On a horizontal rough road, value of coefficient of friction  $\mu$ =0.4. Find the minimum time in which a distance of 400m can be covered. The car starts from rest and finally comes to rest.

 Maximum friction on horizontal rough road, $f_{\max }=\mu mg$

$\therefore$Maximum acceleration or retardation of the car may be

$$\begin{gathered} a_{\max }\text{ or }a=\frac{f_{\max }}{m}=\frac{\mu mg}{m}=\mu g \\ =0.4\times 10=4 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Let, the car accelerates and retards for time ' t' with $4 \mathrm{m}/{\mathrm{s}}^2$.

Then.

$$\begin{gathered} \frac{1}{2}a{t}^2+\frac{1}{2}a{t}^2=400 \mathrm{m} \\ a{t}^2=400 \mathrm{m} \\ 4t^2=400 \\ t=10 \mathrm{s} \end{gathered}$$

or

Therefore, the minimum time is $20 \mathrm{s}$ , 10s of acceleration and $10 \mathrm{s}$ of retardation).