe right parenthesis space d t apostrophe space equals space u t plus 1 half a t squared end style](https://images.topperlearning.com/topper/tinymce/cache/49c6f4113274fb7a7301b00a09bdfb44.png)
where u is initial velocity and a is acceleration
Answered by Shiwani Sawant | 28 May, 2018, 15:09: PM
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