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Given vi and a(v) calculate 1. v(t) 2. time to stop 3. distance to top

  1. Oct 12, 2012 #1
    -->A particle is slowing down from initial velocity v_0=10 with acceleration a=-(1+2v), where v is velocity.
    Find
    a) v(t)
    b) time to stop
    c) stopping distance

    What I have so far
    A) V=Vo+at
    v=10+(-1-2v)t
    v+2vt=10-t
    v(1+2t)=10-t
    v(t)=(10-t)/(1+2t)

    B) V=0m/s at stop
    0=(10-t)/(1+2t)
    0=10-t
    t=10s

    C) This is where I'm stuck, I think I have to integrate V=(10-t)/(1+2t), t=0,10 in order to get my distance covered in t seconds. But I'm not quite sure?
     
  2. jcsd
  3. Oct 12, 2012 #2
    I think I may have gotten the last part

    v=dx/dt

    v dt = dx

    ∫v dt= ∫dx

    ∫(10-t)/(1+2t) dt = ∫dx t=0 to 10 and x=0 to x

    and this is where my integration is so bad I get stuck.
     
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