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This is not my homework! I am just curious!

  1. Sep 23, 2010 #1
    1. The problem statement, all variables and given/known data

    A car starts from rest and accelerates uniformly at 3.0 m/s^2 a second car starts from rest 6 seconds later at the same point and accelerates uniformly at 5 m/s^2. How long does it take the second car to overtake the first car.


    2. Relevant equations
    d=vit+(1/2)at^2 I know that there might be a quadratic equation to solve for x.


    3. The attempt at a solution

    I came across this question at this website: http://answers.yahoo.com/question/i...6cAp8NojzKIX;_ylv=3?qid=20090917181717AAiO4kQ and I have been stuck on it for about 20 minutes.

    Again this is not my homework, but I do need help.
     
  2. jcsd
  3. Sep 23, 2010 #2
    Think about it. The two cars are basically going through the same process. Each starts from rest and begins accelerating. The same equation will apply to both cars. The only difference is when each car starts. Think about how you can write the kinematic equation for the second car relative to the time of the first car.
     
  4. Sep 23, 2010 #3
    This is what I have so far 1.5t^2=2.5t^2-30t+90 is this right?
     
  5. Sep 23, 2010 #4
    So you want to know when their distances are equal

    ∆s=vit+.5at^2

    ∆s1=1.5t^2
    ∆s2=2.5(t-6)^2 because the second driver is 6 seconds behind

    then just set the two to be equal

    t^2-30t+90=0

    looks like you got it, nice work!

    t=26 by my calculation
     
  6. Sep 24, 2010 #5


    Thank you, when you say ∆s=vit+.5at^2 do you mean d=vit+(1/2)at^2?
     
  7. Sep 24, 2010 #6
    I suppose you could call it d. The kinematic equation is really

    s=si+vi5+.5at^2

    where s is the ending point and si is the starting point. Of course, for this case we can define the start at 0, and both have the same start. However, if the other car was 10m ahead then we would have to use the kinematic equation I listed.
     
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