Question

Derive an expression for position-time relationship​

Answer 1

Consider the graph shown in the given attachment. Obviously, the total distances travelled by the object in time t is given by are under v-t graph. Obviously, OABGC is a trapezium, whose area is (OA + BC/2) × OC. Therefore, this equation is theposition velocity relation for uniformly accelerated motion.

Answer 2

Explanation:

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) x OC.

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

Distance travelled,

S = (aube) x oc = ("x") xt =

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

S = (") × (",*) = 2a

v2 - u? = 2as

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

Hope it helps you!