Question

A car starts from rest. It has to cover a distance of $500\text{m}$ and come to rest at the end of the journey. The coefficient of friction between the road and the tyre is $\frac{1}{2}$. The minimum time in which the car can cover this distance is $(g=10{\text{m/s}}^2)$

• A. $20\text{seconds}$
• B. $10\text{seconds}$
• C. $30\text{seconds}$
• D. $15\text{seconds}$

Answer 1

The correct option is A $20\text{seconds}$

Maximum acceleration and maximum retardation of the car can be
$\mu g$ or $\frac{1}{2}\times 10=5{\text{m/s}}^2$
The corresponding velocity-time graph is as shown in the figure.
Let $t_0$ be the time of acceleration and retardation. Then
$v_m=\mu g{t}_0=5t_0$
Now, displacement = area under $v-t$ graph
$\therefore 500=\frac{1}{2}\times 2t_0\times 5t_0$
or $t_0=10\text{seconds}$
$\therefore$ Total time of journey is $2t_0$ or $20\text{seconds}$.
Hence, the correct answer is option (a).