Question

A driver having a definite reaction time is capable of stopping his car over a distance of 30 m on seeing a red traffic signal, when the speed of the car is 72 km/hr and over a distance of 10 m when the speed is 36 km/hr. What is the reaction time of the driver?

A. 1s
B. 0.25s
C. 0.5s
D. 2s

Answer 1

Hint: Reaction time is that time in which a driver realizes that he should apply brakes and he does it. After application of the brakes anyhow due to the deceleration of the car it stops at the signal. It is to be assumed that the deceleration is constant and reaction time is constant.

Formula used:
$\eqalign{
  & s = ut \cr
  & {v^2} - {u^2} = 2as \cr} $

Complete step by step answer:
Usually in the ideal case of the problems we solve them by taking reaction time as zero. That means as soon as the driver sees the traffic signal he will apply the brakes and the time taken to stop is only stopping time and the distance travelled will be completely stopping distance due to application of brakes.
But in this case we consider that after the driver sees the signal he takes time ‘t’ and applies the brakes and then after sometime the car will stop completely. During this time ‘t’ there will be no deceleration and the car moves with uniform velocity. Let the distance travelled during this time be${s_1}$. After application of brakes it moves with deceleration ‘a’ and the distance travelled be ${s_2}$.
We have the formula
$s = ut$
Here ‘u ‘is the velocity and $u = 72km/hr = 20m/s$.
$s = ut$
$ \Rightarrow {s_1} = 20t$
After application of brakes we have ${v^2} - {u^2} = 2as$ where ‘v’ is the final velocity which is zero in this case and ‘a’ is the deceleration and ‘u’ is 20m/s
${v^2} - {u^2} = 2as$
$ \Rightarrow {s_2} = \dfrac{{{{20}^2}}}{{2a}}$
So ${s_1} + {s_2} = 30m$
$ \Rightarrow 20t + \dfrac{{{{20}^2}}}{{2a}} = 30$
   …eq 1
For the second case we have
We have the formula
$s = ut$
Here ‘u ‘is the velocity and $u = 36km/hr = 10m/s$.
$s = ut$
$ \Rightarrow {s_1} = 10t$
After application of brakes we have ${v^2} - {u^2} = 2as$ where ‘v’ is the final velocity which is zero in this case and ‘a’ is the deceleration and ‘u’ is 10m/s
${v^2} - {u^2} = 2as$
$ \Rightarrow {s_2} = \dfrac{{{{10}^2}}}{{2a}}$
So ${s_1} + {s_2} = 10m$
$ \Rightarrow 10t + \dfrac{{{{10}^2}}}{{2a}} = 10$ … eq 2
Hence by solving equation 1 and equation 2 we will get t=0.5 sec

So, the correct answer is “Option C”.

Note:
Actually the reason that they gave us two cases is because we will get an equation after solving the first case which contains unknown deceleration and stopping time. So when there are two unknowns then obviously we should have two equations to solve them.