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Time to a driver catch another

  1. Jul 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Two drivers met at a station and they discover that both have the same destination.
    The driver A has a car which goes from 0 to 100km/h in 16s and driver B has a car which goes from 0 to 100km/h in 5s. The maximum velocity of each is 100km/h
    How much time after driver A starts driving should driver B start driving to be sure that he catches driver A?

    2. Relevant equations
    Should I use constant acceleration equations?


    3. The attempt at a solution
    I tried the following:
    100km/h = 27.8m/s

    Driver A
    a = 27.8/16 = 1.73m/s^2

    Driver B
    a = 27.8/5 = 5.56m/s^2

    Now I cant continue because I dont know if I can use constant acceleration equations
     
  2. jcsd
  3. Jul 22, 2013 #2

    tiny-tim

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    yes of course :smile:

    start by finding how long it takes the first one to reach 100 km/h (after which it stays at that speed while the other one catches up) :wink:
     
  4. Jul 22, 2013 #3
    To reach 100km/h it takes 16s and the car B takes 5s
    But I still cant figure how to get how long after driver A starts driving should driver B start to reach it :confused:
     
  5. Jul 22, 2013 #4

    tiny-tim

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    write an equation for the first car after it reaches 100 km/hr, and an equation for the second car while it is still accelerating

    start one car at t = 0, and the other at t = to :wink:
     
  6. Jul 22, 2013 #5
    [itex]x-x_{0}=\frac{1}{2}(1.73)*16^{2}[/itex]
    [itex]x-x_{0}=\frac{1}{2}(5.56)*t^{2}[/itex]

    Something like this?
     
  7. Jul 22, 2013 #6

    tiny-tim

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    yes :smile:
    no …
    also …
     
  8. Jul 22, 2013 #7
    [itex]27.8^{2}=2(1.73)(x-x_{0})[/itex]
    [itex]x-x_{0}=\frac{1}{2}(5.56)*t^{2}[/itex]

    This? :confused:
     
  9. Jul 22, 2013 #8

    tiny-tim

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    ?? :confused:

    after it reaches 100 km/hr, it has constant speed
     
  10. Jul 22, 2013 #9
    Yes, but with this the acceleration will be 0, right?
     
  11. Jul 22, 2013 #10

    tiny-tim

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    yes :confused:
     
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