Question

Consider a car moving along a straight horizontal road with a speed of $72$ km/h. If the coefficient of static friction between the tyre and the road is $0.5$, the shortest distance in which the car can be stopped is?(Take $g=10m{s}^{-2}$)

• A. $30$m
• B. $40$m
• C. $72$m
• D. $20$m

Answer 1

The correct option is B $40$m

$\text{Step 1: Calculation of acceleration due to friction force}$

For stopping the car in shortest distance without skidding(sliding), the maximum static friction should act in opposite direction.

So, $f=\mu N$

$\Rightarrow$ $f=\mu mg$ $($ From figure, $N=mg)$

Now, Applying Newton's second law on car along x-direction (Rightwards Positive)

$\Sigma F_x=m{a}_x$

$\Rightarrow$ $-f=ma$

$\Rightarrow$ $-\mu mg=ma$

$\Rightarrow$ $a=-\mu g$ $=-0.5\times 10$ $=-5$ $m{s}^{-2}$

Negative means left direction (retardation)

$\text{Step 2: Calculation of Stopping distance}$

Since Retardation is constant, so applying kinematics equation of motion:

$v²=u²+2as$

Put $a=-5$ $m{s}^{-2}$ and initial velocity $u$ $=72\times 5/18=20m/s$

$\Rightarrow$ $0={20}^2-2\times 5\times s$

$\therefore$ $s=40m$

Shortest distance in which car can be stopped is $40m$

Option B is correct.