Question

A car starts from rest to cover a distance s.
The coefficient of friction between the road
and the tyres is u. The minimum time in
proportional to -
which the car can cover the distance is
(B) VH
(C)
1
H
(D)​

Answer 1

First of all , we need to draw the FBD of the car and road system .

we can say that in Limiting Conditions :

Applied force = Frictional force

$= > f = \mu mg$

$= > \cancel ma = \mu \cancel mg$

$= > a = \mu g$

Now applying the equations of Kinematics , we get :

$s = ut + \frac{1}{2} a {t}^{2}$

$= > s = (0 \times t) + \frac{1}{2} a {t}^{2}$

$= > s = \frac{1}{2} \times ( \mu g) \times {t}^{2}$

$= > {t}^{2} = \dfrac{ 2s}{ \mu g}$

$= > t = \sqrt{\dfrac{ 2s}{ \mu g} }$

Considering s and g to be constant ;

$= > t \propto \sqrt{\dfrac{ 1}{ \mu } }$

$= > t \propto \dfrac{1}{ \sqrt{ \mu} }$

So final answer :

$\boxed{ \red{ \huge{ \bold{ t \propto \dfrac{1}{ \sqrt{ \mu} } }}}}$