Algebra
posted by Jim B .
I don't understand can someone help me on how to figure this out please !!
Trains A and B are traveling in the same direction on parallel tracks. Train A is traveling at 80 miles per hour an train B is traveling at 100 miles per hour. Train A passes a station at 2:25 pm. If train B passes the same station at 2:37 pm,. at what time will train B catch up to train A?

why did train b which is traveling faster then train a, need to catch up to train a? doesn't make sense.did train b stop somewhere along the way, or did train be have a different starting point?

That's the question asked on the homework.

not enough facts present or the there's a misprint.
a formula useful for similar equations, though, is distance = rate x time 
Let the station be at x = 0. After 2:37 pm, train A is at
x(A) = (2:372:25)/(60 minutes per hour)*80 mph + 80 * t,
or
x(A) = (12 min / 60 min per hour) * 80 mph + 80 * t
x(A) = 16 + 80 * t
where t is hours after 2:37
for train B,
x(B) = 100*t
Set x(A) = x(B) or
100*t = 16 + 80*t
20*t = 16
t = 3/4 hour
Add this time to 2:37 to find the time when they catch up with each other