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Question

Derive the equation for displacement-time relation. Graphically.

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Solution

Derivation of the second equation of motion:

Graph of a particle whose initial velocity is u and final velocity is v. Total time taken is t.


Distance covered by the object in the given time t is given by the area of the trapezium ABDOE

Let in the given time, t the displacement covered by the moving object = s

The area of trapezium, ABDOE

Displacement (s) = Area of △\triangleABD + Area of rectangle ADOE

Or,

s=12×AB×AD+AE×OEs=\frac 12 \times AB\times AD +AE\times OEs=21×AB×AD+AE×OE

Since, AB=DC=atAB=DC=atAB=DC=at, AD=OE=tAD=OE=tAD=OE=t and AE=uAE=uAE=u therefore we get,

s=12×at×t+u×ts=\frac 12 \times at \times t + u\times ts=21×at×t+u×t

Or,

s=ut+12at2

Hence the equation is derived.


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