Question

A man is crossing a river of width $ 60 m$ flowing with a velocity of $ 5 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$. He reaches a point B from point A as shown in the figure in $ 5 sec$ This velocity in still water should be

• A. $13\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$
• B. $12\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$
• C. $5\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$
• D. $10\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Answer 1

The correct option is A

$13\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Step 1. Given data:

width of river = $60m$

Velocity of the river $\left(v_r\right)$ = $5\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Time to reach from point B from point A = $5sec$

Step 2. Formula used:

Velocity = Distance/Time.

According to Pythagoras's theorem, Hypoteneous $v_{br}$ = $\sqrt{\left({v_b}^{{}_2}\right)+\left({v^2}_r\right)}$ in our case.

Step 3. Calculation:

1. Component of velocity in x-direction should counter the flow so as to reach the point across,$v_r=5\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ This is the velocity of the river.
2. Component of velocity in y should be such that he reaches $60m$ in $5$ seconds, this is the velocity of the man with respect to the ground.
3. So Component of velocity in the y direction, $v_b$= distance/time.=$\raisebox{1ex}{$60$}\!\left/ \!\raisebox{-1ex}{$5$}\right.=12\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$
4. Now, the man will swim to other side of the river by making an angle with both the velocities.
5. So, the resultant of these two velocities will give us our required velocity.
6. Total velocity (Velocity of the man with respect to the river) $\left(v_{br}\right)$ = $\sqrt{{v_r}^2+{v_b}^2}=\sqrt{{\left(5\right)}^2+{\left(12\right)}^2}=13\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Thus, option A is the correct option.