A motorcyclist drives from place A to B with a uniform speed of $30\ km h^{-1}$ and returns from place B to A with a uniform speed of $20\ km h^{-1}$. Find his average speed.


Let $D$ be the distance from $A$ to $B$.

As given, Speed of the motorcyclist from $A$ to $B$ $v_1=30\ kmh^{-1}$

Therefore, time taken from $A$ to $B$ $t_1=\frac{distance}{speed}=\frac{d}{30}\ h$

And also given, speed from $B$ to $A$ $v_2= 20\ km h^{-1}$

Therefore, time taken in return $t_2=\frac{d}{v_2}=\frac{d}{20}\ h$

Therefore, total distance travelled $=d+d=2d$

Total time taken $t_1+t_2=\frac{d}{30}+\frac{d}{20}=\frac{50d}{600}=\frac{d}{12}$

Therefore, average speed $=\frac{total\ distance}{total\ time}$

$=\frac{2d}{\frac{d}{12}}$

$=\frac{2d\times12}{d}$

$=24\ kmh^{-1}$

There, the average speed of the motorcyclist is $24\ kmh^{-1}$

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Updated on: 10-Oct-2022

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