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Car and truck velocity question

  1. Feb 17, 2013 #1
    A car is initially 4.0km/h south of a gas station and is moving at a constant speed of 55km/h due north. A truck is initially 6.0km north of the gas station and is moving at a constant speed of 45km/h due south. How far are the vehicles from the gas station when they pass each other?


    Attempt at solution:
    So I said:

    D=4.0km south
    V=55km/h north

    D=6.0km north
    V= 45km/h south

    Now I had the idea of making the times equal,

    So I calculated the time for the car:
    Tcar= d/v = 4.0/55= 0.073h
    Ttruck = d/v = 6.0/45=0.13h

    From here I am stuck I tryed subtracting the times, but could get any further. Help from this point would be awesome.

    I have done a little diagram attached ImageUploadedByPhysics Forums1361134571.226902.jpg
  2. jcsd
  3. Feb 17, 2013 #2


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    Staff: Mentor

    You calculated the time difference as seen at the gas station. This is not relevant for the task here.
    Can you express their position as function of time?
  4. Feb 17, 2013 #3
    Would that be t=d/v=d/v??
  5. Feb 17, 2013 #4


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    That would be the time to travel some specific distance.
    No, what I mean is an equation of the type x(t)=....
    You plug in the time t, and get the position x (relative to the gas station) at that time t.
  6. Feb 17, 2013 #5
    like x(t)=changeinD/change in v
  7. Feb 18, 2013 #6


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    Staff: Mentor

    There is no change in v.

    A simple example: Imagine I start 10km in front of from the gas station (seen by an arbitrary direction) and move with 20km/h. In that case, my position is given by x(t)=-10km + 20km/h*t
    For t=0, this givex x(0)=-10km, as expected.
    1/2 hour later, I am at x(1/2 h)=-10km + 20km/h * 1/2 h = -10km + 10km = 0 - in other words, directly at the gas station.

    Can you find similar equations for your vehicles?
    Afterwards: What means "passing" for their positions?
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