Average Acceleration Question for Bullet Through Board

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SUMMARY

The discussion centers on calculating the average acceleration of a bullet passing through a 5.0 cm thick board. The bullet enters at a speed of 200 m/s and exits at 100 m/s, resulting in an average acceleration of -300,300 m/s² over a time span of 0.000333 seconds. The solution confirms that using the average velocity of 150 m/s to derive the time taken is correct, and the approach aligns with established kinematic equations. The relationship between distance, time, and acceleration is effectively demonstrated through the derivation of the kinematic equation.

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Homework Statement


"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.

Homework Equations


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

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:
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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:\frac{d}{T} = \frac{1}{2}(v + u) \Rightarrow 2d = (v + u)T... using distace d, change in time T, final velocity v, initial velocity u.

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

Soooo... combining them:2d = \big ( (aT + u) + u\big )T = aT^2 +2uT \Rightarrow d = uT + \frac{1}{2}aT^2... 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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