Question

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 $T$. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

• A. $\frac{2 v_{d} v_{u}}{v_{d} + v_{u}}$
• B. $ \sqrt{v_{u}v_{d}}$
• C. $ \frac{v_{d}v_{u}}{v_{d}+v_{u}}$
• D. $ \frac{v_{u}+v_{d}}{2}$

Answer 1

Answer: (A)

Average speed of a body in a given time interval is defined as the ratio of distance travelled to the time taken.
Average speed $=\frac{\text { Distance travelled }}{\text { Time taken }}$
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, $t_{1}+t_{2}=\frac{X Y}{v_{u}}+\frac{X Y}{v_{d}}$
$=X Y\left(\frac{v_{u}+v_{d}}{v_{u} v_{d}}\right)$
Total distance travelled
$=X Y+X Y=2 X Y$
Therefore, average speed of the car for this round trip is
$v_{a v}=\frac{2 X Y}{X Y\left(\frac{v_{u}+v_{d}}{v_{u} v_{d}}\right)}$
or $v_{a v}=\frac{2 v_{u} v_{d}}{v_{u}+v_{d}}$