Question

A motorcyclist moving with uniform retardation takes $10\,s$ and $20\,s$ to travel successive quarter kilometres. How much further will he travel before coming to rest?

Answer 1

We will use second equation of motion i.e.,
$s = ut + \dfrac{1}{2}a{t^2}$
Where,
$s =$ distance travelled by motorcyclist
$u =$ initial velocity of motorcyclist
$a =$ acceleration
$t =$ time duration
Since the motorcycle is retarding hence acceleration will be negative. Further we will use ${v^2} = {u^2} + 2as$ to serve the purpose.

Complete Step by Step Answer
For first quarter kilometer, $s = 250\,m$, $t = 10\,s$
$\therefore \,250\, = \,u\left( {10} \right) - \dfrac{1}{2}\,a{\left( {10} \right)^2}$
$u - 5\,a = 25\,........\,(i)$
For second quarter kilometer, $s = 250\,m$
Hence, total distance travelled in $t = 30\,\sec$ equal to $500\,m$
Now, $s = 500\,m$, $t = 30\,\sec$
$\therefore \,\,\,500 = \,u\left( {30} \right) - \dfrac{1}{2}a{\left( {30} \right)^2}$
$3\,u - 45\,a = 50\,........\,(ii)$
On solving equation $(i)$ and $(ii)$ we get
$3\,u - 15\,a = 75\,$
$3u - 45a = 50$
$-$ $+$ $-$
$30a = 25$
$a = \dfrac{5}{6}\,m/{\sec ^2}$
Now, $3u - 15 \times \dfrac{5}{6} = 75$ (using value of $a$ in $(i)$)
$u = \dfrac{{175}}{6}\,m/s$
Now total distance $s$ before coming to rest.
Given, $u = \dfrac{{175}}{6}\,m/s$, $v = 0$, $a = \dfrac{5}{6}\,m/{\sec ^2}$
Using third equation of motion,
${v^2} - {u^2} = - 2as$
$- {u^2} = - 2as$
s = \dfrac{{{u^2}}}{{2a}}$$$${\left( {\dfrac{{175}}{6}} \right)^2} \times \dfrac{1}{2} \times \dfrac{6}{5} = 510.42\,m
\because \,$$$$500\,m is already travelled. Hence, distance travelled before coming to rest is $\left( {510.42 - 500} \right) = 10.42\,m$

Note:
We know acceleration is the rate of change of velocity. Since, motion is retarding motion, motorcycle. Finally coming to rest means velocity is decreasing continuously. Hence, acceleration is taken negatively.