# When will train B catch up with train A?

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• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?

When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:

Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?

When will train B catch up with train A?

When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.

B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:

$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)

The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:

$60t + 15 = 80t$

Find the value of $t$ (which is time in hours) at that moment, and add it to the time (5.30 pm).
• January 27th 2010, 08:57 PM
bball20
ok, let me see if I got this:

60t +15 = 80t
-60t -60t

15 = 20 t

3/4 = t

so I would add 45 min to 5:30 pm to get 6:15 pm

Is that right?
• January 27th 2010, 09:00 PM
VonNemo19
Quote:

Originally Posted by bball20
ok, let me see if I got this:

60t +15 = 80t
-60t -60t

15 = 20 t

3/4 = t

so I would add 45 min to 5:30 pm to get 6:15 pm

Is that right?

yes
• January 27th 2010, 09:04 PM
bball20
Thank you both for your help!!!!
• January 28th 2010, 04:55 PM
Masterthief1324
Quote:

Originally Posted by fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:

$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)

The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:

$60t + 15 = 80t$

Find the value of $t$ (which is time in hours) at that moment, and add it to the time (5.30 pm).

Quick question behind the reasoning of this problem. Why did you set the question to y = 60t + 15 instead of y = 80t + 15?
• January 28th 2010, 05:59 PM
fishcake
Quote:

Originally Posted by Masterthief1324
Quick question behind the reasoning of this problem. Why did you set the question to y = 60t + 15 instead of y = 80t + 15?

Because when $t = 0$ (at 5.30pm), train A is 15m away from the station, while train B is at the station (0m away from the station).