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Calculate the force to stop a train

  1. Mar 13, 2017 #1
    1. The problem statement, all variables and given/known data

    Superman must stop a 120-km/hr train in 150 m to keep it from hitting a stalled car on the tracks. If the trains mass is 3.6 x 10^5 kg, how much force must he exert?

    Vi = 33 m/s (120 km/h)
    Vf = 0 m/s
    Displacement (Xf - Xi) = 150 m
    M = 3.6 x 10^5 kg

    2. Relevant equations

    v = x / t

    a = v / t

    F = m a

    3. The attempt at a solution

    v = x / t
    33 = 150 / t
    t = 4.55 secs

    a = v / t
    a = 33 / 4.55
    a = 7.26 m/s^2

    F = 3.6 * 10^5 * 7.26
    F = 2613600N

    Ignoring the sign convention for just a second, the answer is half the value of what I calculated. I would like to know why I have to use the formula below, and why my method is incorrect.

    upload_2017-3-14_14-6-23.png
     
  2. jcsd
  3. Mar 13, 2017 #2
    Your formula for acceleration is wrong. You were probably thinking of ##a=\frac {dv} {dt}##. And your equation for velocity only works if the velocity is constant. In a case of changing velocity, like this one, the correct relationship is ##v=\frac {dx} {dt}##. The solution is using the kinematic equation ##v^2=v^2_0+2a\Delta x## to solve for the required acceleration, and plugging it into your equation for force.
     
  4. Mar 13, 2017 #3
    As TJGilb pointed out, v = x/t is only valid for constant velocity - NOT constant acceleration. Therefore, your time calculation of t = 4.55 seconds is how long it would take the train to move 150 meters if it was traveling at a constant velocity of 33 m/s. Since the train is not moving at constant velocity, but decelerating, the time will take longer than 4.55 seconds to cover the 150 m distance. Because the time you calculated was shorter than the actual, your acceleration was greater than actual, which results in a greater required force than actual.

    Just an observation: Superman probably could have moved the car with his little finger. Just sayin'.
     
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