A passenger train of length 60 m travels at a speed of 80 km/hr. Another freight train of length 120

Q: Why do we add the lengths of the trains in both cases?

To completely cross each other, the front of one train must pass the entire length of the other, and its own tail must also clear the other train's front. This total relative displacement is exactly $l_1 + l_2$, regardless of whether they move in the same or opposite directions.

Q: Shouldn't we convert the speeds from km/hr to m/s since the lengths are in meters?

You could, but it's unnecessary work! Since we are asked for a *ratio* of times, the conversion factor (5/18) would appear in both the numerator and denominator and simply cancel out. Always look for ratios to save time.

Q: Why is relative speed $v_1 - v_2$ for the same direction?

Imagine you are in the slower train. The faster train is catching up to you. From your perspective, it's only moving at the 'extra' speed it has over you. Hence, $v_{rel} = v_1 - v_2$.

Q: Why is relative speed $v_1 + v_2$ for opposite directions?

When moving towards each other, both trains are actively closing the gap. The rate at which the gap decreases is the sum of their individual speeds. Hence, $v_{rel} = v_1 + v_2$.

Q: What if the trains were accelerating?

If they were accelerating, we would use the relative acceleration $a_{rel} = a_1 \pm a_2$ and apply the second equation of motion $d = u_{rel}t + \frac{1}{2}a_{rel}t^2$. The problem would become a quadratic equation in $t$!

Question

A passenger train of length $60	ext{ m}$ travels at a speed of $80	ext{ km/hr}$. Another freight train of length $120	ext{ m}$ travels at a speed of $30	ext{ km/hr}$. The ratio of times taken by the passenger train to completely cross the freight train when : (i) they are moving in the same direction and (ii) in the opposite direction is

Options:

(A) $\frac{3}{2}$

(B) $\frac{25}{11}$

(C) $\frac{11}{5}$

(D) $\frac{5}{2}$

Sigma Percentile · Accepted Answer

Welcome to a classic problem of relative kinematics! This question might seem like it requires tedious unit conversions, but as we'll see, a little bit of strategic thinking will save us a lot of time.

Imagine you are standing on a platform, watching two massive trains. One is a swift passenger train, and the other is a heavy freight train. Our goal is to compare the time it takes for them to completely cross each other under two different scenarios.

### Analyzing the Setup

Let's break down the given data. We have a passenger train with length $l_1 = 60 	ext{ m}$ traveling at a speed of $v_1 = 80 	ext{ km/hr}$.

On another track, we have a freight train with length $l_2 = 120 	ext{ m}$ traveling at $v_2 = 30 	ext{ km/hr}$.

The most crucial concept in any "train crossing" problem is understanding the total distance that must be covered. For one train to completely pass the other, the front of the first train must travel the entire length of the second train, and then its own tail must clear the front of the second train.

Therefore, the **total relative distance** $d$ they must cover is always the sum of their lengths:
$$d = l_1 + l_2 = 60 + 120 = 180 	ext{ m}$$

Notice that this distance remains exactly the same whether the trains are moving in the same direction or in opposite directions!

### The Master Equation: Case 1

Let's explore the first scenario: both trains are moving in the **same direction**.

Imagine you are sitting in the slower freight train. When the faster passenger train overtakes you, it doesn't look like it's moving at $80 	ext{ km/hr}$. It only appears to be moving at the *difference* of your speeds.

This is the essence of relative velocity. In the same direction, the relative speed is:
$$v_{	ext{same}} = v_1 - v_2 = 80 - 30 = 50 	ext{ km/hr}$$

Using the fundamental relation $	ext{Time} = \frac{	ext{Distance}}{	ext{Speed}}$, the time $t_1$ taken to cross in the same direction is:
$$t_1 = \frac{d}{v_{	ext{same}}} = \frac{d}{50}$$

### The Master Equation: Case 2

Now, let's look at the second scenario: the trains are moving in **opposite directions**.

This time, they are rushing towards each other. The gap between them closes much faster because both trains are actively contributing to reducing the distance.

Thus, the relative speed is the *sum* of their individual speeds:
$$v_{	ext{opp}} = v_1 + v_2 = 80 + 30 = 110 	ext{ km/hr}$$

The time $t_2$ taken to cross in opposite directions is:
$$t_2 = \frac{d}{v_{	ext{opp}}} = \frac{d}{110}$$

### Final Calculation and The Catch

We need to find the ratio of the times, $\frac{t_1}{t_2}$.

Here is where many students make a silly mistake! They rush to convert the speeds from $	ext{km/hr}$ to $	ext{m/s}$ because the distance is in meters. But wait! Let's look at the ratio:
$$\frac{t_1}{t_2} = \frac{\frac{d}{50}}{\frac{d}{110}}$$

Do you see the magic? The distance $d$ is in the numerator of both fractions. When we divide them, $d$ completely cancels out!

$$\frac{t_1}{t_2} = \frac{110}{50}$$

By simply canceling the zeros, we arrive at our final, elegant answer:
$$\frac{t_1}{t_2} = \frac{11}{5}$$

This means it takes $\frac{11}{5}$ times longer for the trains to cross when moving in the same direction compared to when they move in opposite directions. The beauty of physics lies in these elegant cancellations. Always look for ratios before you start crunching numbers!

Select Answer:

Visualized Solution

Given Data

Total Crossing Distance

Calculating Distance

Case 1: Same Direction

Relative Speed (Same)

Time $t_{1}$

Case 2: Opposite Direction

Relative Speed (Opposite)

Time $t_{2}$

Ratio of Times

Simplifying Ratio

Final Answer

Conclusion

The Sigma Insight: Relative Velocity

Analyzing the Setup

The Master Equation

The Master Equation

Final Calculation and The Catch

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Select Answer:

Visualized Solution

Given Data

Total Crossing Distance

Calculating Distance

Case 1: Same Direction

Relative Speed (Same)

Time .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t1​

Case 2: Opposite Direction

Relative Speed (Opposite)

Time .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t2​

Ratio of Times

Simplifying Ratio

Final Answer

Conclusion

The Sigma Insight: Relative Velocity

Analyzing the Setup

The Master Equation

The Master Equation

Final Calculation and The Catch

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A boat which has a speed of 5 km/h in still water crosses a river of width 1 km along the shortest possible path in 15 min. The velocity of the river water in km/h is

Time $t_{1}$

Time $t_{2}$

A boat which has a speed of $5 km/h$ in still water crosses a river of width $1 km$ along the shortest possible path in $15 min$ . The velocity of the river water in km/h is