Question

A body covers half of its journey with a speed of $60{\mathrm{ms}}^{-1}$ and the other half with a speed of $40{\mathrm{ms}}^{-1}$. Then find the average speed during the whole journey.

Answer 1

Step 1: Given information

1. A body covers half of its journey with a speed of$60{\mathrm{ms}}^{-1}$.
2. The body covers the other half of its journey with a speed of$40{\mathrm{ms}}^{-1}$.

Step 2: Calculating time taken by the body in both halves of the journey

1. Let$\text{'}d\text{'}$ be the total distance covered by the body.
2. According to the question, $\frac{d}{2}$ distance is covered with a speed of$60{\mathrm{ms}}^{-1}$
3. So the time taken by the body in the first half of the journey is$\frac{dis\tan ce}{speed}$
4. $t_1=\frac{d}{2\times 60}=\frac{d}{120}$s
5. Now the time is taken by the body in the second half of the journey is$t_2=\frac{d}{2\times 40}=\frac{d}{80}$

Step 3: Calculating average speed

1. We know that the average speed of a body in motion is a ratio of the total distance covered by the body to the total time taken by the body.

$averagespeed=\frac{totaldis\tan ce}{totaltime}$

$\Rightarrow V_{avg}=\frac{{\displaystyle \frac{d}{2}}+{\displaystyle \frac{d}{2}}}{t_1+t_2}$

$\Rightarrow V_{avg}=\frac{{\displaystyle \frac{d}{2}}+{\displaystyle \frac{d}{2}}}{{\displaystyle \frac{d}{120}}+{\displaystyle \frac{d}{80}}}=\frac{d}{{\displaystyle \frac{2d+3d}{240}}}$

$\Rightarrow V_{avg}=\frac{d}{{\displaystyle \frac{5d}{240}}}$

$\Rightarrow V_{avg}=\frac{d\times 240}{5d}=\frac{240}{5}$

=$48m/s$

Hence,

The average speed of the body is$48m/s$