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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, 01:43: PM
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×t

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×t

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

where u is initial velocity and a is acceleration

Answered by Shiwani Sawant | 28 May, 2018, 03:09: PM

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