Derive the following equations of uniformly accelerated motion with the help of velocity-time graph [a] V=u+at [velocity-time relation] [b]2as=v square-u square [position-velocity relation] [c] S=ut+1/2at square [position-time relation]
Answers
Answer -
Equation of motion for uniformly accelerated motion -
i) \bf 1^{st}1
st
equation of motion -
Let an object travels with uniform acceleration 'a' whose initial velocity 'u'. After time 't' it's velocity becomes v -
From the formula of acceleration -
\implies⟹ \bf a = \frac{v - u}{t}a=
t
v−u
\implies⟹ \bf at = v - uat=v−u
\implies⟹ \bf v = u + atv=u+at
\boxed{\sf\purple{First\: equation \:of \:motion - v = u + at}}
Firstequationofmotion−v=u+at
━━━━━━━━━━━━
ii) \bf 2^{nd}2
nd
equation of motion -
If 's' is the displacement of body during time 't' then -
\bf Displacement = Average \:velocity \times timeDisplacement=Averagevelocity×time
\implies⟹ \bf s =( \frac{u + v}{2} )ts=(
2
u+v
)t
Value of v from 1st equation of motion
\implies⟹ v = u + at
\implies⟹ \bf s = t \times[\frac{u +( u + at)}{2}]s=t×[
2
u+(u+at)
]
\implies⟹ \bf s = \frac{t}{2} [2u + at]s=
2
t
[2u+at]
\implies⟹ \bf s = ut + \frac{1}{2} at^2s=ut+
2
1
at
2
\boxed{\sf\purple{Second\: equation \:of motion - s = ut + \frac{1}{2} at^2}}
Secondequationofmotion−s=ut+
2
1
at
2
━━━━━━━━━━━━
iii) Third equation of motion -
\bf Displacement = Average \:velocity \times timeDisplacement=Averagevelocity×time
\bf s =( \frac{u + v}{2} )ts=(
2
u+v
)t
Put value of t from 1st equation of motion -
\implies⟹ \bf v = u+atv=u+at
\implies⟹ \bf v-u = atv−u=at
\implies⟹ \bf t = \frac{v - u}{a}t=
a
v−u
\implies⟹ \bf s = (\frac{u + v}{2} )( \frac{v - u}{a} )s=(
2
u+v
)(
a
v−u
)
\implies⟹ \bf 2as = v^2 - u^22as=v
2
−u
2
\boxed{\sf\purple{Third \:equation \:of\: motion - v^2 - u^2 = 2as}}
Thirdequationofmotion−v
2
−u
2
=2as
━━━━━━━━━━━━