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Overtaking cars and trucks.

  1. Sep 2, 2007 #1
    A car capable of a constant acceleration of 4.85 m/s^2 is stopped at a traffic light. When the light turns green, the car starts from rest with this acceleration. At the very same moment, a truck traveling with constant velocity 17.2 m/s passes the car. As the car's velocity increases, it will eventually move faster than the truck and later overtake it. How far from the light will the car catch up with the truck? Answer in units of m.

    I am not really sure how to start this problem, but I do know that the two vehicles have time and distance in common. Since I am finding distance, however, from the stoplight, I first need to find out how much time it took for the car to reach the truck. I don't think I'm suppose to, however, find out how much time it took for the car to reach the truck's velocity, because that does not necessarily mean that the car has caught up with the truck. In fact, the car must exceed the truck's velocity, at some point, in order to catch up with it. I do not think that anything behind the stoplight really matters, save for the acceleration of the car given.

    I'm given these equations:

    v = u + at

    displacement = ut + .5a(t^2)

    v^2 = u^2 + 2a(displacement)
     
  2. jcsd
  3. Sep 2, 2007 #2

    learningphysics

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    Homework Helper

    let t = 0 bet the moment the truck is passing the car... ie the moment they're at the same position...

    Write the equation for displacement of the truck in terms of time. It's just going at constant speed.

    Write the equation for displacement of the car in terms of time.
     
  4. Sep 2, 2007 #3
    Well, I used x = ut + .5a(t^2).

    The distance, once the vehicles meet, will be the same, as will the time (assuming that the time started when the car and truck passed the stoplight). So, I can set up the equations and set them equal to one another.

    Car:

    x = 0(t) + .5(4.85)(t^2)

    Truck:

    x = 17.2t + .5(0)(t^2)

    So:

    17.2t = .5(4.85)(t^2)

    t = 0 or about 7.1 seconds, which makes sense, as the truck and car are beside each other at the very very beginning and again when the car catches up with the truck. Since the truck, this entire time, has been traveling at a constant velocity, it is a reliable source from which I can find the distance.

    x = (17.2)(7.1)
    = about 122, or 123, if using the long decimal answer given in the calculator for time.

    Correct?
     
  5. Sep 2, 2007 #4
    or, sorry, 122 is right. I'm losing it. XDDD
     
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