Question

derivation of equation of motion

Answer 1

We know that
a= v-u / t
v-u = at
therefore v = u +at ; This is our first eq of motion
velocity = distance traveled / time taken v =  s/t => s= vt

If u is the initial velocity of an uniformly accelerating body and v is its velocity after a time t, then since the acceleration is UNIFORM, we can find the average velocity of the body as follows average velocity = (u+v)/2 Now, the distance s, traveled in the time t by the body is given by distance traveled = average velocity x time s = [(u+v)/2]t From eq 1 we have v=u+at, substituting this in the above equation for v, we get s = [(u+u+at)/2]t => s = [(2u+at)/2]t => s = [(u + (1/2)at)]t               => s =  ut + (1/2)at2 ; This is our 2nd eq of motion

   
We start with squaring eq 1 Thus we have v2 = (u+at)2                  =>    v2 = u2 + a2t2 + 2uat            =>    v2 = u2 + 2uat + a2t2              =>    v2 = u2 + 2a(ut + (1/2)at2) now, using eq 2  we have =>   v2 = u2 + 2as
That was the third eq of motion.

HOPE THAT IT HELPED YOU...          

Answer 2

Answer:

Explanation:

draw a velocity time graph for a body moving with uniform acceleration.

1. a = v - u/ t

v - u/ t = a

at = v - u

at + u = v

v = u + at

 position - time relation

for that find the area of the graph

we will get

s = 1/2 x t x (v-u) + ut

substituting (v-u) as at

from, a = v - u / t

at = v - u

since  s = 1/2 x t x at x ut

s = 1/2 at² + ut

position - velocity relation

find the area

1/2 x t (u + v )

substituting for t = v-u / a

from , a =v - u/ t

t = v - u / a

since

s = 1/2 (v - u /a) (u + v )

s= 1 / 2a ( v - u ) ( v + u)

s = 1 / 2a (v² - u²)

2as = v² - u²

hope it helps.....

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