Calculating Train Speeds and Crossing Times

This problem involves understanding the concept of relative speed and applying it to calculate the time it takes for trains to cover certain distances.

Step 1: Understanding the First Scenario (Train Overtaking a Man)

We are given the length of the train ($L_T$), the speed of the man ($S_M$), and the time it takes for the train to overtake the man ($t_1$). The man is moving in the same direction as the train.

• Train Length ($L_T$): 490 m
• Man's Speed ($S_M$): 9 km/hr
• Time to overtake ($t_1$): 12 seconds

First, we need to convert the man's speed from km/hr to m/s because the train's length and time are in meters and seconds, respectively. The conversion factor is $1 \text{ km/hr} = \frac{5}{18} \text{ m/s}$.

Man's Speed in m/s: $S_M = 9 \times \frac{5}{18} = \frac{45}{18} = 2.5 \text{ m/s}$

When a train overtakes a moving object (like a man) in the same direction, the relative speed is the difference between the train's speed ($S_T$) and the object's speed ($S_M$).

Relative Speed = $S_T - S_M$

The distance covered during overtaking is equal to the length of the train. We use the formula: Distance = Speed × Time.

Here, Distance = $L_T$, Speed = Relative Speed ($S_T - S_M$), Time = $t_1$.

So, $L_T = (S_T - S_M) \times t_1$.

Substituting the values:

$490 \text{ m} = (S_T - 2.5 \text{ m/s}) \times 12 \text{ s}$

Now, we solve for $S_T$:

$S_T - 2.5 = \frac{490}{12}$

$S_T - 2.5 = \frac{245}{6}$

$S_T = \frac{245}{6} + 2.5 = \frac{245}{6} + \frac{5}{2} = \frac{245 + 15}{6} = \frac{260}{6} = \frac{130}{3} \text{ m/s}$

So, the speed of the train is $\frac{130}{3}$ m/s.

Step 2: Calculating the Time to Cross the Second Train

Now, the first train needs to cross a second train moving in the opposite direction. We know:

• Length of the first train ($L_{T1}$): 490 m
• Speed of the first train ($S_{T1}$): $\frac{130}{3}$ m/s
• Length of the second train ($L_{T2}$): 500 m
• Speed of the second train ($S_{T2}$): 42 km/hr

Convert the speed of the second train to m/s:

$S_{T2} = 42 \times \frac{5}{18} = \frac{210}{18} = \frac{35}{3} \text{ m/s}$

Since the two trains are moving in opposite directions, their relative speed is the sum of their individual speeds.

Relative Speed = $S_{T1} + S_{T2}$

Relative Speed = $\frac{130}{3} + \frac{35}{3} = \frac{165}{3} = 55 \text{ m/s}$

To completely cross each other, the total distance the trains need to cover relative to each other is the sum of their lengths.

Total Distance = $L_{T1} + L_{T2} = 490 \text{ m} + 500 \text{ m} = 990 \text{ m}$

We use the formula: Time = $\frac{\text{Total Distance}}{\text{Relative Speed}}$

Time to cross ($t_2$) = $\frac{990 \text{ m}}{55 \text{ m/s}}$

Calculating the time:

$t_2 = \frac{990}{55}$

$t_2 = 18 \text{ seconds}$

Summary of Calculations

Therefore, the train will take 18 seconds to completely cross the second train.