Question

A driver, having a definite reaction time, is capable of stopping over a distance of $30\text{m}$ on seeing a red traffic light when speed of the car is $72\text{km/hr}$ and over a distance of $10\text{m}$ when the speed is $36\text{km/hr}$. Assuming that his reaction time and the acceleration of the car remain the same in all cases, find the distance over which he can stop the car when he is going at $54\text{km/hr}$.

• A. $11.75\text{m}$
• B. $13.75\text{m}$
• C. $15.75\text{m}$
• D. $18.75\text{m}$

Answer 1

The correct option is D $18.75\text{m}$

Let the reaction time of driver be $t$
So, the distance covered before application of brake $=20t\text{m}$
For the first case, speed of the car is $72\text{km/hr}=20\text{m/s}$
So, we have
$v^2-u^2=2as\Rightarrow 0^2-{20}^2=2a(30-20t)$...(i)
Similarly, in the second case, speed of the car is $36\text{km/hr}=10\text{m/s}$
$\therefore 0-{10}^2=2a(10-10t)$...(ii)
On solving (i) and (ii) we get
$\frac{400}{100}=\frac{2a(30-20t)}{2a(10-10t)}$
$\Rightarrow 4=\frac{30-20t}{10-10t}$
or, $40-40t=30-20t\Rightarrow 20t=10$
or, $t=0.5\text{sec}$
and, putting the value of $t$ in any equation we get $a=-10{\text{m/s}}^2$
So, in the third case, let the stopping distance be $s$
Thus, speed of the car is $54\text{km/hr}=15\text{m/s}$
$\Rightarrow 0^2-{15}^2=-2\times 10(s-15\times 0.5)$
$\Rightarrow -225=-20(s-7.5)\Rightarrow 11.25=s-7.5$
$\Rightarrow s=18.75\text{m}$