A man goes to market with a speed of $ 30 \mathrm{~km} / \mathrm{h} $ and returns back with thespeed of $ 40 \mathrm{~km} / \mathrm{h} $. Find his average speed.

Given: 

Speed, $S_1$ = 30 km/h    (while going to the market)

Speed, $S_2$ = 40 km/h   (while returning from the market)

To find: Average speed, $S$.

Solution:

Let $t_1$ is the time taken by the man while going to the market, and it is given as-

$t_1=\frac {d}{30}$    $[\because t=\frac {D}{S}]$

and $t_2$ is the time taken by the man while returning from the market, and it is given as-

$t_2=\frac {d}{40}$    $[\because t=\frac {D}{S}]$

Let $d$ is the distance from the initial position of the man and the market.

Thus, 

The total distance travelled, $D=d+d=2d$   $(\because the\ distance\ is\ same\ while\ going\ to\ the\ market\ and\ returning\ from\ the\ market)$

Now, we know that average speed is the ratio of total distance travelled and the total time taken.

It is given as-

$Average\ Speed\ (S)=\frac{Total\ distance\ \ travelled\ (d)}{Total\ time\ \ taken\  (t)}$

$S=\frac {2d}{\frac {d}{30}+\frac {d}{40}}$

$S=\frac {2d}{\frac {4d+3d}{120}}$

$S=\frac {2d}{\frac {7d}{120}}$

$S=\frac {2d\times {120}}{7d}$

$S=\frac {240d}{7d}$

$S=34.28km/h$

Therefore, the average speed, $S$ is 34.28km/h.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

47 Views

Kickstart Your Career

Get certified by completing the course

Get Started