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Kinematics Problem

  1. Feb 24, 2008 #1
    1. The problem statement, all variables and given/known data

    A student throws a stone vertically downward with an initial velocity of 5.0 m/s from the top of a 165 m building. How far does the stone travel during its 4th second of travel?

    2. Relevant equations

    d= vit + 1/2at2


    3. The attempt at a solution

    Vi = -5.0 m/s
    t = 4.0 s
    a = -9.81 m/s2
    d = ?

    I tried it in the kinematics equation d= vit + 1/2at2 and the answer is wrong.
     
  2. jcsd
  3. Feb 24, 2008 #2
    Well, d=Vot+.5at^2
    so d should be 98.4m. Did you get that?
     
  4. Feb 24, 2008 #3
    The answer given is 39 m.
     
  5. Feb 24, 2008 #4
    Wait, I think it was the d for the 4th second, from 3-4s. Because the first second is 0-1s, second second is for 1-2s, etc.

    So
    Vf=vo+at
    V@3=5+9.8*3 and V@4=5+9.8*4
    Vf^2=Vo^2+2ad
    (V@4)^2=(V@3)^2+2*9.8*d
    Find d
     
    Last edited: Feb 24, 2008
  6. Feb 24, 2008 #5
    d = displacement/distance.
     
  7. Feb 24, 2008 #6
    Yes, i know.
     
  8. Feb 24, 2008 #7
    Is anyone going to help solve this problem please?
     
  9. Feb 24, 2008 #8
    I just gave you the answer.
     
  10. Feb 24, 2008 #9
    As silvashadow has already pointed out, you are not reading the question carefully. It does not ask for the distance after 4 seconds, it asks for the displacement during its 4th second

    Displacement during any given time interval=
    (Position at final time)-(Position at initial time).

    What is the final time? What is the initial?
     
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