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Kinematics bus problem

  1. Nov 26, 2009 #1
    1. The problem statement, all variables and given/known data
    Question reads: A bus travelling at 15.0m/s [E], started to slow down with an acceleration of 0.5m/s^2[W] as soon as it was abreast of a sports car. At the same instant, the sports car had a velocity of 6m/s [E] and was accelerating at 2.5m/s^2 [E]. At what time does the sports car catch up to the bus?

    2. Relevant equations
    vf^2 = vi^2 + 2ad
    d = Vit + 1/2at^2


    3. The attempt at a solution

    d = 15t + 1/2(-0.5)t^2 <-- bus
    d = 6t + 1/2(2.5)t^2 <-- car

    these are the two equations that i have and will i use these two equations to solve for time ? or distance?

    im very confused at this point. ( putting two different object's speed together)

    please help me further guide me with this problem.
     
  2. jcsd
  3. Nov 26, 2009 #2
    Re: Kinematics

    Hi ya,

    The equation you have used is correct - remember, that if the car has caught up with the bus, then d will be the same for both. Hope this helps :)
     
  4. Nov 26, 2009 #3
    Re: Kinematics

    The question sounds so confusing to me :S If the bus and the car are abreast, then the car has already caught up with the bus..

    Anyway, usually for this kind of thing, you have 2 equations, use the variable that is in common with both of them, equate them to each other and solve
     
  5. Nov 27, 2009 #4
    Re: Kinematics

    d = 15t + 1/2(-0.5)t^2 <<--- for the bus

    d = 6t + 1/2(2.5)t^2 <<-- for the sport car.


    Solve the simultaneous equation and you had it.

    I hope i'm right @_@.

    My answer is 6seconds.
     
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