Problem 80 - Master Resource Book in JEE Main Physics

Question

Two trains \(A\) and \(B\) of length \(400 \mathrm{~m}\) each are moving on two parallel tracks with a uniform speed of \(72 \mathrm{~km} / \mathrm{h}\) in the same direction, with \(A\) ahead of \(B\). The driver of \(B\) decides to overtake \(A\) and accelerates by \(1 \mathrm{~m} / \mathrm{s}^{2}\). If after \(50 \mathrm{~s}\), the guard of \(B\) just brushes past the driver of \(A\), what was the original distance between them? (a) \(1250 \mathrm{~m}\) (b) \(1350 \mathrm{~m}\) (c) \(1450 \mathrm{~m}\) (d) None of these

Step-by-Step Solution

Answer

The original distance between the trains was 1250 m (Option a).

Step 1: Convert Speeds

Both trains are moving at a speed of \(72\text{ km/h}\) initially. First, convert this speed into meters per second since acceleration is given in meters per second squared. Use the conversion: \(1\text{ km/h} = \frac{5}{18} \text{ m/s}\).\[72\text{ km/h} = 72 \times \frac{5}{18} = 20\text{ m/s}.\]

Step 2: Calculate Distance Covered by Train B

Since train \(B\) accelerates, first calculate the distance it covers during this accelerating period. The formula for the distance covered under accelerated motion is:\[s = ut + \frac{1}{2}at^2,\]where- \(u = 20 \text{ m/s}\) is the initial velocity,- \(a = 1 \text{ m/s}^2\) is the acceleration,- \(t = 50 \text{ s}\) is the time.Substituting the given values:\[s = 20 \times 50 + \frac{1}{2} \times 1 \times 50^2\]\[s = 1000 + 1250 = 2250\text{ m}.\]

Step 3: Calculate Distance Covered by Train A

Train \(A\) moves at uniform speed, without acceleration. Calculate the distance covered using:\[s = v \times t,\]where \(v = 20\text{ m/s}\).\[s = 20 \times 50 = 1000\text{ m}.\]

Step 4: Determine Original Distance Between Trains

After \(50\text{ seconds}\), the driver of train \(A\) and guard of train \(B\) are at the same point. Train \(B\) needed an extra \(400\) meters to actually overtake the driver of \(A\) completely.Therefore, the initial distance between the two trains is:\[\text{Distance Advanced by B} - \text{Distance Covered by A} - 400\text{ m (length of train \(A\))} = 2250 - 1000 - 400 = 850\text{ m.}\]Thus, add this gap (850 m) to the 400 m of train A:\[\text{Original distance} = 850 + 400 = 1250\text{ m}.\]

Key Concepts

Relative Motion

Relative motion is the study of how objects move in relation to one another. In this exercise, we analyze the motion of Train B with respect to Train A, both of which are initially moving in the same direction at the same speed.

When Train B decides to overtake Train A, it needs to move faster than Train A. This is where relative velocity comes into play as Train B accelerates, changing the distance between them.

• This change in distance helps calculate how much faster Train B needs to travel.
• Understanding relative motion aids in analyzing scenarios where multiple objects interact like this one.

Realizing the relative speed is essential as it dictates when Train B will finally catch up and pass Train A. Taking into account the distance traveled by both trains, we can measure their relative locations over time.

Uniform Acceleration

Uniform acceleration involves an object moving at a constant acceleration. In this problem, Train B accelerates at a rate of 1 m/s² to overtake Train A. Understanding uniform acceleration is crucial because:

• It allows for predicting future motion using known equations.
• The acceleration value affects how quickly Train B can reduce the gap.

Uniform acceleration follows straightforward laws described by the equation:
\[ s = ut + \frac{1}{2}at^2 \]
where \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time. By substituting the values, we calculate the distance Train B covers as it overtakes Train A.
Understanding this concept helps predict how much extra distance the train covers during the period of acceleration.

Distance Calculation

Distance calculation is central to understanding motion problems. It helps us know how far an object has traveled over a given time. For Train B, it combines the effects of initial motion and added acceleration.
In this exercise:

• The distance Train B travels is calculated using the formula for uniformly accelerated motion:
  \[ s = ut + \frac{1}{2}at^2 \]
• Train A, moving at a steady speed, covers distance as:
  \[ s = vt \]

The difference between distances traveled by Train A and Train B, plus the additional length of Train A, results in the initial separation between the two.
Understanding distance calculation clarifies how the acceleration impacts overall motion.

Velocity Conversion

Velocity conversion converts speed from one unit to another to match the problem's required units. Here, Train A and Train B both start with a velocity of 72 km/h, which needs conversion to meters per second for consistency:

• The conversion factor used is \( 1\text{ km/h} = \frac{5}{18} \text{ m/s} \).
• Using this, 72 km/h becomes 20 m/s.

This step is crucial since acceleration is measured in m/s², and time in seconds.
Handling different units wisely allows one to correctly apply formulas and avoid potential calculation errors. Being adept in converting velocity ensures accuracy when solving physics problems, facilitating seamless calculation across diverse units.