A driver, having a definite reaction time is capable of stopping his car over a distance of $30m$ on seeing a red traffic signal, when the speed of the car is $72km/hr$ and over a distance of $10m$ when the speed is $36km/hr$. Find the distance (in$m$) over which he can stop the car if it were running at a speed of $54km/hr$. Assume that his reaction time and the deceleration of the car remains same in all the three cases.

Let $t$ second be the reaction time of the driver to be able to put brakes and let $a$ be the acceleration of the car, when full brakes are applied.

For Case $-I:$
Given,
$v=72km/hr=72\times \frac{5}{18}=20m/s$
Distance travelled by the car, after the driver has seen the signal and before the driver has just applied the brakes (i.e., due to reaction time)
$=\text{velocity}\times t=20tm/s$
Thus the brakes stopped the car within the distance of $(30-20t)m$.
From,
$v^2=u^2+2as$
$a=\frac{0^2-{20}^2}{2\times (30-20t)}=\frac{-{20}^2}{2\times (30-20t)}$
$a=-\frac{20}{(3-2t)}$
Similarly for Case $-II:$
$v=36km/hr=36\times \frac{5}{18}=10m/s$
Distance travelled by the car, after the driver has seen the signal and before the driver has just applied the brakes (i.e., due to reaction time)
$=\text{velocity}\times t=10tm/s$
Thus the brakes stopped the car within the distance of $(10-10t)m$
From,
$v^2=u^2+2as$
$a=\frac{0-{10}^2}{2(10-10t)}=\frac{-5}{1-t}$
Similarly for Case $-III:$
$v=54km/hr=54\times \frac{5}{18}=15m/s$
Distance travelled by the car, after the driver has seen the signal and before the driver has just applied the brakes (i.e., due to reaction time)
$=\text{velocity}\times t=15tm/s$
Thus the brakes stopped the car within the distance of $(x-15t)m$.
where $x:$ distance travelled by car in third case
From $v^2=u^2+2as$
$a=\frac{0-{15}^2}{2(x-15t)}=-\frac{{15}^2}{2(x-15t)}$
As acceleration is same in all the cases.
$\frac{20}{3-2t}=\frac{5}{1-t}=\frac{{15}^2}{2(x-15t)}$
Reaction time $t=\frac{1}{2}$ sec
$⟹x=18.75m$
Final answer:18.75 $m$