Question

Assume that a car is able to stop with a retardation of $8{\text{m/s}}^2$ and that a driver can react to an emergency in $0.5\text{sec}$. Calculate the overall stopping distance of the car for a speed of $72\text{km/hr}$ of the car

• A. $25\text{m}$
• B. $10\text{m}$
• C. $35\text{m}$
• D. $30\text{m}$

Answer 1

The correct option is C $35\text{m}$

Given, reaction time of driver $t=0.5\text{sec}$
retardation of the car $8{\text{m/s}}^2$ and initial velocity of the car $u=72\text{km/hr}=20\text{m/s}$
Distance covered by the car before application of brake $S_1=20t=20\times 0.5=10\text{m}$
Now, car begins to move with the retardation
So, distance covered by car before coming to rest is
$v^2-u^2=2as\Rightarrow 0^2-{20}^2=-2\times 8\times S_2$
$S_2=\frac{400}{16}=25\text{m}$
$\therefore$ Overall stopping distance of the car is
$S=S_1+S_2=10+25=35\text{m}$