Derived the position, time relation
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Explanation:
Consider the graph shown in the given attachment. obviously, the total distances travelled by the object in the 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,
s = (oa + bc / 2) × oc = (u×v/2) × t
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 = (u + v / 2) × ( v - u / a ) = v² - u²/ 2a
= v² - u² = 2as
Therefore, this equation is the position velocity
relation for uniformly accelerated motion.
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