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Old Sep 28th 2009, 06:09 PM   #1
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Bullet Problem

A bullet is fired through a board 10.0 cm. thick in such a way that the bullet's line of motion is perpendicular to the face of the board. If the initial speed of the bullet is 400 m/s and it emerges from the other side of the board with a speed of 300 m/s, find (a) the acceleration of the bullet as it passes through the board and (b) the total time the bullet is in contact with the board.

I have tried to use the x=xo+vot+1/2at^2 and v=vo+at equations but that still leaves me with two unknowns. I am not really sure how to do this. Please help!
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Old Sep 28th 2009, 07:38 PM   #2
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Bullet problem

USE THE FOLLOWING TO FIND THE ACCELERATION

v^2 - v0^2= 2ax
and convert x=10cm =0.1 m

a= [ (300x3000) - ( 400x 400)]/(2 x0.1)= - 35 X 10^4 m/s^2

the acceleration is negative because it is losing speed

V= vo + at , thus t= [V- Vo]/a

[ 300- 400]/( -35 x10^4) = 2.857 x 10^(-4) seconds
after you find a, then use the relation v= v0 + at to find the time

Last edited by joephos; Sep 28th 2009 at 07:46 PM.
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Old Sep 28th 2009, 07:52 PM   #3
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Bullet problem

V^2- Vo^2 = 2aX thus a= [ V^2- Vo]/2x

v= 300 m/s Vo = 400m/s

X= 10 cm = 0.1m

a= [ (300 x300)- (400 x400)]/ 2x 0.1

a= - 35 x 10^4 m/s^2

then V= Vo + at , thus t= (V-Vo)/a

t= (300-400)/(-35x10^4) = 2.857 x 10^(-4) seconds

I hope that this help
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