Train Crossing Time Calculation

This problem involves calculating the time when two trains, traveling in opposite directions between two cities, will cross each other.

Train Travel Details

• Train R: Departs Chennai at 5 a.m., Arrives Vijayawada at 12 p.m.
• Train S: Departs Vijayawada at 7 a.m., Arrives Chennai at 5:30 p.m.

Calculating Speeds and Time

Let the distance between Chennai and Vijayawada be $D$.

• Train R:
  • Travel time = 12 p.m. - 5 a.m. = 7 hours.
  • Speed of Train R ($V_R$) = $\frac{D}{7}$ units per hour.
• Train S:
  • Travel time = 5:30 p.m. - 7 a.m. = 10.5 hours.
  • Speed of Train S ($V_S$) = $\frac{D}{10.5}$ units per hour.

Determining Crossing Time

Let the time of crossing be $t$ hours after 5 a.m. (the earliest departure time).

• At time $t$, Train R has traveled for $t$ hours. Distance covered by R = $V_R \times t = \frac{D}{7} \times t$.
• Train S departs at 7 a.m., which is 2 hours after 5 a.m. So, at time $t$, Train S has traveled for $(t-2)$ hours (assuming $t \geq 2$).
• Distance covered by S = $V_S \times (t-2) = \frac{D}{10.5} \times (t-2)$.
• When they cross, the sum of the distances covered equals the total distance $D$: $ \frac{D \times t}{7} + \frac{D \times (t-2)}{10.5} = D $
• Divide by $D$: $ \frac{t}{7} + \frac{t-2}{10.5} = 1 $
• Multiply by 21 (LCM of 7 and 10.5) to clear denominators: $ 21 \times \frac{t}{7} + 21 \times \frac{t-2}{10.5} = 21 \times 1 $ $ 3t + 2(t-2) = 21 $ $ 3t + 2t - 4 = 21 $ $ 5t = 25 $ $ t = 5 \text{ hours} $
• The crossing time is 5 hours after 5 a.m. $ 5 \text{ a.m.} + 5 \text{ hours} = 10 \text{ a.m.} $

The two trains cross each other at 10 a.m.