Question

A boat crosses a river with a velocity of $8{\mathrm{Kmhr}}^{-1}$. If the resulting velocity of boat is $10{\mathrm{Kmhr}}^{-1}$, then the velocity of river water is:

• A. $4\mathrm{km}/\mathrm{hr}$
• B. $6\mathrm{km}/\mathrm{hr}$
• C. $8\mathrm{km}/\mathrm{hr}$
• D. $10\mathrm{km}/\mathrm{hr}$

Answer 1

The correct option is B

$6\mathrm{km}/\mathrm{hr}$

Step 1: Given data and drawing the vector diagram of the given situation

Velocity of boat ${\mathrm{v}}_{\mathrm{b}}=8{\mathrm{Kmhr}}^{-1}$

Resultant velocity of boat ${\mathrm{V}}_{\mathrm{B}}=10{\mathrm{kmhr}}^{-1}$

Let velocity of river $=\mathrm{V}$

The vector diagram is given below:

Step 2: Calculating the velocity of river

By observing the vector diagram we can write

$$\begin{gathered} \mathrm{Resultant}\mathrm{velocity}=\sqrt{{\left({\mathrm{V}}_{\mathrm{b}}\right)}^2+{\mathrm{V}}^2+2{\mathrm{V}}_{\mathrm{b}}\mathrm{Vcos}{\left(90\right)}^0} \\ \Rightarrow 10=\sqrt{8^2+{\mathrm{V}}^2+2\times 8\times \mathrm{V}\times 0} \\ \Rightarrow 10=\sqrt{64+{\mathrm{V}}^2} \end{gathered}$$

Now squaring on both sides we get

$$\begin{gathered} 100=64+{\mathrm{V}}^2 \\ \Rightarrow {\mathrm{V}}^2=100-64 \\ \Rightarrow {\mathrm{V}}^2=36 \\ \Rightarrow \mathrm{V}=6{\mathrm{Kmhr}}^{-1} \end{gathered}$$

Therefore the velocity of river water is equal to $6\raisebox{1ex}{$\mathrm{km}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{hr}$}\right.$.

Hence, option(B) is the correct answer.