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Basic Avg Speed/Constant Acceleration train question

  1. Jul 3, 2012 #1
    1. The problem statement, all variables and given/known data

    A train starts from rest. The train (including the locomotive and all the freight cars) is 120 m long. A person standing still next to the front train (not in front of it -- she is NOT run over!) notices that the end of the last freight car passes her exactly 2 minutes after the train started moving.

    (i) What is the average speed of the train during the 2-minute time interval?
    (ii) Assuming that the train accelerates at a constant rate, what is the speed of the train when the end of the last freight car passes the observer?



    I can figure out i) just fine as average speed is 120m/120sec to get 1 m/s.

    I don't know why ii) is confusing to me, but it is. It seems straightforward but I don't know how to get the speed of the train when it passes the person. Wouldn't it also be 1 m/s? Speed is just a scalar so how does acceleration come into play?
     
  2. jcsd
  3. Jul 3, 2012 #2
    Your part (i) is right.

    For part (ii), the train starts at rest. So that means at its velocity is zero.
    Therefore, if it averages 1 m/s, and it starts at 0 m/s, it has to be moving faster that 1 m/s at some point in the interval to bring the average up to 1 m/s.

    What kinematic formulas do you know for uniform acceleration?
     
  4. Jul 3, 2012 #3
    ii) Asks you to find the instantaneous velocity at 2 minutes, whereas i) asks you to find the average velocity during the 2 minutes.
     
  5. Jul 3, 2012 #4
    If the instant. velocity at t=120 sec is the slope of the tangent line on the position/time graph and the position = 120m, then wouldn't the instantaneous velocity at 120 sec be 120/120 or 1 m/s? Am I missing something on this point?

    However, I think this is correct:

    If Vinitial is zero and acceleration is constant then, using X-Xo = 1/2 (Vo +V)t gives me:
    120 = 1/2 (0 + v) * 120 sec
    120 = 60v
    v=2 m/s
     
  6. Jul 4, 2012 #5

    lewando

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    Gold Member

    The graph of position vs time is not a straight line-- it is parabolic due to constant acceleration. The slope of the graph at t=0 is 0m/s. At t = 120, the slope is 2 m/s
     
    Last edited: Jul 4, 2012
  7. Jul 4, 2012 #6
    Thanks. That further confirms what I posted after thinking on it, indicating my answer as 2 m/s.
     
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