Q. A police car is travelling in a straight line with a constant speed $v.$ A truck travelling in the same direction with constant velocity 3v/2 passes, the police car at $t = 0.$ The police car starts acceleration 10 s after passing the truck, at a constant rate of $3\ {ms}^{- 2},$ while truck continues to move at constant speed. If he police car takes 10 s further to catch the truck, find the value of $v$
A) $10\ {ms}^{- 1}$
B) $15\ {ms}^{- 1}$
C) $20\ {ms}^{- 1}$
D) $30\ {ms}^{- 1}$
Explanation
(b) [image:3_image174.emf]
The diagram is showing the position of car and truck at various instants.
$v \times 20 + \frac{1}{2} \times 3 \times 10^{2} = \frac{3v}{2} \times 20$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{3}{2} \times 100 = \frac{v}{2} \times 20$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = 15\ {ms}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.