Question

Derive The Velocity Position Equation of motion (v2=u2+2aS ) graphically (with the help of a velocity time graph)

Answer 1

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Here's your answer!!
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Consider the velocity time graph for a uniformly accelerated motion along a straight line, which is shown in the above attachment. Obviously, the total distance travelled by the object in time t is given by area under v-t graph.

∴ Distance travelled, s = Area OABGC

Obviously, OABGC is a trapezium, whose area is...

$= ( \frac{oa + bc}{2} ) \times oc$

But, OA = u, BC = v and OC = t

∴ Distance travelled,

$s = ( \frac{oa + bc}{2} ) \times oc = ( \frac{u + v}{2} ) \times t$

From the velocity time relation, we have

at = v - u or t = v - u/a

On substituting this value of t in the equation, we get,

$s = ( \frac{u + v}{2} ) \times ( \frac{v - u}{a} ) = \frac{v {}^{2} - u {}^{2} }{2a}$

➡ v²- u² = 2as

∴ v² = u² +2as

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