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?

Hint:
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.