Question

A boat crosses a river of width $1\mathrm{Km}$ by shortest path in $15\min$. If the speed of boat in still water is $5{\mathrm{Kmh}}^{-1}$, then what is the speed of the river?

• A. $5{\mathrm{Kmh}}^{-1}$
• B. $12{\mathrm{Kmh}}^{-1}$
• C. $3{\mathrm{Kmh}}^{-1}$
• D. $4{\mathrm{Kmh}}^{-1}$

Answer 1

The correct option is C

$3{\mathrm{Kmh}}^{-1}$

Step 1: Given data and drawing the vector diagram

Shortest path that is, Width of river $=1\mathrm{Km}$

Speed of boat in still water $\mathrm{V}=5{\mathrm{Kmhr}}^{-1}$

Time taken by boat to cross the river $\mathrm{t}=15\min =\frac{1}{4}\mathrm{hr}$

Let the speed of river $=\mathrm{U}$

The vector diagram is shown below:

Step 2: Calculating the velocity of river

By observing the vector diagram we can calculate the velocity along the shortest path as below:

Speed in the direction of shortest path$=\frac{shortestpath}{time}$

And also we know that

$$\begin{gathered} \mathrm{time}=\frac{\mathrm{shortest}\mathrm{path}}{\sqrt{{\mathrm{V}}^2-{\mathrm{U}}^2}} \\ \Rightarrow \frac{1}{4}=\frac{1}{\sqrt{25-{\mathrm{U}}^2}} \\ \Rightarrow {\left(\frac{1}{4}\right)}^2=\frac{1}{25-{\mathrm{U}}^2} \\ \Rightarrow 25-{\mathrm{U}}^2=16 \\ \Rightarrow {\mathrm{U}}^2=9 \\ \Rightarrow \mathrm{U}=3{\mathrm{Kmh}}^{-1} \end{gathered}$$

Therefore the speed of river is equal to $3{\mathrm{Kmh}}^{-1}$.

Hence, option(C) is the correct answer.