Question

A swimmer at a point A on the river bank wants to cross it. River is flowing with speed $v_0$. (Refer the diagram below). If $v_{sr}$ denotes velocity of swimmer in still water, then

• A. In order to reach directly at point B, $v_{sr}$ must be greater than $v_0$
• B. In order to reach directly at point B', $v_{sr} \ge (\frac{4v_0}{5})$
• C. If $v_{sr} = 2v_0$, then swimmer must head upstream at an angle, $\theta = 30^\circ$ with the line AB to reach directly at point B
• D. If swimmer reaches directly at point B' with minimum possible $v_{sr}$ then time taken by swimmer to cross the river of width w will be $(\frac{25w}{12v_0})$

Answer 1

Answer: (A), (B), (C), (D)

(1), (2), (3), (4)

Answer 2

The problem involves concepts of relative velocity, specifically in the context of river-swimmer problems. We will use a coordinate system where the x-axis is along the river flow and the y-axis is perpendicular to the river flow (from A to B).

Let:

• $v_0$ be the speed of the river. So, $\vec{v}_r = v_0 \hat{i}$.
• $v_{sr}$ be the speed of the swimmer in still water (velocity of swimmer with respect to river).
• $\vec{v}_s$ be the velocity of the swimmer with respect to the ground.

The relation is $\vec{v}_s = \vec{v}_{sr} + \vec{v}_r$.

Let's analyze each option:

Option (1): In order to reach directly at point B, $v_{sr}$ must be greater than $v_0$.

To reach point B, the swimmer must have a net velocity perpendicular to the river flow, i.e., the x-component of $\vec{v}_s$ must be zero. Let the swimmer head upstream at an angle $\alpha$ with the y-axis (line AB). Then, $\vec{v}_{sr} = -v_{sr} \sin \alpha \hat{i} + v_{sr} \cos \alpha \hat{j}$. The resultant velocity is $\vec{v}_s = (v_0 - v_{sr} \sin \alpha) \hat{i} + v_{sr} \cos \alpha \hat{j}$. For the swimmer to reach point B, the x-component must be zero: $v_0 - v_{sr} \sin \alpha = 0 \implies v_{sr} \sin \alpha = v_0$. Since the maximum value of $\sin \alpha$ is 1, for a solution to exist, we must have $v_0 \le v_{sr} \times 1$, i.e., $v_{sr} \ge v_0$. If $v_{sr} = v_0$, then $\sin \alpha = 1$, which implies $\alpha = 90^\circ$. In this case, the swimmer heads directly upstream. The velocity component across the river is $v_{sy} = v_{sr} \cos 90^\circ = 0$. This means the swimmer cannot cross the river. Therefore, for the swimmer to successfully reach point B, $v_{sr}$ must be strictly greater than $v_0$. So, option (1) is correct.

Option (2): In order to reach directly at point B', $v_{sr} \ge (\frac{4v_0}{5})$.

Point B' is at an angle of $37^\circ$ with the line AB. This means the net velocity vector $\vec{v}_s$ must be directed at an angle of $37^\circ$ with the y-axis (perpendicular to the river flow), towards the downstream side. Let $v_{sx}$ and $v_{sy}$ be the x and y components of $\vec{v}_s$. Then, $\frac{v_{sx}}{v_{sy}} = \tan 37^\circ = \frac{3}{4}$ (since $\tan 37^\circ \approx 3/4$). For the swimmer to reach B', the swimmer must head upstream. Let the swimmer head upstream at an angle $\alpha$ with the y-axis. $\vec{v}_{sr} = -v_{sr} \sin \alpha \hat{i} + v_{sr} \cos \alpha \hat{j}$. $\vec{v}_s = (v_0 - v_{sr} \sin \alpha) \hat{i} + v_{sr} \cos \alpha \hat{j}$. So, $v_{sx} = v_0 - v_{sr} \sin \alpha$ and $v_{sy} = v_{sr} \cos \alpha$. Using the condition $\frac{v_{sx}}{v_{sy}} = \frac{3}{4}$: $\frac{v_0 - v_{sr} \sin \alpha}{v_{sr} \cos \alpha} = \frac{3}{4}$ $4(v_0 - v_{sr} \sin \alpha) = 3 v_{sr} \cos \alpha$ $4v_0 = 3 v_{sr} \cos \alpha + 4 v_{sr} \sin \alpha$ $4v_0 = v_{sr} (3 \cos \alpha + 4 \sin \alpha)$ $v_{sr} = \frac{4v_0}{3 \cos \alpha + 4 \sin \alpha}$.

To find the minimum value of $v_{sr}$, the denominator $(3 \cos \alpha + 4 \sin \alpha)$ must be maximum. The maximum value of $a \cos \theta + b \sin \theta$ is $\sqrt{a^2+b^2}$. Here, $a=3, b=4$. So, the maximum value is $\sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5$. Thus, the minimum $v_{sr}$ is $v_{sr, min} = \frac{4v_0}{5}$. Therefore, for the swimmer to reach B', $v_{sr}$ must be greater than or equal to $\frac{4v_0}{5}$. So, option (2) is correct.

Option (3): If $v_{sr} = 2v_0$, then swimmer must head upstream at an angle, $\theta = 30^\circ$ with the line AB to reach directly at point B.

From the analysis of Option (1), to reach point B, we need $v_{sr} \sin \alpha = v_0$, where $\alpha$ is the angle with the line AB (y-axis). Given $v_{sr} = 2v_0$. $(2v_0) \sin \alpha = v_0$ $\sin \alpha = \frac{1}{2}$ $\alpha = 30^\circ$. So, the swimmer must head upstream at an angle of $30^\circ$ with the line AB. So, option (3) is correct.

Option (4): If swimmer reaches directly at point B' with minimum possible $v_{sr}$ then time taken by swimmer to cross the river of width w will be $(\frac{25w}{12v_0})$.

From Option (2), the minimum possible $v_{sr}$ to reach B' is $v_{sr, min} = \frac{4v_0}{5}$. This minimum occurs when $3 \cos \alpha + 4 \sin \alpha = 5$. This implies $\cos \alpha = 3/5$ and $\sin \alpha = 4/5$. (This corresponds to $\alpha \approx 53^\circ$). The time taken to cross the river of width $w$ depends only on the component of velocity perpendicular to the river flow, which is $v_{sy}$. $v_{sy} = v_{sr} \cos \alpha$. Substitute the values for minimum $v_{sr}$: $v_{sy} = \left(\frac{4v_0}{5}\right) \times \left(\frac{3}{5}\right) = \frac{12v_0}{25}$. The time taken $T = \frac{\text{width}}{v_{sy}} = \frac{w}{12v_0/25} = \frac{25w}{12v_0}$. So, option (4) is correct.

All four given options are correct.

The final answer is $\boxed{\text{(1), (2), (3), (4)}}$

Explanation of the solution:

1. Option (1): To reach point B (straight across), the swimmer's velocity component parallel to the river must cancel the river's velocity. This requires $v_{sr} \sin \alpha = v_0$. For a real angle $\alpha$, $\sin \alpha \le 1$, so $v_{sr} \ge v_0$. If $v_{sr} = v_0$, $\alpha = 90^\circ$, meaning the swimmer heads directly upstream, resulting in zero velocity perpendicular to the flow, thus unable to cross. Hence, $v_{sr} > v_0$ is necessary.
2. Option (2): To reach point B' at $37^\circ$ downstream, the ratio of the swimmer's net velocity components must be $v_{sx}/v_{sy} = \tan 37^\circ = 3/4$. If the swimmer heads upstream at angle $\alpha$ with the perpendicular, $v_{sx} = v_0 - v_{sr} \sin \alpha$ and $v_{sy} = v_{sr} \cos \alpha$. Substituting these into the ratio and solving for $v_{sr}$ gives $v_{sr} = \frac{4v_0}{3 \cos \alpha + 4 \sin \alpha}$. The minimum $v_{sr}$ occurs when the denominator is maximum, which is $\sqrt{3^2+4^2}=5$. Thus, $v_{sr, min} = \frac{4v_0}{5}$.
3. Option (3): For reaching point B, we have $v_{sr} \sin \alpha = v_0$. Given $v_{sr} = 2v_0$, this becomes $2v_0 \sin \alpha = v_0$, leading to $\sin \alpha = 1/2$, so $\alpha = 30^\circ$.
4. Option (4): When $v_{sr}$ is minimum to reach B', $v_{sr} = \frac{4v_0}{5}$. This occurs when $\cos \alpha = 3/5$ and $\sin \alpha = 4/5$. The time to cross the river of width $w$ is $T = w/v_{sy}$. The component $v_{sy} = v_{sr} \cos \alpha = (\frac{4v_0}{5})(\frac{3}{5}) = \frac{12v_0}{25}$. Substituting this, $T = \frac{w}{12v_0/25} = \frac{25w}{12v_0}$.

Answer: The correct options are (1), (2), (3), (4).