Question

Derive the position time relation.

Answer 1

Hey mate!

Here's your answer!!

Consider the graph shown in the given attachment. Obviously, the total distance s travelled by the object in time t is given by are under v-t graph.

Therefore, distance travelled, s = Area OABGC

Obviously, OABGC is a trapezium, whose area is (OA + BC/2) × OC.

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

Distance travelled,



From the velocity - time relation, we have

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

On substituting this value of t in equation, we get



➡ v² - u² = 2as

Therefore, this equation is the position velocity relation for uniformly accelerated motion.

Hope it helps you!
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Answer 2

$\textbf{Answer is in Attachment !}$