Question

A swimmer swims in a $90m$ long pool. He covers the distance of $180m$ by swimming from one end to other end and back along the same path. If he covers the first $90m$ at a speed of $2m{s}^{-1}$, then how fast does he swim such that his average speed is $3m{s}^{-1}$?

• A. $2m{s}^1$
• B. $3m{s}^{-1}$
• C. $5m{s}^{-1}$
• D. $6m{s}^{-1}$

Answer 1

The correct option is D $6m{s}^{-1}$

Length of the pool $=90m$
Speed to cover first $90m=2m{s}^{-1}$
$Speed=\frac{distance}{time}$
$\Rightarrow 2m{s}^{-1}=\frac{\left(90m\right)}{t}$
$\Rightarrow t=45s$
Time taken to cover first $90m=\frac{90}{2}=45s$
Time taken to travel the next $90m$ back $=t_2$
Total distance $=180m$
Total time to cover $180m=45+t_2$
Average speed to travel $180m=3m{s}^{-1}$
$Averagespeed=\frac{Totaldistancecovered}{Totaltimetaken}$
=> $3m{s}^{-1}$ = $\frac{180m}{45+t_2}$
$t_2=15s$
Now, his speed for next $90m$:
$\frac{90m}{15s}$ = $6m{s}^{-1}$
Hence , he must swim with a speed of $6m{s}^{-1}$ to make his average speed to be $3m{s}^{-1}$