Two motorboats that can move with velocities 4.0 m/s and 6.0 m/s relative to water are going up-stre

Q: Why do both boats reach the buoy at the exact same time?

In the river frame, the buoy is stationary. Each boat travels outwards at a constant speed $v$ for time $t_0$, covering a distance $v t_0$. To return this same distance at the same speed $v$, it takes exactly time $t_0$ again. Thus, both return at $t = 2t_0$.

Q: Does the river velocity affect the time it takes for the boats to return?

No! This is the beauty of relative motion. The river current carries the boats and the buoy together. The relative velocities between the boats and the water remain unchanged, so the time intervals are independent of the river's speed.

Q: Why is the maximum separation exactly when they turn back?

While moving outwards, the faster boat continuously pulls away from the slower boat, increasing their separation. The moment they turn back, they start moving towards each other, and the separation begins to decrease.

Q: What does "distance between the places" mean?

It refers to the distance measured on the stationary ground. The first place is where the buoy was dropped. The second place is where the boats catch up to the buoy. The buoy has drifted between these two places.

Q: Why didn't we need to calculate the ground velocities of the boats?

Because the events (dropping the buoy and meeting it again) are defined by the intersection of the boats and the buoy. By finding the total time in the river frame, we only needed the buoy's ground velocity to find the displacement between these events.

Question

Two motorboats that can move with velocities 4.0 m/s and 6.0 m/s relative to water are going up-stream in a river. When the faster boat overtakes the slower boat, a buoy is dropped from the slower boat. After lapse of a time interval, both the boats turn back simultaneously and move at the same speeds relative to the water as before. Their engines are switched off when they reach the buoy again. If the maximum separation between the boats is 200 m after the buoy is dropped and water flow velocity in the river is 1.5 m/s, find distance between the places where the faster boat passes by the buoy.

Options:

(A) 75 m

(B) 150 m

(C) 300 m

(D) 350 m

Sigma Percentile · Accepted Answer

### The Illusion of Complexity

When you first read this problem, your instinct might be to set up a coordinate system on the riverbank. You'd write down the ground velocities of the boats—taking into account the upstream and downstream river flow—and try to solve a system of equations. While this brute-force method works, it is mathematically exhausting and prone to errors.

In physics, the right frame of reference can turn a nightmare into a walk in the park. Let's abandon the ground and jump into the river!

### The Magic of the River Frame

Imagine you are floating on a raft, moving perfectly in sync with the river current. From your perspective, the water around you is completely still.

What happens to the buoy in this frame? Since the buoy simply floats with the water, it has **zero velocity**. It is completely stationary!

What about the boats? The problem states their velocities relative to the water are $4.0 	ext{ m/s}$ and $6.0 	ext{ m/s}$. In our river frame, this is exactly how fast they are moving away from the stationary buoy.

### The Outward Journey and Maximum Separation

Let's say the boats travel outwards for a time $t_0$ before turning back. During this time, the faster boat is constantly pulling ahead of the slower boat. The rate at which their separation increases is simply the difference in their speeds:
$$v_{	ext{rel}} = v_2 - v_1 = 6.0 - 4.0 = 2.0 	ext{ m/s}$$

The separation will be at its absolute maximum right at the instant they decide to turn around. The problem tells us this maximum separation is $200 	ext{ m}$. We can easily find $t_0$:
$$S_{	ext{max}} = v_{	ext{rel}} 	imes t_0$$
$$200 = 2.0 	imes t_0 \implies t_0 = 100 	ext{ s}$$

### The Symmetry of the Return

Now, both boats turn around and head back towards the buoy. They maintain their same engine speeds relative to the water.

Think about the symmetry here. If a boat travels away from a stationary point at a speed $v$ for $100 	ext{ s}$, it covers a distance of $100v$. To cover that exact same distance back at the same speed $v$, it will take exactly another $100 	ext{ s}$!

This beautiful symmetry means **both boats will arrive back at the buoy at the exact same instant**. The total time elapsed from dropping the buoy to meeting it again is:
$$T_{	ext{total}} = 2 	imes t_0 = 200 	ext{ s}$$

### Translating Back to the Ground

The question asks for the distance between the places where the faster boat passes the buoy. 
1. **Place 1:** The exact spot where the buoy was dropped into the water ($t = 0$).
2. **Place 2:** The spot where the faster boat catches up to the buoy ($t = 200 	ext{ s}$).

Notice that both of these events happen exactly where the buoy is! Therefore, the distance between these two places on the ground is simply the distance the buoy drifted down the river during those $200 	ext{ s}$.

Since the river flows at $v_r = 1.5 	ext{ m/s}$, the buoy's displacement is:
$$d = v_r 	imes T_{	ext{total}}$$
$$d = 1.5 	imes 200 = 300 	ext{ m}$$

By choosing the right perspective, we bypassed all the messy algebra and solved the problem using pure physical intuition.

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Visualized Solution

$T h e P h y s i c a l S e t u p$

$T h e R i v er F r am e$

$O u tw a r d J o u r n ey$

$M a x im u m S e p a r a t i o n$

$T h e R e t u r n J o u r n ey$

$G r o u n d F r am e T r an s l a t i o n$

$F ina l C a l c u l a t i o n$

$T h e W a y F or w a r d$

The Sigma Insight: Relative Velocity

The Illusion of Complexity

The Magic of the River Frame

The Outward Journey and Maximum Separation

The Symmetry of the Return

Translating Back to the Ground

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Physical Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The River Frame

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Outward Journey

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Maximum Separation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Return Journey

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ground Frame Translation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward