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

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20 km h^{- 1}

25 km h^{- 1}

24 km h^{- 1}

10 km h^{- 1}

answer is C.

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Detailed Solution

Given data in the problem,
Uniform speed of motorcyclist from A to B,

v_{1} = 30 km h^{- 1}

Time taken,

t_{1} = \frac{s}{v_{1}}

= > t_{1} = \frac{s}{30} h

Uniform speed of motorcyclist from B to A,

v_{2} = 20 km h^{- 1}

Time taken,

t_{2} = \frac{s}{v_{2}}

{= > t}_{2} = \frac{s}{20} h

By using the below formula,
Average speed,

v_{av} = \frac{Total distance travelled}{Total time taken}

{= > v}_{av} = \frac{s + s}{t_{1} + t_{2}}

= > v_{av} = \frac{s + s}{\frac{s}{30} + \frac{s}{20}}

= > v_{av} = \frac{2 s}{\frac{5}{60} s}

= > v_{av} = 2 \times \frac{60}{5}

= > v_{av} = 24 km h^{- 1}

So, the average speed is

24 km h^{- 1}

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