Question

A car starts from rest to cover distance $s$. The coefficient of kinetic friction between the road and tire is $\mu .$ The minimum time in which the car cover the distance s is proportional to
A. $\mu$
B. $\sqrt \mu$
C. $\dfrac{1}{\mu }$
D. $\dfrac{1}{{\sqrt \mu }}$

Answer 1

In this question, it is asked to determine the minimum time taken to cover a distance by a car when it starts from the rest position. To solve this question determine the frictional force acting and then evaluate the maximum acceleration. Finally, use the equation of motion to solve this problem. By substituting the value of evaluated acceleration and rearranging the equation the final answer can be obtained.

Complete step by step solution:
For a car to change its speed, a horizontal force must be applied to it. As a result of Newton’s third law, a force with the same amount of magnitude will be applied to it. If a force greater than the frictional force $\mu mg$ is applied, the car would slip.
From Newton’s second law, the acceleration can be calculated by dividing the force by the mass. So in this case the maximum acceleration will be
$a = \dfrac{{\mu mg}}{m}$
$\Rightarrow a = \mu g$
Here m is the mass of the car and g is the acceleration due to gravity.
From the equation $s = ut + \dfrac{1}{2}a{t^2}$
We can determine the minimum time
As the object is starting from rest, its initial velocity u will be equal to zero.
Therefore we get,
$s = \dfrac{1}{2}a{t^2}$
By rearranging and substituting the value of acceleration in the equation we get,
$s = \dfrac{1}{2}a{t^2}$
$\Rightarrow s = \dfrac{1}{2}\mu g{t^2}$
$\Rightarrow t = \sqrt {\dfrac{{2s}}{{\mu g}}}$
From this, we have obtained that the minimum time, $t \propto \sqrt {\dfrac{1}{\mu }}$
The minimum time, $t \propto \sqrt {\dfrac{1}{\mu }}$
Hence, Option D is correct answer.

Note:
From the above discussion, we have determined the minimum time it takes for a car to travel a distance s from its rest position. In addition to the equation of option employed in this question, there are other equations of motion too. One must be careful to use the correct equation to solve the problem otherwise the problem will become unnecessarily lengthy.