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Motion problem

  1. Sep 29, 2009 #1
    Motion problem!!

    1. Objects A and B both start from rest. They both accelerate at the same rate. However, object A accelerates for twice the time as object B. What is the distance traveled by object A compared to that of object B?


    2. Vf = Vo + at, X - Xo = Vot + 1/2at2, V2= Vo2 + 2a(X - Xo)


    3. 2aA= aB
    => VAo2 + 4a(X - Xo) = VBo2 + 2a(X - Xo)
    => distance traveled by object A four times as far of object B??

    Am i correct?
     
  2. jcsd
  3. Sep 30, 2009 #2
    Re: Motion problem!!


    Shouldn't be ta=2tb?
     
  4. Sep 30, 2009 #3
    Re: Motion problem!!

    yes, the distance traveled by A at the time A stops accelerating is 4 times the distance
    traveled by B at the time B stops accelerating.

    I can't see how you get there however. you never use the fact that the time that a
    accelerates is twice as long. I think it's easiest to use x = v_0 t + (1/2) a t^2, and

    substitue t_A = T and t_B = 2T in it
     
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