Question

Prove that:
(i) v=u + at
(ii) s=ut+1/2at2
(iii) v2=u2 + 2as

Answer 1

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Prove that:-

• $\sf{v = u + at}$
• $\sf{s = ut + \dfrac{1}{2}a{t}^{2}}$
• $\sf{ {v }^{2} = {u}^{2} + 2as}$

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Let us assume that an article has a initial velocity 'u' and a uniform acceleration 'a' for time 't' so that its final velocity becomes 'v'.

And let the distance travelled by the article be 's' . The distance travelled by a moving body in time 't' can be found out by considering its average velocity.

$\\\sf{\therefore{ \bar{v} = \dfrac{s}{t}}}$

$\\ \sf{\implies{\bold{\blue{s = \bar{v}t }}- - - - - - - - - - - (1)}}$

Since the initial velocity of the body is 'u' and its final velocity is 'v', the average velocity is given by :-

$\begin{gathered}\begin{gathered} \\\sf{\blue{\bold{\bar{v} = \frac{u + v}{2} }}} - - - - - - - - (2)\\\\\end{gathered}\end{gathered}$

Putting the value of equation 2 in equation 1:-

• $\sf{\bold{\blue{s = (\dfrac{u + v}{2})t}} - - - - - - - - - - (3)}$

Again, rate at which velocity changes with time, in terms of both speed and direction is called acceleration. Which means:-

$\\ \bold{a = \dfrac{v - u}{t}}$

$\\ \sf{\implies{v - u = at}}$

$\\ \sf{\implies{\red{\bold{v = u+ at}}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\rm{Proved}}$

Putting the value of 'v' in equation 3,

$\begin{gathered}\begin{gathered} \\ \sf{\bold{s = ( \frac{u + u + at}{t} )t}}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\\\sf{\implies{\sf{\bold{s} = ( \frac{2u + at}{2})t}}} \end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\\\sf{\implies{\bold{s} = \frac{2ut + a {t}^{2} }{2} }}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\\\sf{\implies{\bold{s} = \frac{\cancel{2}ut}{\cancel{2}} + \frac{a {t}^{2} }{2}}} \end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\\\sf{\implies{\bold{s} = ut + \frac{at {}^{2} }{2}}} \end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\\\sf{\therefore{\bold{\red{s = ut + \frac{1}{2} a {t}^{2} }}}}\end{gathered} \end{gathered} \: \: \: \: \: \: \: \: \: \boxed{\rm{Proved}}$

Again,

Again, rate at which velocity changes with time, in terms of both speed and direction is called acceleration. Which means:-

$\\ \bold{a = \dfrac{v - u}{t}}$

$\\ \sf{\implies{\blue{\bold{{t = \dfrac{v - u}{a}}}}}}$

Putting the value of 't' in equation 3,

$\\ \bold{s = ( \dfrac{u + v}{2} ) \times ( \dfrac{v - u}{a} )}$

$\\ \sf{\implies{s = \frac{ {v}^{2} - {u}^{2} }{2a}}}$

$\\ \sf{\implies{ {v}^{2} - {u}^{2} = 2as}}$

$\\ \sf{\implies{ {v}^{2} = 2as + {u}^{2} }}$

$\\ \sf{\therefore{\red{\bold{ {v}^{2} = {u}^{2} + 2as}}}} \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\rm{Proved}}$

Where,

• v = Final velocity
• u = Initial velocity
• a = acceleration
• s = Distance
• t = Time