The stream of a river is flowing with a speed of 2 km/h. A swimmer can swim at a speed of 4 km/h. Wh

Q: Why can't the swimmer just aim straight across?

If the swimmer aims straight across, they will have no velocity component opposing the river. The river's current will push them, and their actual path will be diagonal, causing them to drift downstream.

Q: What if the river's speed is greater than the swimmer's speed?

If $v_r > v_s$, the equation $\sin\theta = \frac{v_r}{v_s}$ yields a value greater than 1, which is impossible. This means the swimmer can never generate enough upstream velocity to cancel the river's flow, and they will always drift downstream.

Q: Why do we add $90^\circ$ to the calculated angle $\theta$?

The angle $\theta$ we calculated is the angle between the swimmer's direction and the straight path across the river. The question asks for the angle with respect to the river's flow. Since the straight path is $90^\circ$ to the river flow, the total angle is $90^\circ + \theta$.

Q: Does swimming at an angle increase the time it takes to cross?

Yes. Only the vertical component of the swimmer's velocity ($v_s \cos\theta$) contributes to crossing the river. Since $\cos\theta$ is less than 1, the effective crossing speed is reduced, taking more time than if they aimed straight across.

Q: How would you find the shortest time to cross the river?

To cross in the shortest time, you want the maximum possible velocity component directed across the river. This happens when you aim straight across ($\theta = 0^\circ$), making the crossing speed equal to $v_s$. However, you will drift downstream in this case.

Question

The stream of a river is flowing with a speed of $2	ext{ km/h}$. A swimmer can swim at a speed of $4	ext{ km/h}$. What should be the direction of the swimmer with respect to the flow of the river to cross the river straight ?

Options:

(A) $60^\circ$

(B) $120^\circ$

(C) $90^\circ$

(D) $150^\circ$

Sigma Percentile · Accepted Answer

### The River-Boat Dilemma: Crossing Straight

Imagine you are standing on the bank of a river that is flowing steadily. You want to swim across to the exact opposite point on the other bank. If you just aim straight across, the river's current will sweep you downstream, and you'll end up far from your target. To counter this, you intuitively know you must aim slightly upstream. But exactly how much?

This classic relative velocity problem requires us to break down the swimmer's velocity into two perpendicular components: one parallel to the river flow (horizontal) and one perpendicular to it (vertical).

### Analyzing the Setup

Let the speed of the river be $v_r = 2 	ext{ km/h}$ and the speed of the swimmer in still water be $v_s = 4 	ext{ km/h}$.

To cross the river straight, the net horizontal velocity must be zero. This means the swimmer must swim at an angle $	heta$ upstream relative to the straight path (the vertical axis). By doing so, the swimmer creates a horizontal velocity component that directly opposes the river's flow.

### The Master Equation

The upstream horizontal component of the swimmer's velocity is given by $v_s \sin	heta$. For the swimmer to not drift downstream, this component must perfectly cancel out the river's velocity $v_r$.

$$v_s \sin	heta = v_r$$

Now, we simply substitute the given values into our equation:

$$4 \sin	heta = 2$$

### Final Calculation

Solving for $\sin	heta$, we get:

$$\sin	heta = \frac{2}{4} = \frac{1}{2}$$

We know from basic trigonometry that the angle whose sine is $\frac{1}{2}$ is $30^\circ$. Therefore, $	heta = 30^\circ$.

However, we must be careful! The question asks for the direction of the swimmer **with respect to the flow of the river**. The river is flowing horizontally, and the straight path is at $90^\circ$ to it. The swimmer is aiming an additional $30^\circ$ upstream from that straight path.

So, the total angle $\alpha$ is:

$$\alpha = 90^\circ + 30^\circ = 120^\circ$$

**The swimmer must swim at an angle of $120^\circ$ with respect to the river flow.**

### A Thought Experiment

What if the river was flowing at $5 	ext{ km/h}$ while the swimmer could still only swim at $4 	ext{ km/h}$? If we plug these numbers into our equation, we get $\sin	heta = \frac{5}{4} = 1.25$. Since the sine of an angle can never exceed $1$, this tells us it is physically impossible for the swimmer to cross straight! They will inevitably be swept downstream, no matter how hard they try.

Animated Solution for Physics - Kinematics: The stream of a river is flowing with a speed of $2 km/h$ . A swimmer can swim at a speed of $4 km/h$ . What should be the direction of the swimmer with respect to the flow of the river to cross the river straight ?

Select Answer:

Visualized Solution

\text{Visualizing the River Crossing}

\text{Condition for Straight Crossing}

\text{Substituting Values}

\text{Calculating } \theta

\text{Finding the Total Angle } \alpha

\text{What if } v_r > v_s \text{?}

The Sigma Insight: Relative Velocity

The River-Boat Dilemma

Analyzing the Setup

The Master Equation

Final Calculation

A Thought Experiment

Similar Questions

A river is flowing from West to East at a speed of $5 m/min$ . A man on the South bank of the river, capable of swimming at $10 m/min$ in still water, wants to swim across the river in the shortest time. He should swim in a direction

A person is swimming with a speed of $10 m/s$ at an angle of $12 0^{\circ}$ with the flow and reaches to a point directly opposite on the other side of the river. The speed of the flow is $x m/s$ . The value of $x$ to the nearest integer is ……… .

A boy crosses a river twice on a straight path at an angle $ϕ = 3 0^{\circ}$ with the downstream direction, first time in two minutes and second time in four minutes. If his speed relative to river current is $v_{b / w} = 3 m/s$ in both the attempts, find speed of the river current.

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