Two boys enter a running escalator at the ground floor of a shopping mall. The first boy repeatedly

Q: Why do we calculate an average relative speed instead of just using 50 cm/s?

Because the boys are not moving continuously in one direction. They take steps up and down, creating a cycle. Their effective progress is the net steps taken over the total steps in that cycle, so their average speed is a fraction of their actual stepping speed.

Q: Why is Boy 1's average relative speed negative?

Boy 1 takes 1 step up and 2 steps down in each cycle. This means he is effectively moving backwards (downwards) relative to the escalator. The negative sign simply indicates this downward direction.

Q: How can Boy 1 reach the top if his relative speed is negative?

Even though Boy 1 is moving downwards relative to the escalator, the escalator itself is moving upwards fast enough to carry him to the top. His net absolute velocity (escalator speed minus his relative downward speed) is still positive.

Q: Why do we equate the two distance equations?

Both boys start at the ground floor and end at the first floor. Regardless of how long it takes them or what their relative speeds are, the total physical distance they cover in the building's frame of reference is exactly the same.

Q: Does the length of a single step matter in this problem?

No, it cancels out! The relative speed is given in cm/s, and the fraction of net steps to total steps is a dimensionless ratio. The actual physical length of a step isn't needed to find the average relative speed.

Question

Two boys enter a running escalator at the ground floor of a shopping mall. The first boy repeatedly follows a cycle of $p_1 = 1$ step up and then $q_1 = 2$ steps down whereas the second boy repeatedly follows a cycle of $p_2 = 2$ steps up and then $q_2 = 1$ step down. Both of them move relative to escalator with a speed $v_r = 50 	ext{ cm/s}$. If the boys take $t_1 = 250 	ext{ s}$ and $t_2 = 50 	ext{ s}$ respectively to reach the first floor in complete numbers of cycles, how fast is the escalator running?

Sigma Percentile · Accepted Answer

The problem of the two boys on the escalator is a classic test of your understanding of relative motion and average velocity. It might seem confusing at first—how can we find the speed of the escalator when the boys are moving in such bizarre, cyclic patterns? But as we'll see, breaking down their motion cycle by cycle reveals a beautiful symmetry.

### Analyzing the Setup

Imagine you are standing at the bottom of a grand shopping mall escalator. The escalator is moving upwards at a constant, unknown speed, which we will call $v_e$. The total vertical distance from the ground floor to the first floor is $H$.

Now, two boys step onto this escalator, but instead of just standing still or walking normally, they decide to play a game. They move relative to the escalator at a constant speed of $v_r = 50 	ext{ cm/s}$, but they do so in specific cycles of stepping up and down.

To solve this, we must rely on the core principle of relative motion: the actual distance covered by either boy in the building's frame of reference is the sum of the distance the escalator carries them and the net distance they cover by their own stepping.

$$H = v_e t + d_{	ext{relative}}$$

### The First Boy's Bizarre Strategy

Let's focus on the first boy. His strategy is to take $p_1 = 1$ step up, and then $q_1 = 2$ steps down. This constitutes one complete cycle.

In one cycle, he takes a total of $p_1 + q_1 = 3$ steps. However, his net progress is $p_1 - q_1 = -1$ step. Because he takes more steps down than up, he is effectively moving backwards relative to the escalator!

His average relative speed is not $50 	ext{ cm/s}$. It is a fraction of that, determined by his net progress per cycle:

$$v_{r1} = v_r \left( \frac{p_1 - q_1}{p_1 + q_1} ight)$$

Substituting the values, we get:

$$v_{r1} = 50 	imes \left( \frac{1 - 2}{1 + 2} ight) = -\frac{50}{3} 	ext{ cm/s}$$

The negative sign perfectly captures his downward relative motion. We are told he takes $t_1 = 250 	ext{ s}$ to reach the top. We can now write the master equation for his journey:

$$H = v_e (250) + \left( -\frac{50}{3} ight) (250)$$

### The Second Boy's Approach

Now, let's look at the second boy. He takes $p_2 = 2$ steps up and $q_2 = 1$ step down.

In his three-step cycle, his net progress is $p_2 - q_2 = 1$ step upwards. His average relative speed will be positive:

$$v_{r2} = 50 	imes \left( \frac{2 - 1}{2 + 1} ight) = \frac{50}{3} 	ext{ cm/s}$$

Because he is effectively moving upwards relative to the escalator, he reaches the top much faster, in just $t_2 = 50 	ext{ s}$. His master equation is:

$$H = v_e (50) + \left( \frac{50}{3} ight) (50)$$

### The Master Equation

Here is where the magic happens. Both boys started at the ground floor and ended at the first floor. This means the total physical distance $H$ they covered is exactly the same. We can equate our two expressions for $H$:

$$250 v_e - \frac{12500}{3} = 50 v_e + \frac{2500}{3}$$

Look at this equation! We have successfully eliminated the unknown height $H$, leaving us with a simple linear equation where the only unknown is the speed of the escalator, $v_e$.

### Final Calculation

Let's solve for $v_e$. We group the $v_e$ terms on the left side and the constants on the right:

$$250 v_e - 50 v_e = \frac{2500}{3} + \frac{12500}{3}$$

$$200 v_e = \frac{15000}{3}$$

$$200 v_e = 5000$$

Dividing both sides by 200, we arrive at our final answer:

**$$v_e = 25 	ext{ cm/s}$$**

The escalator is running at a constant speed of $25 	ext{ cm/s}$. This problem beautifully demonstrates how complex, cyclic relative motion can be tamed by calculating average relative velocities and anchoring them to a shared physical constraint—in this case, the total height of the floor.

Visualized Solution

$V i s u a l i z in g t h e S e t u p$

$C o n ce pt o f R e l a t i v e M o t i o n$

$A na l y z in g B oy 1^{'} s M o t i o n$

$B oy 1^{'} s A v er a g e R e l a t i v e S p ee d$

$A na l y z in g B oy 2^{'} s M o t i o n$

$B oy 2^{'} s A v er a g e R e l a t i v e S p ee d$

$E q u a t in g t h e D i s t an ces$

$S o l v in g f or E sc a l a t or S p ee d$

$F ina l A n s w er$

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

The Sigma Insight: Relative Velocity

Analyzing the Setup

The First Boy's Bizarre Strategy

The Second Boy's Approach

The Master Equation

Final Calculation

Similar Questions

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A boy crosses a river twice on a straight path at an angle $ϕ = 3 0^{\circ}$ with the downstream direction, first time in two minutes and second time in four minutes. If his speed relative to river current is $v_{b / w} = 3 m/s$ in both the attempts, find speed of the river current.

Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Boy 1′s Motion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Boy 1′s Average Relative Speed

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Boy 2′s Motion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Boy 2′s Average Relative Speed

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

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

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

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