Question

A motorcyclist drives from A to B with a uniform speed of $30\raisebox{1ex}{$km$}\!\left/ \!\raisebox{-1ex}{$h$}\right.$ and returns back with a speed of $20\raisebox{1ex}{$km$}\!\left/ \!\raisebox{-1ex}{$h$}\right.$. Find its average speed.

Answer 1

Step 1. Given data:

Motorcyclist speed from A to B = $30{\mathrm{kmh}}^{-1}$

Motorcyclist speed from B to A = $20{\mathrm{kmh}}^{-1}$

Let the distance from A to B is D km.

Distance for the entire journey is $2D$ km.

Step 2. Formula used:

Time taken $\left(T\right)$ = Distance$\left(D\right)$/ Speed $\left(v\right)$

Step 3. Calculations:

So, the total time taken $T$ is

$$\begin{gathered} T=\left(\frac{D}{30}\right)+\left(\frac{D}{20}\right) \\ =\frac{2D+3D}{60}=\frac{5D}{60}=\frac{D}{12}Hrs. \end{gathered}$$

Average speed = Total distance/Total time.

Average speed = $\frac{2D}{{\displaystyle \frac{D}{12}}}=\frac{2D\times 12}{D}=24{\mathrm{kmh}}^{-1}$

Hence, the average speed of the motorcycle is $24{\mathrm{kmh}}^{-1}$.