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Math Help - When will train B catch up with train A?

  1. #1
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    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?
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by bball20 View Post
    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?
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  3. #3
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    Ok, so I am still lost?
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  4. #4
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    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).
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  5. #5
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    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?
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by bball20 View Post
    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
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  7. #7
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    Thank you both for your help!!!!
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  8. #8
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    Quote Originally Posted by fishcake View Post
    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?
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  9. #9
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    Quote Originally Posted by Masterthief1324 View Post
    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).
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