Question

derive position - time relation by graphical method

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,

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

From the velocity - time relation, we have

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

On substituting this value of t in equation, we get

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

➡ v² - u² = 2as

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

Hope it helps you!
✌ ✌ ✌

#BE BRAINLY

Answer 2

Answer: S = \frac{1}-{2} [u+v]t

               v = u+at

             v-u = at

                 t = \frac{v-u}-{a}

                 S = \frac{1}-{2} [u+v]t

                    = \frac{1}-{2} [u+v][\frac{u-v}-{a} ]

               as = \frac{1}-{2} [u+v][u-v]

             2as = [u+v][u-v]

                    = v^{2} - u^{2}

Explanation: