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Constant acceleration of rocket car

  1. Sep 12, 2011 #1
    A "rocket car" is launched along a long straight track at t=0s. It moves with constant acceleration a1=1.2m/s2. At t=2s a second car is launched with constant acceleration a2=7.2m/s2.

    At what time does the second car catch up with the first?

    How far down the track do they meet?

    I always have problems with these problems because you need to use the kinematics equations and i get confused. Any help would be greatly appreciated.
     
  2. jcsd
  3. Sep 13, 2011 #2

    wukunlin

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    Gold Member

    at constant acceleration, the distance travelled with respect to time is given by:
    [tex]d = \frac{1}{2} a t^2[/tex]
    note that sometimes the equation needs to be modified to fit the context of your problem :wink:

    basically your question requires you to produce two of these equations and solve for t and d
     
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