Question

A river is flowing with a velocity of 4 m/s towards east direction. When the boat runs with a velocity of 3 m/s relative to the river in the direction of the river flow, the flag on the boat flutter in north direction. If the boat runs with the same speed but in north direction relative to river, the flag flutter towards north-east direction. The actual velocity of the wind should be (Take $\hat{i}$ for east and $\hat{j}$ for north direction)

• A. 7$\hat{i}$ + 6$\hat{j}$
• B. 4$\hat{i}$ – 6$\hat{j}$
• C. 6$\hat{i}$ – 4$\hat{j}$
• D. 7$\hat{i}$ – 6$\hat{j}$

Answer 1

Answer: (A)

7$\hat{i}$ + 6$\hat{j}$

Answer 2

Let $\vec{v}_W$ be the actual velocity of the wind, $\vec{v}_R$ be the velocity of the river, $\vec{v}_B$ be the actual velocity of the boat (relative to ground), $\vec{v}_{B/R}$ be the velocity of the boat relative to the river, and $\vec{v}_{W/B}$ be the velocity of the wind relative to the boat.

The fundamental relations are:

1. $\vec{v}_B = \vec{v}_{B/R} + \vec{v}_R$
2. $\vec{v}_{W/B} = \vec{v}_W - \vec{v}_B$

We are given $\vec{v}_R = 4\hat{i}$ m/s (East). Let the wind velocity be $\vec{v}_W = v_x\hat{i} + v_y\hat{j}$.

Case 1: The boat runs with $\vec{v}_{B/R1} = 3\hat{i}$ m/s (East, relative to the river). The actual velocity of the boat is: $\vec{v}_{B1} = \vec{v}_{B/R1} + \vec{v}_R = 3\hat{i} + 4\hat{i} = 7\hat{i}$ m/s.

The flag flutters in the North direction, which means the wind velocity relative to the boat, $\vec{v}_{W/B1}$, is in the North direction. $\vec{v}_{W/B1} = \vec{v}_W - \vec{v}_{B1} = (v_x\hat{i} + v_y\hat{j}) - 7\hat{i} = (v_x - 7)\hat{i} + v_y\hat{j}$. For $\vec{v}_{W/B1}$ to be in the North direction, its East component must be zero, and its North component must be positive. $v_x - 7 = 0 \implies v_x = 7$. $v_y > 0$. So, the wind velocity is $\vec{v}_W = 7\hat{i} + v_y\hat{j}$, with $v_y > 0$.

Case 2: The boat runs with $\vec{v}_{B/R2} = 3\hat{j}$ m/s (North, relative to the river). The actual velocity of the boat is: $\vec{v}_{B2} = \vec{v}_{B/R2} + \vec{v}_R = 3\hat{j} + 4\hat{i} = 4\hat{i} + 3\hat{j}$ m/s.

The flag flutters in the North-East direction, which means the wind velocity relative to the boat, $\vec{v}_{W/B2}$, has equal positive components. $\vec{v}_{W/B2} = \vec{v}_W - \vec{v}_{B2} = (7\hat{i} + v_y\hat{j}) - (4\hat{i} + 3\hat{j}) = (7-4)\hat{i} + (v_y-3)\hat{j} = 3\hat{i} + (v_y-3)\hat{j}$. For $\vec{v}_{W/B2}$ to be in the North-East direction, its components must be equal and positive. $3 = v_y - 3$. Also, $3 > 0$ and $v_y - 3 > 0$. From $3 = v_y - 3$, we get $v_y = 6$. This value $v_y = 6$ satisfies the conditions $v_y > 0$ (from Case 1) and $v_y - 3 > 0$ (since $6-3=3>0$).

Therefore, the actual velocity of the wind is $\vec{v}_W = v_x\hat{i} + v_y\hat{j} = 7\hat{i} + 6\hat{j}$ m/s.