Question

Trains A and B are traveling in the same direction on parallel tracks. Train A is traveling at 40 miles per hour and train B is traveling at 48 miles per hour. Train A passes a station at 1:15AM. If train B passes the same station at 1:30am, at what time will train B catch up to train A?

Answer 1

Train B will catch up to Train A at 2:45am

Explanation:

Let's build the equations of motion for both train A and train B, starting at 1:30 AM.

These equations have the following format:

`S(t) = S(0) + vt`

In which S is the position after t minutes, S(0) is the initial position and v is the velocity, in miles per hour.

Train A

Traveling at 40 mph, so
`v = 40`.

At 1:30, 15 minutes have passed. 15 minutes is 0.25 hours. So
`S(0) = 0.25*40 = 10`

The equation of motion for train A is

`S_(a) = 10 + 40t`

Train B

Starts at the position 0.

48 miles per hour, so
`v = 48`

The equation of motion for train B is

`S_(b) = 48t`

If train B passes the same station at 1:30am, at what time will train B catch up to train A?

We first have to find how long it takes for train B to catch up to train A.

This is when

`S_(b) = S_(a)`

`48t = 10 + 40t`

`8t = 10`

`t = 1.25`

It is going to take 1.25 hours for the train to catch up. 1.25 hours is 60 + 0.25*60 = 75 minutes.

So train B will catch up train A 75 minutes after 1:30 am, so at 2:45am