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 the tyre and the road is 0.5, the shortest distance in which the car can be stopped is? (taking \[g = 10\,m{s^{ - 2}}\] )
A. 30 m
B. 40 m
C. 72 m
D. 20 m

Answer 1

Hint:To solve this question we should know about the coefficient of static friction. It is the ratio of the greatest static friction force (F) between the surfaces in contact before movement begins to the normal (N) force is the coefficient of static friction.So, in this problem we have to apply the deceleration concept. Here, fraction force works as retardation force which brings motion to an end. We will apply Newton's law of motion.

Complete step by step answer:
Given in the question: Car moving with speed= $72\,km{h^{ - 1}}$
Change speed from $72\,km{h^{ - 1}} = 72 \times \dfrac{5}{{18}}\,m{s^{ - 1}}= 4 \times 5\,m/s =20\,m/s$
The coefficient of static friction between the tyre and the road = 0.5
So, the acceleration will be $ = 10 \times 0.5\,m{s^{ - 2}} = 5\,m{s^{ - 2}}$
This is due to friction force. So, it will be negative in nature.
Hence, $a = - 5\,m{s^{ - 2}}$
Initial velocity, u $ = 20\,m{s^{ - 1}}$
Final velocity, v $ = 0$

As we have to find distance travel between initial and final when it will stop.So, we will apply a kinematic equation.That is,
${v^2} - {u^2} = 2as$
Keeping value in it,
${0^2} - {(20)^2} = 2( - 5)s$
Solving it,
$ \Rightarrow 400 = 10s$
Doing cross multiplication:
$ \Rightarrow \dfrac{{400}}{{10}} = s$
\[ \therefore s = 40\,m\]
So, the shortest distance in which the car can be stopped is 40 m.

Hence, the correct option is B.

Note:We must keep in mind that there will be negative acceleration due to friction force. So, it will stop moving. When we will solve such a problem then take it as negative.Always correctly choose initial and final speed because our whole solution depends upon it.