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 a speed of \[54{\text{ }}km/h\] and the brakes cause a deceleration of \[6.0{\text{ }}m/{s^2}\] , find the distance traveled by the car after he sees the need to put the brakes on.

Hint- In this question, first we will determine the distance travelled by car during reaction time, then we will proceed further by using the third equation of motion to find the distance covered by the car in the deceleration period. We will consider the final velocity 0 as the will finally come to halt.
Formula used- \[Distance = speed \times time,{v^2} - {u^2} = 2as\]

Complete step-by-step answer:

Given that
Initial velocity \[ = u = 54km/hr = 54 \times \dfrac{5}{{18}}m/s = 15m/s\]
Reaction time=t=0.2sec
Final velocity \[ = v = 0m/s\]
Deceleration \[6.0{\text{ }}m/{s^2}\]
Now, Distance travelled by car during reaction time:
\[
   \Rightarrow Distance = speed \times time \\
   \Rightarrow Distance = 15m/s \times 0.2s \\
   \Rightarrow Distance = 3m \\
 \]
As we know that the third equation of motion gives the final velocity of an object under uniform acceleration given the distance traveled and an initial velocity:
\[{v^2} - {u^2} = 2as\]
Substitute the value of v, u and a in this equation, so we have
$
   \Rightarrow {0^2} - {15^2} = 2\left( { - 6} \right)s \\
   \Rightarrow - 225 = - 12s \\
   \Rightarrow s = \dfrac{{ - 225}}{{ - 12}} \\
   \Rightarrow s = \dfrac{{225}}{{12}} = 18.75m \\
 $
Therefore, the total distance travelled by the driver after he sees the need to put the brakes ON
\[ = 18.75 + 3 = 21.75m\]
Hence, distance traveled by the car after he sees the need to put the brakes on is 21.75m

Note- Newton's laws of motion are three physical laws that, together, laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. Students must remember all the three laws of motion in order to solve the problem related to kinematics easily.