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Physics word problem- velocity, acceleration, time

  1. Sep 27, 2008 #1
    1. The problem statement, all variables and given/known data

    two cars are traveling along a straight road, one behind the other. the first is traveling at a constant velocity of 12 m/s. the second, approaching from the rear, is traveling at 25 m/s. when the second car is 200m behind the first, the driver applies the brakes, producing a constant acceleration of -0.20 m/s^2. will the cars crash and if so, where and when?

    2. Relevant equations
    acceleration = (change in velocity)/(change in time)
    change in dispacement = (Initial velocity) (time) + (1/2) (acceleration) (time sqaured)
    velocity = (acceleration) (time)

    3. The attempt at a solution
    i got they woud crash after 50 meters? but i dont think thats right :(
  2. jcsd
  3. Sep 27, 2008 #2


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    Welcome to PF!

    Hi meredith ! Welcome to PF! :smile:

    Show us what you did, and then we can see where it went wrong. :smile:
  4. Sep 27, 2008 #3
    If the second car had a velocity of 12m/s^2, then they wouldn't crash, so we try to see if they do;

    25 - 0,2t = 12 => t=65s

    in 65 seconds we see that the cars actually crash. So its time to see where they actually crash..

    so in "t" seconds after the break, the second cars velocity will be "25-0,2t" and first velocity was "25". Time past is "t" and the length passed is; "12t+200m". So if we equalize these, we have;


    then we find t=13(approximately) (At the 13th second they crash) so then we have a velocity of "22,4m/s^2" of the second car when they crash and this hapens after "346" meters passed...

    I hope I am correct!
  5. Sep 27, 2008 #4


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    Yes … but how did you get "t=13(approximately)" ?

    This is a quadratic equation … find the exact solution! :smile:
  6. Sep 27, 2008 #5
    I think you're wrong.
    First I think we need to set an equality of the distance, having 25t-.2t2+200=12t.
    So we solve for t, getting t1=-12.85s, and t2=77.85s, so t2 is out real answer.
    Then you plug t in a distance formula. So based on the first car you have d = 12m/s(77.85s)=934.2m.
  7. Sep 28, 2008 #6
    Sakha, after 65 seconds second car's velocity becomes 12m/s^2, and they are not supposed to crash with same velocities. So that your t2=77 second is impossible...

    The right answer is 13 seconds(I could'nt solve but you can use a program to get the exact number). After 13 secs they crash...
  8. Sep 28, 2008 #7


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    Yes you can! :rolleyes:

    It's 0.1 t2 - 13t + 200 = 0, so the answer is … ? :smile:
  9. Sep 28, 2008 #8
    I mean't i didn't work for it:P Just used a program.
  10. Sep 28, 2008 #9
    Re: Welcome to PF!

    i just registered for this site today and i was wondering how i go about asking questions?
  11. Sep 28, 2008 #10


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    Welcome to PF!

    Hi rmax58! Welcome to PF! :smile:

    Go to the Homework & Coursework Questions forum, choose the appropriate sub-forum, and click the "NEW TOPIC" button. :smile:

    (and read the homework FAQs! :wink:)
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