Q. A car is moving along a straight road with uniform, acceleration. It passes through two points $P$ and $Q$ separated by a distance with velocities 30$kmh^{- 1}$ and 4$0kmh^{- 1}$ respectively. The velocity of car midway between $P$ and $Q$ is
A) 33.3$km^{- 1}$
B) 1$\ km^{- 1}$
C) 25$\sqrt{2}km^{- 1}$
D) 35.35$\ km^{- 1}$
Explanation
(d) $40^{2} - 30^{2} = 2aS,v^{2} - 30^{2} = 2a\frac{S}{2}$
or $2\left( v^{2} - 30^{2} \right) = 2aS$
Comparing, $2\left( v^{2} - 900 \right) = 1600 - 900 = 700$
or $v^{2} = 900 + 350 = 1250$ or $v = 35.35\ kmh^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.