Question

derive the equation of motion

Answer 1

Derivation..

1. v=u+at

We know,

a=v-u/t

at+u=v

Therefore,v=u+at

2.s=ut+1/2at^2

Average speed=average velocity=u+v/2

D=s*t

S=average speed *t

S=(u+v/2)*t

Let it be equation no. 1

V=u+at

In equation no.1

S=(u+u+at/2)*t

S=(2u+at/2)*t

S=2ut+at^2/2

S=2ut/2+at^2/2

Therefore, s=ut+1/2at^2

3.v^2=u^2+2as

Average speed =average velocity =u+v/2

S=average speed*t

S=(u+v/2)*t

Let it be equation no.1

We know,

a=v-u/t

Therefore, t=v-u/a

Let it be equation no.2

Put equation 2in1

S=(u+v/2)(v-u/a)

S=(u+v)(v-u)/2a

S=u^2-v^2/2a(by applying formula a^2-b^2)

2as=u^2-v^2

Therefore, v^2=u^2+2as

Thus derived

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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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