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Homework Help: Motion in One Dimension - Finding two separate changes in time

  1. Jan 19, 2009 #1
    1. The problem statement, all variables and given/known data

    A commuter train travels between two downtown stations. Because the stations are only 1.30 km apart, the train never reaches its maximum possible cruising speed. During rush hour the engineer minimizes the travel interval Δt between the two stations by accelerating for a time interval Δt1 at a1 = 0.100 m/s2 and then immediately braking with acceleration a2 = -0.300 m/s2 for a time interval Δt2. Find the minimum time interval of travel Δt and the time interval Δt1.

    2. Relevant equations

    X=Xi + Vit + .5at^2
    V=Vi + at

    3. The attempt at a solution

    I am not sure how I am supposed to start this problem. I set up a diagram that divided the two time intervals, showed the starting and ending position, and the total distance. I tried to find the velocity using the above equations but I can't figure out how to do so without knowing a value for "t". I am assuming that I have to split the problem into two parts, one with positive acceleration and the other with negative acceleration. However I am always missing a variable, regardless of how I attempt to solve it. Any clues about how to go about this problem would be greatly appreciated! =)
  2. jcsd
  3. Jan 19, 2009 #2
    FinalVelocity^2 = InitialVelocity^2 + 2*Acceleration*Distance

    TotalDistance = AccelerationDistance + BrakingDistance
  4. Jan 19, 2009 #3
    I still can't seem to get this problem. I used those equations but I can't figure out how I am supposed to find any new variables because I am always missing two (as far as I can see, of course). Are there certain equations I am supposed to set equal to each other or something? Thanks!
  5. Jan 20, 2009 #4


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    Homework Helper

    You know x = 1/2*a*t2

    You have an acceleration distance that takes t1 and a deceleration distance that takes t2. Total T is what you want to minimize.

    You also know that the total of acceleration distance and deceleration is 1300m.

    1300 = 1/2(.1)t12 + 1/2(.3)t22

    You also know that for T to be minimized then acceleration to Vmax and down again from Vmax must be at maximum acceleration and deceleration rates.

    Vmax = a*t = .1*t1 = .3*t2
  6. Jan 20, 2009 #5
    Ah ok, I got it now! I should have seen that from the start. Thanks! =)
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