Two cyclists, Mike and Josh, simultaneously started towards each other from two towns d = 24 km apar

Q: Why is the fly's displacement equal to Josh's displacement?

The fly starts its journey at Josh's initial position. It stops flying exactly when and where Josh and Mike meet. Therefore, its initial and final positions are identical to Josh's, making their net displacements equal.

Q: Why do we add and subtract the wind speed?

Velocities are vectors. When the fly moves towards Mike, it moves in the same direction as the wind, so their speeds add up ($v_F + u$). When it returns towards Josh, it flies against the wind, so the wind speed subtracts from its air speed ($v_F - u$).

Q: Can we solve this by calculating an infinite geometric series?

Yes, you can calculate the distance of each individual leg of the fly's journey and sum them as an infinite geometric progression. However, using the displacement constraint is a much faster and more elegant shortcut.

Q: What if the wind was blowing towards Josh instead?

The approach remains exactly the same, but the ground speeds would swap. The fly would travel at $v_F - u$ towards Mike and $v_F + u$ towards Josh. You would just update the velocity values in the displacement equation.

Q: Does the fly's turning time affect the answer?

In classic kinematics problems like this, we assume the fly turns instantaneously, meaning the turning time is zero. If turning took time, we would have to subtract it from the total flying time.

Question

Two cyclists, Mike and Josh, simultaneously started towards each other from two towns $d = 24 	ext{ km}$ apart. Josh rode at $v_J = 25 	ext{ km/h}$, and Mike at $v_M = 15 	ext{ km/h}$. The moment they start, a fly also starts from Josh towards Mike and after reaching Mike, immediately returns towards Josh. The fly continues back and forth motion between the cyclists till the cyclists meet. Air speed of fly is $v_F = 30 	ext{ km/h}$ and the wind blows always towards Mike with a constant velocity $u = 10 	ext{ km/h}$. Find the total distance $s$ flown by the fly.

Sigma Percentile · Accepted Answer

### The Setup: A Classic Chase with a Twist

Imagine the scene: two towns, $A$ and $B$, separated by a distance $d = 24 	ext{ km}$. Josh starts from Town $A$ riding at $v_J = 25 	ext{ km/h}$, and Mike starts from Town $B$ riding at $v_M = 15 	ext{ km/h}$. They are heading straight for each other.

If this were a standard relative motion problem, it would be over in a flash. But there's a twist! A highly energetic fly takes off from Josh's nose the moment he starts, zipping towards Mike at an air speed of $v_F = 30 	ext{ km/h}$. Once it reaches Mike, it instantly turns around and flies back to Josh, repeating this back-and-forth dance until the two cyclists finally meet.

To make matters even more interesting, there is a steady wind blowing towards Mike with a velocity $u = 10 	ext{ km/h}$. Our mission? Find the total distance $s$ the fly travels.

### Step 1: The Ticking Clock

Before we worry about the fly, we need to know how long this entire event lasts. The fly will only be in the air until Josh and Mike meet.

Since the cyclists are moving towards each other, their relative velocity of approach is simply the sum of their individual speeds:
$$v_{rel} = v_J + v_M = 25 + 15 = 40 	ext{ km/h}$$

The time it takes for them to cover the $24 	ext{ km}$ distance is:
$$t_0 = \frac{d}{v_{rel}} = \frac{24}{40} = 0.6 	ext{ hours}$$

So, the fly has exactly $0.6 	ext{ hours}$ to do its zigzagging.

### Step 2: The Fly's Ground Speed

The fly's speed isn't constant relative to the ground because of the wind.

When it flies towards Mike, the wind is at its back, pushing it forward. Its ground speed becomes:
$$v_1 = v_F + u = 30 + 10 = 40 	ext{ km/h}$$

When it turns around to fly back to Josh, it must fight the wind. Its ground speed drops to:
$$v_2 = v_F - u = 30 - 10 = 20 	ext{ km/h}$$

### Step 3: The Master Stroke (The Displacement Insight)

Here is where the magic happens. You might be tempted to calculate the distance of the first leg, then the second leg, and sum an infinite geometric series. While that works, it's a mathematical nightmare. Let's use physics intuition instead.

Think about the fly's **net displacement**. 
Where does the fly start? At Josh's starting position.
Where does the fly finish? At the exact spot where Josh and Mike meet.

This means that despite all the chaotic back-and-forth flying, the fly's overall displacement is exactly the same as Josh's displacement!

Let's calculate Josh's displacement over the $0.6 	ext{ hours}$:
$$s_{fly} = v_J 	imes t_0 = 25 	imes 0.6 = 15 	ext{ km}$$

### Step 4: Setting Up the Equations

Now we have two powerful pieces of information: the total time of flight, and the net displacement. Let's define two variables:
- $t_1$: The total time the fly spends flying towards Mike.
- $t_2$: The total time the fly spends flying towards Josh.

We know the sum of these times is the total duration:
$$t_1 + t_2 = 0.6$$

We also know that the forward distance minus the backward distance must equal the net displacement of $15 	ext{ km}$:
$$v_1 t_1 - v_2 t_2 = 15$$
$$40 t_1 - 20 t_2 = 15$$

### Step 5: Solving for the Distance

We have a simple system of linear equations. From the time equation, we can express $t_2$ as:
$$t_2 = 0.6 - t_1$$

Substitute this into the displacement equation:
$$40 t_1 - 20(0.6 - t_1) = 15$$
$$40 t_1 - 12 + 20 t_1 = 15$$
$$60 t_1 = 27$$
$$t_1 = 0.45 	ext{ hours}$$

Now, find $t_2$:
$$t_2 = 0.6 - 0.45 = 0.15 	ext{ hours}$$

Finally, we want the **total distance** flown, which is the sum of the forward distance and the backward distance (no negative signs here, distance is a scalar!):
$$s = v_1 t_1 + v_2 t_2$$
$$s = (40)(0.45) + (20)(0.15)$$
$$s = 18 + 3 = 21 	ext{ km}$$

### The Elegant General Formula

If we had solved this entirely with variables, we would arrive at a beautiful general formula for the total distance:
$$s = \frac{d(v_F^2 + u v_J - u^2)}{v_F(v_J + v_M)}$$

Plugging our numbers into this formula yields the exact same result of $21 	ext{ km}$. This problem is a brilliant demonstration of how looking at the macroscopic constraints (like net displacement) can completely bypass the need for tedious microscopic calculations!

Visualized Solution

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$T h e D i s pl a ce m e n t I n s i g h t$

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$T o t a l D i s t an ce F l o w n$

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The Sigma Insight: Relative Velocity

The Setup

Step 1

Step 2

Step 3

Step 4

Step 5

The Elegant General Formula

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Meeting Time of Cyclists

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fly′s Ground Velocities

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up Time Equations

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

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

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