Question

A person goes from A to B at a speed of $40{\text{m s}}^{-1}$ and comes back from B to A at a speed of $60{\text{m s}}^{-1}$. What is his average speed?

• A. $34{\text{m s}}^{-1}$
• B. $42{\text{m s}}^{-1}$
• C. $40{\text{m s}}^{-1}$
• D. $48{\text{m s}}^{-1}$

Answer 1

The correct option is D $48{\text{m s}}^{-1}$

We know that,
$\text{Average speed}=\frac{\text{Total distance travelled}}{\text{Total time taken}}$

Let the distance between A and B be $d$.
The total distance travelled $=2d$.
Let the time taken to travel from A to B be $t_1$ and the time taken to travel from B to A be $t_2$

i) Speed from A to B $=\frac{d}{t_1}=40$
$\Rightarrow t_1=\frac{d}{40}$
ii) Speed from B to A $=\frac{d}{t_2}=60$
$\Rightarrow t_1=\frac{d}{60}$

Now, the required average speed $=\frac{2d}{t_1+t_2}$

$=\frac{2d}{\frac{d}{40}+\frac{d}{60}}$

$=\frac{2d}{\frac{(40+60)\times d}{40\times 60}}$

$=\frac{2\times 2400}{100}=48{\text{m s}}^{-1}$