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Linear motion problem

  1. Jun 5, 2009 #1
    1. The problem statement, all variables and given/known data

    A car is stopped at a red traffic light. At the instant the traffic light turns green, the car
    starts to move forward with a constant acceleration of 3.20 m/s2. At the same time as the
    light turns green, a truck traveling in the same direction with a constant speed of 20.0
    m/s, overtakes and passes the car.

    How far beyond its starting point does the car overtake the truck?

    So, for the car we have:

    - initial velocity = 0
    - acceleration = 3.20

    for the truck we got:

    - final velocity = 20.0
    - acceleration = 0

    2. Relevant equations

    x-x. = v. t + 1/2 a t^2
    v = v. + a t
    v^2 = v.^2 + 2 a (x-x.)

    3. The attempt at a solution

    I really have no clue on how to deal with this as i don't have time or final velocity for the car, a lot of missing information here and i really don't know how to approach to the answer.

    I just need help please.
     
  2. jcsd
  3. Jun 5, 2009 #2
    Hi there,

    The car has a constant acceleration, therefore no theoretical upper speed limit. To answer questions like these, you need to put the position equation for the two moving objects. One has a constant speed, and the other a constant acceleration.

    Then, you can guess the rest, the car will catch the truck when the position of both are the same.

    Cheers.
     
  4. Jun 5, 2009 #3
    so according to what you said, my answer going to be 250 meters?

    here is my work so you can check if it's right or wrong.

    Car: X= v. t + 1/2 a t^2

    X = 1.6 t^2

    Truck: X = v. t + 1/2 a t^2

    X=20.0 t

    ----

    then, x = x

    1.6 t^2 = 20.0 t
    finally, t = 12.5 s

    then i substitute t back to any of the previous equations to get X = 250 m


    is that correct?

    Thanks btw :)
     
  5. Jun 5, 2009 #4
    That's what I find to.

    It means that it takes the car 12.5sec to catch the truck. With this, you can also determine the car's speed at this time.

    Cheers
     
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