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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
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Verified Answer
The correct answer is:
$\frac{2 v_d v_u}{v_d+v_u}$
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
$\begin{aligned} v_{\mathrm{av}} & =\frac{2 X Y}{X Y\left(\frac{v_u+v_d}{v_u v_d}\right)} \\ \text { or } \quad v_{\mathrm{av}} & =\frac{2 v_u v_d}{v_u+v_d}\end{aligned}$
Total distance travelled
$=X Y+X Y=2 X Y$
Therefore, average speed of the car for this round trip is
$\begin{aligned} v_{\mathrm{av}} & =\frac{2 X Y}{X Y\left(\frac{v_u+v_d}{v_u v_d}\right)} \\ \text { or } \quad v_{\mathrm{av}} & =\frac{2 v_u v_d}{v_u+v_d}\end{aligned}$
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