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Average Acceleration Question

  1. Aug 21, 2012 #1
    1. The problem statement, all variables and given/known data
    "A bullet is fired through a 5.0 cm thick board of some material. The bullet strikes the board with a speed of 200 m/s, and emerges out the other end at 100 m/s.


    2. Relevant equations
    a= (change in velocity)/(change in time)
    average velocity = .5(initial v + final v)


    3. The attempt at a solution

    a= (-100 m/s)/(.000333 seconds) = -300300 m/s^2
    I found .000333s as the time it took for the bullet to pass through the board, using the average velocity of 150 m/s.


    Is this correct?

    SOLVED.
     
    Last edited: Aug 21, 2012
  2. jcsd
  3. Aug 22, 2012 #2

    Simon Bridge

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    It helps, if you are unsure of your reasoning, to relate what you did with some other representation... eg. from your equations:

    ave speed = distance over change in time:[tex]\frac{d}{T} = \frac{1}{2}(v + u) \Rightarrow 2d = (v + u)T[/tex]... using distace d, change in time T, final velocity v, initial velocity u.

    acceleration is change in speed over change in time:[tex]a = \frac{v - u}{T} \Rightarrow v = aT + u[/tex]

    Soooo... combining them:[tex]2d = \big ( (aT + u) + u\big )T = aT^2 +2uT \Rightarrow d = uT + \frac{1}{2}aT^2[/tex]... which you will recognize as a kinematic equation.
    ... so what you've done is basically the same as assuming a constant acceleration, and is consistent with other physics you know.
     
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