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Relative velocity

  1. May 6, 2010 #1
    on a two lane road, car A is travelling with a speed of 36km/h. two cars B and C approach car A in opposite direction with a speed of 54km/h each. at a certain instant, when distance AB is equal to AC both being 1km, B decides to overtake A before C does. what minimum acceleration of B is required to avoid accident?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. May 6, 2010 #2


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    Gold Member

    Welcome to PF! :)
    Consider a reference frame associated with the car A.
    Solve this problem step by step:
    What is the speed of B and C relative to the A?
    How much time [tex]t_0[/tex] does it take for C to overtake A? (it's easy to calculate because C moves with a constant speed relative to the A)
    Use the equation of motion for B:
    [tex]x(t) = v_{0B} t + a t^2/2[/tex]
    where [tex]v_{0B}[/tex] is a relative speed for B you've found earlier, [tex]a[/tex] -- unknown acceleration (it is the same in both laboratory (that is from your point of view) and moving relative to the A frames of references).
    For a specific moment of time [tex]t=t_0[/tex], where [tex]t_0[/tex] you've already found this equation says:
    [tex]L = v_{0B}t_0 + a t_0^2/2[/tex]
    where L is the distance between A and B and A and C at the moment t=0, that is 1 km.
    From this you'll find the acceleration.
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