Q. A car $A$ is travelling on a straight level road with a uniform speed of 60 km/h. It is followed by another car $B$ which is moving with a speed of $70km/h$. When the distance between them is $2.5\ km,\ $the car $B$ is given a deceleration of $20\ km/h^{2}$.After how much time will $B$ catch up with $A$.
A) $1\ hr$
B) $1/2hr$
C) $1/4hr$
D) $1/8hr$
Explanation
(b) Let car $B$ catches, car $A$ after $'t'$sec, then
$60t + 2.5 = 70t - \frac{1}{2} \times 20 \times t^{2}$
$\Rightarrow 10t^{2} - 10t + 2.5 = 0 \Rightarrow t^{2} - t + 0.25 = 0$
$\therefore t = \frac{\sqrt{1 - 4 \times (0.25)}}{2} = \frac{1}{2}hr$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.