Question

A boat which has a speed of $5km{h}^{-1}$ in still water crosses a river of width $1km$ along the shorter possible path in $15min$. The velocity of the river water in $km{h}^{-1}$ is

• A. $1$
• B. $3$
• C. $4$
• D. $\sqrt{41}$

Answer 1

Answer: (B)

$3$

Let the velocity of the river be $x$ km/hr.

Given that,

Speed of the boat = $5km/h$

Width of the river (AB) = $1km/h$

Time taken to cross the river by boat = 15 minutes

The covered distance by the boat,

$D=vt$

$D=5\times \frac{15}{60}$

$D=1.25km/h$

Now, $AB=1km/h$

$AC=1.25km/h$

And angle $B={90}^0$

Now, apply Pythagoras theorem $\Delta ABC$

${\left(AB\right)}^2+{\left(BC\right)}^2={\left(AC\right)}^2$

${\left(1\right)}^2+{\left(BC\right)}^2={\left(1.25\right)}^2$

${\left(BC\right)}^2=\sqrt{{\left(1.25\right)}^2-{\left(1\right)}^2}$

$BC=0.75kM$

The covered distance by the river water in $15$minutes is $0.75km$.

Now, the velocity of the river

$v=\frac{D}{t}$

$v=\frac{0.75}{\frac{1}{4}}$

$v=3km/h$

Hence, the velocity of the river water is $3km{h}^{-1}$.