Question

Consider a car moving along a straight horizontal road with a speed of ${ 72\, km\, h^{-1}}$. If the coefficient of static friction between road and tyres is $0.5$, the shortest distance in which the car can be stopped is

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

Answer 1

The Correct Option is B

Initial kinetic energy of the car $= \frac{1}{2} mv^2$
Work done against friction = $?mgs$
From conservation of energy
$\mu mgs = \frac{1}{2} m v^2 $ or $s = \left( \frac{v^2}{2 \mu g} \right)$
$\therefore$ Stopping distance, $s = \left( \frac{v^2}{2 \mu g} \right)$
Given, $v = {72 \, km \, h^{-1} = 72 \times \frac{5}{18} = 20 \, m \, s^{-1}}$
$\mu = 0.5$ and $g = {10 \, m \, s^{-2}}$
$ \therefore \:\:\: {s = \frac{20 \times 20}{2 \times 0.5 \times 10} = 40 \, m}$

Answer 2

Given in the question: Car moving with speed = 72kmh^-1

Change speed from 72kmh^-1 = 72×\(\frac{5}{18}\)ms^-1 = 4×5 m/s = 20 m/s

The coefficient value caused by the static friction between the tire and the road = 0.5

So, the acceleration will be = 10×0.5 ms^-2

This is due to frictional force. So, this is negative.

Hence, a = -5 ms^-2

Initial velocity, u = 20 ms^-1

Final velocity, v = 0

We need to find the distance between the initial and final travel points to know when it will stop.  So, we may apply a kinematic equation for this. That is,

v² - u² = 2as

Keeping value in it,

02 - (20)² = 2(-5)s

Solving it,

⇒ 400 =10s

Doing cross multiplication:

⇒ \(\frac{400}{10}\) = s

∴ s = 40m

So, the shortest distance where the car will be stopped is 40 m.

 

Hence, the correct option is B.