A biker is moving with constant velocity v away from a long straight wall at an angle theta with the

Q: Why can we just assume the horizontal velocities are equal? What if they aren't?

By Fermat's principle, sound takes the path of least time. The specific sound ray that actually reaches the biker is the one that was emitted at the exact angle $\alpha$ which makes its horizontal velocity match the biker's. Other rays will miss the biker entirely!

Q: Why is the vertical velocity of sound $c \cos\alpha$?

Velocity is a vector. If the total speed is $c$ and the sound travels at an angle $\alpha$ with the vertical normal, its horizontal component is $c \sin\alpha$ and its vertical component is $c \cos\alpha$. They satisfy $(c \sin\alpha)^2 + (c \cos\alpha)^2 = c^2$.

Q: Does the wall's reflection change the speed of sound?

No, the reflection is perfectly elastic. The sound's vertical velocity simply reverses direction from downwards to upwards, but its magnitude $c \cos\alpha$ remains exactly the same.

Q: Why did we rationalize the denominator at the end?

Rationalizing removes the square root from the denominator, making the expression mathematically cleaner and easier to analyze for limiting cases (like what happens if $v$ approaches $c$).

Q: Could we solve this using the method of images?

Absolutely! You can imagine a 'virtual source' at distance $l$ behind the wall. The distance from this virtual source to the biker at time $t$ must equal $ct$. Setting up that distance formula and squaring both sides will lead to the exact same quadratic equation for $t$.

Question

A biker is moving with constant velocity $v$ away from a long straight wall at an angle $	heta$ with the wall. He honks a short beep of horn when he is at a distance $l$ from the wall. After how long from the instant he has honked, will he again hear an echo of the honking? Speed of the sound in air is $c$.

Sigma Percentile · Accepted Answer

### The Setup and the Challenge

Imagine you are the biker in this problem. You are riding away from a massive, straight wall at a constant speed $v$, cutting a path at an angle $	heta$.

At a specific moment, when you are exactly a distance $l$ from the wall, you honk your horn.

The sound waves explode outward in all directions at speed $c$. Some of these waves travel towards the wall, bounce off it like a billiard ball, and race back towards you. The question is: exactly when will that reflected sound wave—the echo—catch up to your moving bike?

### The Master Insight: Matching Horizontal Velocities

At first glance, this looks like a nightmare of 2D geometry. You are moving diagonally, the sound is moving diagonally, and the intersection point is constantly shifting.

But let's pause and think like a physicist. For you to hear the echo, the sound wave and your bike must arrive at the **exact same point in space** at the **exact same time $t$**.

This means your $x$-coordinates must match, and your $y$-coordinates must match.

What if we force their horizontal motions to be identical? You are moving horizontally with a velocity of $v_x = v \cos	heta$. Let's assume the specific sound ray that reaches you was emitted at an angle $\alpha$ such that its horizontal velocity is $c_x = c \sin\alpha$.

If we set **$c \sin\alpha = v \cos	heta$**, a magical thing happens. Since you both started at the same $x$-position (let's call it $x=0$), you will **always** have the same $x$-coordinate!

### Reducing to 1D Relative Motion

Because the horizontal positions are now permanently locked together, we can completely ignore the $x$-axis. The problem collapses into a beautiful, simple 1D relative motion problem along the $y$-axis.

Let's find the vertical velocity of the sound. We know its total speed is $c$, and its horizontal speed is $c \sin\alpha$. Using the Pythagorean theorem, its vertical speed $c_y$ is:

$$c_y = c \cos\alpha = \sqrt{c^2 - (c \sin\alpha)^2}$$

Substituting our locked horizontal velocity, we get:

$$c_y = \sqrt{c^2 - v^2 \cos^2	heta}$$

In this 1D vertical world, you are simply moving upwards with a constant velocity $v_y = v \sin	heta$. The sound, on the other hand, travels downwards at speed $c_y$, hits the wall at $y=0$, and bounces back upwards at the same speed $c_y$.

### The Final Calculation

Let's track the total vertical distance the sound must cover to catch you.

In time $t$, you have moved upwards by a distance $vt \sin	heta$. Since you started at height $l$, your final height is $y = l + vt \sin	heta$.

The sound had to travel down a distance $l$ to hit the wall, and then travel all the way back up to your height $y$. Therefore, the total vertical distance covered by the sound is:

$$d_{	ext{sound}} = l + (l + vt \sin	heta) = 2l + vt \sin	heta$$

We know that time equals distance divided by speed. So, we can set up our master equation:

$$t = \frac{2l + vt \sin	heta}{c_y}$$

Substituting $c_y = c \cos\alpha$, we get:

$$t = \frac{2l + vt \sin	heta}{c \cos\alpha}$$

Now, we just need to isolate $t$. Multiplying both sides by $c \cos\alpha$:

$$t(c \cos\alpha) = 2l + vt \sin	heta$$

$$t(c \cos\alpha - v \sin	heta) = 2l$$

$$t = \frac{2l}{c \cos\alpha - v \sin	heta}$$

Finally, we substitute our expression for $c \cos\alpha$:

$$t = \frac{2l}{\sqrt{c^2 - v^2 \cos^2	heta} - v \sin	heta}$$

To make this result mathematically elegant and match standard forms, we rationalize the denominator by multiplying the numerator and denominator by the conjugate $\left( \sqrt{c^2 - v^2 \cos^2	heta} + v \sin	heta ight)$.

The denominator becomes $(c^2 - v^2 \cos^2	heta) - v^2 \sin^2	heta$, which beautifully simplifies to $c^2 - v^2(\cos^2	heta + \sin^2	heta) = c^2 - v^2$.

This leaves us with our final, pristine answer:

$$t = \frac{2l \left( v \sin	heta + \sqrt{c^2 - v^2 \cos^2	heta} ight)}{c^2 - v^2}$$

Visualized Solution

\text{The Setup}

\text{The Meeting Condition}

\text{Matching Horizontal Motion}

\text{Vertical Velocity Component}

\text{Reducing to 1D Kinematics}

\text{Vertical Distance Traveled}

\text{Solving for Time } t

\text{Final Expression}

The Sigma Insight: Motion in a Plane

The Setup and the Challenge

The Master Insight

Reducing to 1D Relative Motion

The Final Calculation

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