Physics Help Forum Bullet Problem

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Sep 28th 2009, 06:09 PM #1 Member   Join Date: Sep 2009 Posts: 32 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!
 Sep 28th 2009, 07:38 PM #2 Junior Member   Join Date: Sep 2009 Posts: 4 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.
 Sep 28th 2009, 07:52 PM #3 Junior Member   Join Date: Sep 2009 Posts: 4 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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### a bullet is fired through a board 10 cm thick

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