A train passes a platform with a uniform speed. A boy standing on the platform decides to estimate l

Q: Why is the relative speed $u - v$ and not $v - u$?

If the train were faster than the boy ($v > u$), he would never be able to pass the coach from behind. Since he successfully passes the coach while running in the same direction, his speed must be greater than the train's speed ($u > v$).

Q: How do we know the distance covered relative to the train is exactly $L$?

The problem states he 'passes by a coach'. This means he starts at one end of the coach and finishes at the other end. In the frame of reference of the train, the coach is stationary, and the boy travels exactly its length $L$.

Q: Why do we use time $t = \frac{n l}{u}$?

The boy takes $n$ steps, each of length $l$, so the total distance he runs on the ground is $n \times l$. Since he runs at a constant ground speed $u$, the time taken is simply distance divided by speed.

Q: Can we solve this without using relative velocity?

Yes, you can use ground frame coordinates. In time $t_1$, the boy moves $u t_1$ and the train moves $v t_1$. The difference in their positions is the length of the coach: $u t_1 - v t_1 = L$. This algebraically leads to the exact same equation!

Q: Why didn't we convert $10 \text{ km/h}$ to $\text{m/s}$ immediately?

Notice that in the formula $v = u \frac{n_1 - n_2}{n_1 + n_2}$, the term in the bracket is a dimensionless ratio. So, $v$ will naturally have the same units as $u$. We only need to be careful with units when calculating $L$, where we must ensure $u$ and $t$ are in compatible units.

Question

A train passes a platform with a uniform speed. A boy standing on the platform decides to estimate length of a coach and speed of the train. For this purpose, he first runs with a constant speed of $u = 10 	ext{ km/h}$ in the direction of the motion of the train and passes by a coach in $n_1 = 30$ steps. Then he turns back, runs at the same constant speed and passes by a coach in $n_2 = 20$ steps. If the boy covers a distance $l = 1.0 	ext{ m}$ in each step, answer the following questions.
(a) What is the speed of the train?
(b) What is the length of a coach?

Sigma Percentile · Accepted Answer

### The Setup: A Boy, A Train, and a Clever Idea

Imagine you are standing on a railway platform. A train is passing by at a uniform speed, and you decide to play a little game to calculate both the speed of the train and the length of its coaches. You don't have a stopwatch or a measuring tape, but you know your own running speed and the length of your stride. This is exactly the scenario our clever boy finds himself in.

Let's define our variables to bring order to this physical situation. Let the unknown speed of the train be $v$ and the unknown length of a single coach be $L$. We are given the boy's constant running speed $u = 10 	ext{ km/h}$. We also know that each of his steps covers a distance $l = 1.0 	ext{ m}$.

### The Power of Relative Velocity

When two objects are moving simultaneously, analyzing their motion from the ground can get messy. You have to track the absolute position of the boy and the absolute position of the train. However, physics offers us a beautiful shortcut: **Relative Motion**.

By shifting our perspective to the frame of reference of the train, the train becomes completely stationary. The boy is now moving relative to this stationary train. The distance he needs to cover to "pass by a coach" is simply the length of the coach itself, $L$. This single realization turns a complex tracking problem into a straightforward application of the formula: $	ext{Distance} = 	ext{Speed} 	imes 	ext{Time}$.

### Chasing the Train: The First Run

In the first scenario, the boy runs in the same direction as the train. Because he successfully overtakes the coach from behind, his speed must be greater than the train's speed ($u > v$).

In the train's frame of reference, the boy is approaching with a relative speed of $v_{	ext{rel}} = u - v$. He takes $n_1 = 30$ steps to pass the coach. Since he covers a distance $l$ in each step, the total time he spends running is the total distance he ran on the ground divided by his ground speed: $t_1 = \frac{n_1 l}{u}$.

Using our relative motion formula, the length of the coach is the relative speed multiplied by the time taken:
$$L = (u - v) t_1 = (u - v) \frac{n_1 l}{u}$$

### Running Head-On: The Return Journey

Now, the boy turns around and runs in the opposite direction. The train is still moving forward, so they are now rushing towards each other.

In the train's frame of reference, the boy is now sprinting past with a much higher relative speed: $v_{	ext{rel}} = u + v$. Because he is moving faster relative to the train, it takes him fewer steps to pass the coach. He completes the pass in just $n_2 = 20$ steps. The time taken for this run is $t_2 = \frac{n_2 l}{u}$.

Again, applying our relative motion formula, the length of the coach is:
$$L = (u + v) t_2 = (u + v) \frac{n_2 l}{u}$$

### The Master Equation: Bringing It Together

We now have two different expressions for the exact same physical quantity: the length of the coach, $L$. This is our golden ticket. By equating these two expressions, we can eliminate $L$ and solve directly for the train's speed, $v$.

$$(u - v) \frac{n_1 l}{u} = (u + v) \frac{n_2 l}{u}$$

Notice how beautifully the term $\frac{l}{u}$ appears on both sides. We can divide it out, leaving us with a pure, elegant algebraic relationship:

$$(u - v) n_1 = (u + v) n_2$$

Let's expand the brackets and group the terms containing $v$ on one side and the terms containing $u$ on the other:

$$u n_1 - v n_1 = u n_2 + v n_2$$
$$u(n_1 - n_2) = v(n_1 + n_2)$$

Isolating $v$, we get our master formula for the train's speed:

$$v = u \frac{n_1 - n_2}{n_1 + n_2}$$

### The Final Reveal: Calculating Speed and Length

Now comes the satisfying part—plugging in the numbers. We know $u = 10 	ext{ km/h}$, $n_1 = 30$, and $n_2 = 20$.

$$v = 10 	imes \frac{30 - 20}{30 + 20} = 10 	imes \frac{10}{50} = 2 	ext{ km/h}$$

The train is moving at a leisurely **$2 	ext{ km/h}$**.

To find the length of the coach, we substitute our expression for $v$ back into our first equation for $L$:

$$L = \left(u - u \frac{n_1 - n_2}{n_1 + n_2}ight) \frac{n_1 l}{u}$$

Watch what happens when we simplify the terms inside the parenthesis. We take a common denominator:

$$L = u \left( \frac{n_1 + n_2 - (n_1 - n_2)}{n_1 + n_2} ight) \frac{n_1 l}{u}$$
$$L = u \left( \frac{2 n_2}{n_1 + n_2} ight) \frac{n_1 l}{u}$$

The boy's speed, $u$, completely cancels out! This is a profound physical insight. The length of the coach depends only on the number of steps and the length of each step, completely independent of how fast the boy was running (as long as he maintained the same speed in both directions).

$$L = \frac{2 n_1 n_2 l}{n_1 + n_2}$$

Substituting our values ($n_1 = 30$, $n_2 = 20$, $l = 1.0 	ext{ m}$):

$$L = \frac{2 	imes 30 	imes 20 	imes 1.0}{30 + 20} = \frac{1200}{50} = 24 	ext{ m}$$

The length of the coach is exactly **$24 	ext{ m}$**. Through the elegant application of relative velocity, we've solved what initially seemed like an impossible puzzle with minimal information.

Visualized Solution

$Visualizing the Setup$

$Case 1: Running in the Same Direction$

$Case 2: Running in the Opposite Direction$

$Equating the Lengths$

$Solving for Train Speed v$

$Calculating Train Speed$

$Solving for Coach Length L$

$Calculating Coach Length$

$Final Conclusion$

The Sigma Insight: Relative Velocity

The Setup

The Power of Relative Velocity

Chasing the Train

Running Head-On

The Master Equation

The Final Reveal

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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; } Case 1: Running in the Same Direction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 2: Running in the Opposite Direction

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

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

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

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

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

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