Q. A car moves from $X\ to\ Y$ with a uniform speed $v_{u}$ and returns to $Y$ with a uniform speed $v_{d}.\ $The average speed for this round trip is
A) $\frac{2v_{d}v_{u}}{v_{d} + v_{u}}$
B) $\sqrt{v_{u}v_{d}}$
C) $\frac{v_{d}v_{u}}{v_{d} + v_{u}}$
D) $\frac{v_{u} + v_{d}}{2}$
Explanation
(a) Let $t_{1}\ and\ t_{2}$ be times taken by the car to go from $X\ to\ Y$ and then from $Y\ to\ X$ respectively.
Then, $t_{1} + t_{2} = \frac{XY}{v_{u}} + \frac{XY}{v_{d}} = XY\ \left( \frac{v_{u} + u_{d}}{v_{u}v_{d}} \right)$
Total distance travelled
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = XY + XY = 2\ XY$
Therefore, average speed of the car for this round trip is
$Average\ speed = \frac{distance\ travelled}{time\ taken}$
$v_{av} = \frac{2\ XY}{XY\ \left( \frac{v_{u} + v_{d}}{v_{u}v_{d}} \right)}or\ v_{av} = \frac{2v_{u}v_{d}}{v_{u} + v_{d}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.