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Distance Covered While Acc, and Dec

  1. Aug 2, 2012 #1
    A subway train starts from rest at a station and accelerates at a rate of 1,20 m/s^2
    for 20,0s. It runs at constant speed for 50,0 s and slows down at -3,00 m/s^2
    until it stops at the next station. Find the total distance covered.

    I just need to know the formulas, because i studied this a long time ago, and forgot them!
     
  2. jcsd
  3. Aug 2, 2012 #2
    Ive been using this formula but somehow it dose not check out i THINK. (acc) D= 0.5 * 1.20 * 20^2
     
  4. Aug 2, 2012 #3
    Break the problem into several parts.

    1. Use your formula above to determine how far it went when accelerating.
    2. Use formula V=a*t to determine the speed attained in 1 above.
    3. Use formula X = V*t to determine how far it went in the 50 seconds.
    4. Use formua X = V*t + .5*a*t^2 to determine how far it went during the slowing down portion of the problem. Note that 'a' is -3.0 in this equation.
    5. Add all the distances up.
     
  5. Aug 2, 2012 #4
    I still need the formula to calculate the distance traveled while acc. to 24m/s^2. Or is mine right D= 1/2 * a * t^2
     
  6. Aug 3, 2012 #5
    Recall my first statement:
    1. Use your formula above to determine how far it went when accelerating.

    You have the proper formula to determine distance for the acceleration portion.

    Secondly your statement above does not make sense. The last part of it says "while acc. to 24m/s^2". You are implying velocity so the units are wrong.

    V = a*t = (m/s^2)*s = m/s
     
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