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Question: The brakes of a car moving at 20m/s along a horizontal road are suddenly applied and it comes to res...

The brakes of a car moving at 20m/s along a horizontal road are suddenly applied and it comes to rest after travelling some distance. If the coefficient of friction between the tyres and the road is 0.90 and it is assumed that all four tyres behave identically, find the shortest distance the car would travel before coming to a stop.
A. 2.22m2.22m
B. 11.35m11.35m
C. 22.2m22.2m
D. 4.54m4.54m

Explanation

Solution

We know that when a car applies the brake the car stops due to the frictional force which acts on the tyres of the car in the direction opposite to the movement of the car. The force of the mass of the car will be acting in the downward direction. Taking all these forces into account we will find the distance travelled by the car before it stops.

Complete Step-By-Step answer:
We have given that,
u=20m/su=20m/s, Initial velocity
v=0m/sv=0m/s, Final velocity
μ=0.90\mu =0.90, Coefficient of friction

We will now start by finding out the acceleration of the car which was caused due to friction (uniform retardation).
We will be finding this by using the formula,
α=−μg\alpha =-\mu g

Now we will substitute the values in the above equation, we get,
α=−(0.9×10)m/s2\alpha =-(0.9\times 10)m/{{s}^{2}}
α=−9m/s2\alpha =-9m/{{s}^{2}}

We have the values of initial velocity, final velocity and acceleration, using these we will use the equation of motion formula to find out the distance travelled by the car before stopping.
The Formula we will be using is,
v2=u2+2as{{v}^{2}}={{u}^{2}}+2as

Now substituting the values, we get,

& {{(0)}^{2}}={{(20)}^{2}}+2(-9)s \\\ & 18s=400 \\\ & s=22.2m \\\ \end{aligned}$$ **Hence, option C is the correct answer.** **Note:** It is important to know the different forces acting on the car when it is in motion and when brake is being applied. However, if the four wheels were not braking optimally then the car would have travelled for more distance. Depending on how the force is being acted upon the car, the signs of the forces varies, which is to be observed carefully. When the value of acceleration due to gravity is not mentioned then the students are supposed to take it as, $$g=10m/{{s}^{2}}$$ .