A train is moving at 39.1 miles per hour. Another train is moving at 45.5 miles per hour. Both trains are to stop at a designated area. The first train at 12pm, the latter 7pm. If the trains must both stop for fuel every 15 minutes, at what time will they cross paths?

The time for the second train to catch up with the first train = T₂

T₂ = where x is the time it takes for each refueling, hence when x = 5 minutes then the second train meets up wirh the first train in 42.75 Hrs or at  3:45 a.m. on the next two days

Step-by-step explanation:

Velocity of first train = 39.1 miles per hour

Velocity of second train = 45.5 miles per hour

Time of depature of the first train = 12 pm

Time of depature of the secon train = 7 pm

Refueling = every 15 minutes

Therefore

From the formula for velocity we have, velocity = Displacement/ time

Therefore Displacement =Velocity × time

If the refueling time = x seconds then total time for each train = T₁ + 4·x·T₁

Where T is the total time  hence

v₁ × (T₁ + 4·x·T₁) = v₂ × (T₂ + 4·x·T₂)

However the first train is ahead by 7 hours hence

v₁ × (T₁ + 4·x·T₁ + 7) = v₂ × (T₂ + 4·x·T₂)

or 39.1×(T₁ + 4·x·T₁ + 7) = 45.5×(T₂ + 4·x·T₂)

That is T₁ = T₂ +7

hence 39.1×(T₂ +7 + 4·x·(T₂ +7)) = 45.5×(T₂ + 4·x·T₂)

= 39.1×((4·x+1)×T₂+28×x+7) = 45.5×(T₂ + 4·x·T₂)

Therefore T₂ = Hence if they stop for 5 minutes which is equal to 5/60 of 1 hour or 1/12 hour we have

T₂ = 42.75 Hrs or since there are 24 hours in a day we have

1 day 8 3/4 hours  or 3:45 a.m. on the next two days