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Homework Help: Physics Problem; help me please

  1. Sep 25, 2005 #1
    Help, I just can't seem to get this.

    A freight train has a mass of 15000000 kg. If the locomotive can exert a constant pull (what is a constant pull?) of 750000 N, how long would it take to increase the speed of the train from rest to 85 km/h? (Disregard friction.)
  2. jcsd
  3. Sep 25, 2005 #2
    newtons second law: f=ma, a=f/m. You can use this to derive the acceration and use some kinematics formulas to find out how long it will take to reach a certain velocity with a known acceleration and a known initial velocity.
    Last edited: Sep 25, 2005
  4. Sep 25, 2005 #3
    F = ma

    750000 N = (15000000 kg) * a
  5. Sep 25, 2005 #4
    Well, you have a fairly basic question that comes from Netwon's second law of motion. By constant pull, the system has a constant force that is pulling the train along. This means that the system also has constant acceleration, which is given by Netwon's Laws.

    F = M A

    You can easily find this acceleration by using the values already given.
    You now can use this in a kinematics equation. You have acceleration,initial velocity,final velocity, and you need to find time.

    I suggest using this equation:


    By the way, make sure you watch your units. The final velocity that the problem gives is in kilometers per hour. Convert this speed to standard units.

    Hope this helps
  6. Sep 25, 2005 #5
    Ok... What is the answer? The book says 470 seconds. I did 750000N/15000000 and got .05m/s squared for the acceleration. Then, If I divide the final speed, 85 km/h by this acceleration, I get 1700 seconds which is wrong. How do they get 4.7 times 10 to the second (470 seconds)? Help please. I appreciate it.
  7. Sep 25, 2005 #6
    conversion from m/s to km/h
  8. Sep 25, 2005 #7
    They are looking for time in this problem.
  9. Sep 25, 2005 #8
    yeah, but the conversion from m/s to km/h is essential in arriving at the proper time. You are dividing 85 km/h by m/s. The units do not cancel out.
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