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Superman problem

  1. Oct 26, 2004 #1
    I'm back once more because of another problem that is driving me nuts. I've been trying to do this one for a long time. I'm posting the previous problem because it has relevant information and then bold typing the problem I'm confused on.

    ----Find the speed at which Superman (mass = 79 kg) must fly into a train (mass = 16351 kg) traveling at 80 km/hr to stop it. Calculate the time it takes Superman to stop the train, if the passengers experience an average horizontal force of 0.55 times their own weight. ------

    And the hint they give for the problem is: Hint: The average horizontal force tells you the maximum acceleration of a passenger. With the change in speed in a given time, you can find the acceleration. Careful with units. Be sure to differentiate between weight and mass.

    Ok....so I need to find the time it takes for Superman to stop the train. What I did was find the weight of the train...16351 kg x 9.8 m/s^2....and then multiplied that answer by 0.55 to the get the horizontal force, which I calculated to be 88131.89 N. I know that F=ma...so 88131.89 N = 16531 x A.
    Acceleration is the change in velocity over the change in time. So I solved for A and got 5.33 m/s^2....the change in velocity is 1.66E+04 km/hr (which I found in the previous problem) - 80 km/hr = 16520 km/hr.....
    So 5.33 = 16520/T (change in time)....and I get 3099.44 s or hr...that I'm not sure of. Either way, the answer is wrong and I really don't know what I'm doing wrong.
    If anyone can make any sense of what I wrote and see what I did wrong, I'd be grateful to know. Thanks so much. :)
     
  2. jcsd
  3. Oct 26, 2004 #2

    ehild

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    Gold Member

    The change of velocity for the train is -80 km/h as it stops (v2=0, v1=80 km/h).
    The value you got for the deceleration is a bit inaccurate because of the rounding errors of the lot of unnecessary calculation steps you performed.
    The horizontal force is 0.55*mg, so the deceleration is

    [tex] -a= \frac{0.55 m*g}{m} = 0.55 g= 5.39 \mbox{ m/s2. }[/tex].

    At the end, you mixed units when calculting the time. The unit of the acceleration is m/s^2, you get the correct time from the relation t=(v2-v1)/a only if you use m/s for the velocity. (Or km/h^2 for the acceleration) You were confused about the unit of the result you got. The best way to avoid confusion is to include the units in the formulas you use. 80 km/h= 80 * (1000 m) / (3600 s) = 22.22 m/s, and
    [tex]t= \frac{22.22 m/s} {5.39 m/s^2} = 4.12 \frac{m/s}{m/s^2}=4.12\frac {ms^2}{ms}=4.12\mbox{ s. }[/tex]

    ehild
     
  4. Oct 27, 2004 #3
    Ohhh....yes, I knew I had messed up my units somehow, but it was escaping me. Thank you sooo much...that makes so much more sense! :eek:)
     
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