A large number of pedestrians are walking in the same direction in queues on each side of a road of

Q: Why can't the boy just walk straight across the road?

If he walks straight across ($\theta = 90^\circ$), his forward speed is zero. The pedestrians will keep moving forward, and he will miss the next person entirely. He must angle his walk to keep up as much as possible.

Q: Why does he target the pedestrian behind him instead of the one ahead?

Because his total speed is $v$, any diagonal path means his forward component $v\cos\theta$ is strictly less than $v$. The pedestrians are moving at $v$, so they are faster than his forward progress. He is effectively moving backwards relative to them!

Q: How did we get the relative displacement as $-0.5a \hat{i} - b \hat{j}$?

In the pedestrian frame, the boy is moving backwards. The closest pedestrian on the other side (distance $b$) is staggered by $0.5a$. Since he falls behind, he intercepts the one at $-0.5a$ relative to his starting x-coordinate.

Q: Why use the half-angle formula for $\cos\theta$?

We derived $\tan(\theta/2) = a/2b$ from the velocity ratios. The half-angle identity $\cos\theta = (1-\tan^2(\theta/2))/(1+\tan^2(\theta/2))$ is the most elegant way to convert this directly into the forward velocity factor without messy square roots.

Q: Does the boy ever reach the end of the queue?

The problem assumes an infinitely long queue ('large number of pedestrians'). He will keep zig-zagging indefinitely, covering a net distance of 144 meters along the road every 2 minutes.

Question

A large number of pedestrians are walking in the same direction in queues on each side of a road of width $b = 3.0$ m. Distance between two adjacent pedestrians on either side of the road is $a = 2.0$ m and pedestrians on one side are displaced by a distance $0.5a$ with respect to pedestrians on the other side as shown in the figure, depicting the pedestrians by small circles. A boy distributing advertisement leaflets bypasses all the pedestrians. The boy and the pedestrians all are walking with the same constant speed $v = 1.5$ m/s. Starting from a pedestrian if the boy handovers leaflets to all the pedestrians he comes across, how much length $l$ of the road will he cover in $2.0$ minutes?

Sigma Percentile · Accepted Answer

The Zig-Zagging Leaflet Boy: A Masterclass in Relative Velocity

Have you ever tried to walk across a moving walkway or a river and realized that to go straight, you have to aim diagonally? This problem takes that classic relative motion concept and turns it up to eleven. We have a boy trying to distribute leaflets to two queues of pedestrians moving at speed $v$. The catch? The boy is also moving at exactly the same speed $v$.

Let's dive into the physics of this beautiful chase and see why the obvious path is a trap.

### The Illusion of Catching Up

At first glance, you might think the boy should just aim for the next pedestrian ahead of him on the opposite side of the road. But let's think about the kinematics.

If the boy walks diagonally at an angle $	heta$ to the road, his velocity vector is split into two components. His forward speed along the road becomes $v \cos	heta$, and his crossing speed becomes $v \sin	heta$.

Because $	heta$ is greater than zero, $\cos	heta$ is strictly less than $1$. This means his forward speed $v \cos	heta$ is **less than** the pedestrians' speed $v$. He is literally slower than the crowd in the forward direction! If he aims for someone ahead of him, they will just walk away from him. He can never catch them.

To successfully hand over a leaflet, he must accept that he is falling behind and target a pedestrian who is currently **behind** him.

### Shifting Perspectives: The Pedestrian Frame

To make the math elegant, let's jump into the frame of reference of the pedestrians. Imagine you are one of the people walking. From your perspective, you and everyone else in the queue are standing completely still.

What does the boy's motion look like to you? 
His relative velocity is given by:
$$ \vec{v}_{bp} = \vec{v}_b - \vec{v}_p $$

If we set the road along the x-axis, the pedestrians have velocity $\vec{v}_p = v \hat{i}$. The boy's velocity is $\vec{v}_b = v \cos	heta \hat{i} - v \sin	heta \hat{j}$ (assuming he crosses downwards).

Subtracting these, his relative velocity becomes:
$$ \vec{v}_{bp} = (v \cos	heta - v) \hat{i} - v \sin	heta \hat{j} $$

Notice that the x-component $(v \cos	heta - v)$ is negative. This confirms our earlier realization: in the pedestrian frame, the boy is moving backwards!

### The Geometry of the Chase

Now, where is his target? The road has a width $b$. The pedestrians on the other side are staggered by a distance of $0.5a$. Since the boy is moving backwards relative to the crowd, he must aim for the pedestrian located at a relative displacement of:
$$ \Delta \vec{r} = -0.5a \hat{i} - b \hat{j} $$

For the boy to intercept this person, his relative velocity vector must point exactly in the same direction as this relative displacement vector. We can equate the ratios of their components:
$$ \frac{v \cos	heta - v}{-v \sin	heta} = \frac{-0.5a}{-b} $$

The $v$ cancels out beautifully, and the negatives drop away, leaving us with:
$$ \frac{1 - \cos	heta}{\sin	heta} = \frac{a}{2b} $$

This is where a touch of trigonometry saves the day. The left side is the classic half-angle identity for tangent. So, we get:
$$ 	an\left(\frac{	heta}{2}ight) = \frac{a}{2b} $$

### The Final Sprint

We don't actually need the angle $	heta$ itself; we need the boy's forward speed $v_x = v \cos	heta$ to find out how far he travels along the road. We can express $\cos	heta$ directly in terms of $	an(	heta/2)$ using the identity:
$$ \cos	heta = \frac{1 - 	an^2(	heta/2)}{1 + 	an^2(	heta/2)} $$

Substituting $	an(	heta/2) = a/2b$, we get:
$$ \cos	heta = \frac{1 - (a/2b)^2}{1 + (a/2b)^2} = \frac{4b^2 - a^2}{4b^2 + a^2} $$

Now, the total distance $l$ the boy covers along the road in time $t$ is simply his forward speed multiplied by time:
$$ l = v_x t = v t \left( \frac{4b^2 - a^2}{4b^2 + a^2} ight) $$

All that's left is to plug in the numbers! We are given $v = 1.5$ m/s, $t = 120$ s (which is 2.0 minutes), $a = 2.0$ m, and $b = 3.0$ m. 
$$ l = (1.5)(120) \left( \frac{4(3)^2 - 2^2}{4(3)^2 + 2^2} ight) $$
$$ l = 180 \left( \frac{36 - 4}{36 + 4} ight) = 180 	imes 0.8 = 144 	ext{ m} $$

**The boy covers exactly 144 meters along the road.**

This problem is a fantastic reminder that in physics, sometimes you have to move backwards (relatively speaking) to reach your goal!

Enter Numerical Value:

Visualized Solution

$Setup$

$The Kinematic Constraint v_{x} < v$

$Relative Velocity v_{b p}$

$Relative Displacement Δ r$

$Equating Directions$

$Simplifying with tan (θ /2)$

$Calculating Forward Speed v_{x}$

$Final Distance l$

The Sigma Insight: Relative Velocity

The Illusion of Catching Up

Shifting Perspectives

The Geometry of the Chase

The Final Sprint

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Kinematic Constraint vx​<v

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relative Velocity vbp​

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying with tan(θ/2)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Forward Speed vx​

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