Q. Two cars $A$ and $B$ at rest at same point initially. If $A\ $starts with uniform velocity of $40\ m/sec$ and $B\ $starts in the same direction with constant acceleration of $4m/s^{2}$, then $B$ will catch $A\ $after how much time
A) 10 $\sec$
B) $20\ sec$
C) $30\ sec$
D) $35\ sec$
Explanation
(b) Let $A\ $and $B$ will meet after time $t\ sec$. it means the distance travelled by both will be equal
$S_{A} = ut = 40t$ and $S_{B} = \frac{1}{2}at^{2} = \frac{1}{2} \times 4 \times t^{2}$
$S_{A} = S_{B} \Rightarrow 40t = \frac{1}{2}4t^{2} \Rightarrow t = 20\ sec$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.