Question

A driver takes 0.20 s to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he is driving a car at the speed of 54km/h and the brakes cause a deceleration of $6.0m/{s^2}$, find the distance travelled by the car after he sees the need to put the brakes on.
$A. 25 \\\ B. 21.75 \\\ C. 15 \\\ D. 11.75 \\\$

Answer 1

Hint- Here we will proceed by using the formula of $Dis\tan ce = speed \times time$ and ${v^2} - {u^2} = 2as$ i.e. the Newton’s third law of motion. Then by substituting the values of initial velocity, distance and deceleration, we will get the required result.

Formulas used-
$Dis\tan ce = speed \times time$
${v^2} - {u^2} = 2as$( newton’s third law of motion)

Complete step-by-step answer:
We are given that-
Initial velocity is u = 54km/h
Time t = 0.2 sec
Final velocity is v = 0m/s
Deceleration is $d = 6.0m/{s^2}$
Now, we will calculate distance travelled by car during reaction time-
$Dis\tan ce = speed \times time$
Firstly, we will convert the initial velocity in km/h to m/s
u = 54km/h
or $u = 54 \times \dfrac{5}{{18}}$
or u = 15m/s (here u is distance)
Now, substituting the given values of speed i.e. 15m/s and time = 0.2 s
We get-
$Dis\tan ce = 15 \times 0.2$
So, the distance is 3m.
Using newton’s third law of motion i.e. For every action in nature, there is an equal and opposite reaction.
We get-
${v^2} - {u^2} = 2as$
Substituting the given values of v=0, u=15 in the third law of motion,
We get-
${0^2} - 15 \times 15 = 2*6*s$
$\Rightarrow s = \dfrac{{ - 15 \times 15}}{{ - 12}}$
$\Rightarrow s = \dfrac{{225}}{{12}}$
$\Rightarrow$ s = 18.75m
So, the total distance travelled by the driver after he sees that there is need to put the brakes on-
18.75 + 3 = 21.75m
Therefore, the option B is correct.

Note- In order to solve this type of question, we must know the three laws of motion as here we one of its laws i.e. for every action in nature, there is an equal and opposite reaction. Also we should make sure that we write SI units along with the answer.