A swimmer can swim with velocity of 12 km/h in still water. Water flowing in a river has velocity 6

Q: Why must the resultant velocity be perpendicular to the river flow?

To reach the point *exactly* opposite, the swimmer must not move left or right relative to the banks. This means their net motion (resultant velocity) must be purely across the river.

Q: What happens if the swimmer just aims straight across?

If they aim straight across, the river's current will push them downstream. They will reach the other bank, but at a point further down the river, experiencing a 'drift'.

Q: Why did we add $90^\circ$ to $\theta$?

$\theta$ is the angle the swimmer makes with the vertical (the line straight across the river). The question asks for the angle with respect to the *river flow*, which is horizontal. The angle between the horizontal flow and the vertical is $90^\circ$, so the total angle is $90^\circ + \theta$.

Q: Can a swimmer always reach the exactly opposite point?

No! If the river's velocity ($v_R$) is greater than the swimmer's velocity in still water ($v_{MR}$), the equation $\sin\theta = v_R / v_{MR}$ gives a value greater than 1, which is impossible. In this case, they will always drift downstream.

Q: Does the width of the river affect the angle?

Interestingly, no. The condition for zero drift ($v_{MR} \sin\theta = v_R$) only depends on the velocities, not the distance to be covered. The width only affects the *time* it takes to cross.

Question

A swimmer can swim with velocity of $12 	ext{ km/h}$ in still water. Water flowing in a river has velocity $6 	ext{ km/h}$. The direction with respect to the direction of flow of river water he should swim in order to reach the point on the other bank just opposite to his starting point is ………$^{\circ}$ (in degree).
(Round off to the nearest integer)

Sigma Percentile · Accepted Answer

### The River Crossing Dilemma

Imagine you are standing on the bank of a steadily flowing river. You want to reach the exact opposite point on the other side. If you just dive in and swim straight across, the river's current will relentlessly push you downstream, and you'll end up far away from your target.

To conquer the current, you must swim at an angle, aiming upstream. This is a classic problem of relative velocity in two dimensions. Let's break down the physics of this scenario.

### The Physics of Zero Drift

Let the velocity of the river flow be $\vec{v}_R$, pointing horizontally to the right. Let your swimming velocity in still water be $\vec{v}_{MR}$. The actual path you take relative to the ground is your resultant velocity, $\vec{v}_M$.

For you to reach the exactly opposite point, your resultant velocity $\vec{v}_M$ must point straight across the river. This means your horizontal motion, or "drift", must be exactly zero.

How do we achieve zero drift? We must ensure that the horizontal component of your swimming velocity perfectly cancels out the river's velocity. If you swim at an angle $	heta$ with respect to the vertical (the line straight across), your upstream horizontal component is $v_{MR} \sin	heta$.

Therefore, the master equation for zero drift is:
$$v_{MR} \sin	heta = v_R$$

### Crunching the Numbers

Now, let's plug in the values given in our problem. We know your speed in still water is $v_{MR} = 12 	ext{ km/h}$, and the river's speed is $v_R = 6 	ext{ km/h}$.

Substituting these into our condition:
$$12 \sin	heta = 6$$

Dividing both sides by 12, we get:
$$\sin	heta = \frac{6}{12} = \frac{1}{2}$$

From our standard trigonometric angles, we know that the angle whose sine is $\frac{1}{2}$ is $30^\circ$. 
$$	heta = 30^\circ$$

### The Final Catch

We have found $	heta = 30^\circ$, but we must be very careful! This is the angle with respect to the **vertical**. The question specifically asks for the direction with respect to the **direction of flow of river water**.

Let's call this total angle $\alpha$. The angle between the river flow (horizontal) and the vertical is $90^\circ$. Therefore, the total angle you must make with the river flow is:
$$\alpha = 90^\circ + 	heta$$

Substituting our value of $	heta$:
$$\alpha = 90^\circ + 30^\circ = 120^\circ$$

So, you must swim at an angle of $120^\circ$ with respect to the river flow to reach the exact opposite point. This is a beautiful example of how vectors help us navigate the physical world!

Enter Numerical Value:

Visualized Solution

\text{Visualizing the River Crossing}

\text{Condition for Zero Drift}

\text{Substituting the Given Values}

\text{Calculating the Angle } \theta

\text{Angle with the River Flow}

\text{Extensions and Variations}

The Sigma Insight: Relative Velocity

The River Crossing Dilemma

The Physics of Zero Drift

Crunching the Numbers

The Final Catch

Similar Questions

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 ?

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