Animated Solution for Physics - Kinematics: A boy crosses a river twice on a straight path at an angle $\varphi =30^{\circ }$ with the downstream direction, first time in two minutes and second time in four minutes. If his speed relative to river current is $v_{b/w}=\sqrt{3}\text{ m/s}$ in both the attempts, find speed of the river current.

You want to reach a specific point on the opposite bank, and the straight-line path to that point makes a $30^{\circ }$ angle with the downstream direction. You jump in and swim at a constant effort, meaning your speed relative to the water, $v$, is constant.

Surprisingly, you find that there are two completely different ways to angle your body (your heading) that will keep you exactly on this same path over the ground! One heading gets you across in 2 minutes, and the other takes 4 minutes. How is this physically possible?

This phenomenon occurs when the river is flowing faster than you can swim ($u>v$). If you try to swim directly across, the river sweeps you away. But if you angle your path downstream, the river actually helps you. You can either point your body slightly upstream (fighting the current, resulting in a slower crossing) or point your body more downstream (riding the current, resulting in a faster crossing). Both headings, when added to the river's velocity, produce a resultant velocity vector that points exactly along your desired path!

Your velocity relative to the ground, $\vec{V}$, is the vector sum of your velocity relative to the water, $\vec{v}$, and the river's velocity, $\vec{u}$:

These three vectors form a triangle. We know the angle $\varphi =30^{\circ }$ between the ground velocity $\vec{V}$ and the river velocity $\vec{u}$. By applying the Law of Cosines to this vector triangle, we can relate their magnitudes:

Let's rearrange this into a standard quadratic equation in terms of your ground speed $V$:

This equation is the mathematical heart of the problem. The fact that it's a quadratic equation perfectly explains why there are two possible crossing times! The two roots of this equation, $V_1$ and $V_2$, represent the two possible ground speeds along the exact same path.

From the properties of quadratic equations, we can write down the sum and the product of these two roots:

We are given that the two crossing times are $t_1=2$ minutes and $t_2=4$ minutes. Since the distance $L$ across the river along the path is identical for both trips, the ground speed is inversely proportional to the time taken ($V=L/t$).

This tells us that the faster ground speed is exactly twice the slower ground speed:

We know $\varphi =30^{\circ }$, so $\cos 30^{\circ }=\frac{\sqrt{3}}{2}$:

Next, we substitute both $V_1=2V_2$ and our new expression for $V_2$ into the product of roots equation. We are also given that your swimming speed $v=\sqrt{3}$ m/s:

We are now one step away from the solution. Let's isolate $u$:

Taking the square root, we find the speed of the river current:

The math perfectly unravels the mystery of the river, revealing the hidden symmetry of the two crossings!