The engine of a motorcycle can produce a maximum acceleration of 5 ms^-2. Its brakes can produce a maximum retardation of 10 ms^-2. The minimum time in which it can cover a distance of 1.5 km is

Let v be the final velocity acquired when accelerated for time t₁ and the initial velocity(u) will be 0, in the first case.

Let t₂ be the time for which it retards and v will be the initial velocity and the final velocity will be 0, in the second case. 

Let ${\mathrm{s}}_1$ and ${\mathrm{s}}_2$ be the distances covered in these two cases. 

then

$\begin{array}{rcl}{\mathrm{\nu }}^2-0^2& =& 2{\mathrm{a}}_1{\mathrm{s}}_1\end{array}$

$\begin{array}{rcl}{\mathrm{s}}_1& =& \frac{{\mathrm{\nu }}^2}{2{\mathrm{a}}_1}=\frac{{\mathrm{\nu }}^2}{10} [\mathrm{Given} {\mathrm{a}}_1=5\mathrm{m}/{\mathrm{s}}^2]\end{array}$

Also,

$0^2-{\mathrm{\nu }}^2=2{\mathrm{a}}_2{\mathrm{s}}_2$

${\mathrm{s}}_2=\frac{-{\mathrm{\nu }}^2}{2{\mathrm{a}}_2}=\frac{{\mathrm{\nu }}^2}{20} [\mathrm{Given} {\mathrm{a}}_2=-10\mathrm{m}/{\mathrm{s}}^2]$

Here,  

$$\begin{gathered} {\mathrm{s}}_1+{\mathrm{s}}_2=1500\mathrm{m} \\ \Rightarrow \frac{{\mathrm{\nu }}^2}{10}+\frac{{\mathrm{\nu }}^2}{20}=1500 \\ \Rightarrow \mathrm{\nu }=100{\mathrm{ms}}^{-1} \end{gathered}$$

Now,

 $\begin{array}{rcl}\mathrm{\nu }& =& 0+{\mathrm{a}}_1{\mathrm{t}}_1\\ 100& =& 5{\mathrm{t}}_1\\ {\mathrm{t}}_1& =& 20\mathrm{s}\end{array}$

$\begin{array}{rcl}0& =& \mathrm{v}+{\mathrm{a}}_2{\mathrm{t}}_2\\ -100& =& -10{\mathrm{t}}_2\\ {\mathrm{t}}_2& =& 10\mathrm{s}\end{array}$ 

and  ${\mathrm{t}}_1+{\mathrm{t}}_2=30\mathrm{s}$