Train Speed Calculation from Overtaking Man Scenario

First, we need to determine the speed of the train. The problem states that the train, with a length of 195 m, overtakes a man moving at a speed of 5 km/hr in the same direction. This overtaking happens in 13 seconds.

When a train overtakes a person moving in the same direction, the relative speed is the difference between the train's speed and the person's speed. The distance the train needs to cover relative to the man is its own length.

• Train Length ($L_1$): 195 m
• Man's Speed ($v_m$): 5 km/hr
• Time to Overtake ($t_1$): 13 s

Let the speed of the train be $v_1$ (in km/hr). We need to convert speeds to meters per second (m/s) for consistency since the time is given in seconds and length in meters. Recall that 1 km/hr = $\frac{5}{18}$ m/s.

• Man's speed in m/s: $v_m = 5 \times \frac{5}{18} = \frac{25}{18}$ m/s.

The relative speed ($v_{rel1}$) when overtaking is: $v_{rel1} = v_1 - v_m$ (where $v_1$ is also in m/s)

Using the formula: Distance = Relative Speed $\times$ Time $L_1 = v_{rel1} \times t_1$ $195 \text{ m} = (v_1 - \frac{25}{18} \text{ m/s}) \times 13 \text{ s}$

Now, solve for the train's speed ($v_1$ in m/s): $v_1 - \frac{25}{18} = \frac{195}{13}$ $v_1 - \frac{25}{18} = 15$ $v_1 = 15 + \frac{25}{18}$ $v_1 = \frac{15 \times 18}{18} + \frac{25}{18}$ $v_1 = \frac{270 + 25}{18}$ $v_1 = \frac{295}{18}$ m/s.

Crossing Time Calculation with Opposite Trains

Next, we need to find the time it takes for this train (Train 1) to cross another train (Train 2) moving in the opposite direction.

• Train 1 Length ($L_1$): 195 m
• Train 1 Speed ($v_1$): $\frac{295}{18}$ m/s
• Train 2 Length ($L_2$): 255 m
• Train 2 Speed ($v_2$): 76 km/hr

First, convert the speed of Train 2 ($v_2$) from km/hr to m/s: $v_2 = 76 \times \frac{5}{18} = \frac{380}{18}$ m/s.

When two trains move in opposite directions, their relative speed is the sum of their individual speeds. Relative Speed ($v_{rel2}$) = $v_1 + v_2$ $v_{rel2} = \frac{295}{18} \text{ m/s} + \frac{380}{18} \text{ m/s}$ $v_{rel2} = \frac{295 + 380}{18}$ $v_{rel2} = \frac{675}{18}$ 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 ($D_{total}$) = $L_1 + L_2$ $D_{total} = 195 \text{ m} + 255 \text{ m}$ $D_{total} = 450 \text{ m}$.

Now, we can calculate the time ($t_2$) taken to cross: Time = Total Distance / Relative Speed $t_2 = \frac{D_{total}}{v_{rel2}}$ $t_2 = \frac{450 \text{ m}}{\frac{675}{18} \text{ m/s}}$ $t_2 = \frac{450 \times 18}{675}$

Simplify the fraction: $t_2 = \frac{450}{675} \times 18$ Notice that $675 = 1.5 \times 450$. So, $\frac{450}{675} = \frac{1}{1.5} = \frac{2}{3}$. Alternatively, divide both by 225: $450 / 225 = 2$, $675 / 225 = 3$. So, $\frac{450}{675} = \frac{2}{3}$. $t_2 = \frac{2}{3} \times 18$ $t_2 = 2 \times 6$ $t_2 = 12$ seconds.

Therefore, the train will take 12 seconds to completely cross the other train moving in the opposite direction.