One-Dimensional Motion, Bullet-through-a-board-type question

[88elephants]

Homework Statement

36 g bullet, speed of 350 m/s strikes a 8 cm fence post. It is retarded by an average Force is 3.6 x 10^{6} N while going through the post.

a. Speed of bullet when it emerges?
b. How many boards could the bullet penetrate?

Homework Equations

v^{2}=v^{2}_{0}+2a\Deltax
F=ma
collision equations, maybe?

The Attempt at a Solution

I took the equation F=ma and plugged in (3.6x10^{6})=(.03 kg)a and then got a=1.0x10^{8} m/s^{2}.
I then plugged that into this equation, v^{2}=v^{2}_{0}+2a\Deltax and got
v^{2}= (350 m/s)^{2} + 2(1.0x10^{8} m/s^{2})(.08 m)
but the v I calculated is a bigger velocity than when the bullet started.
so...what is going wrong here? thank you so much in advance :)

 

[Siracuse]

Homework Statement

It is retarded by an average Force is 3.6 x 10^{6} N while going through the post.

Given an x-axis,

-------------------------------->x
Vector v, that is the speed of the bullet, goes along with x. The force F that retards(sp?) the bullet is in the opposite direction. In your equations, you're assuming F is positive.

 

[88elephants]

Ah, it's always the positives and negatives. Thanks.

for part b, I am just going to use the same equation, with the delta x as the variable to be found, setting the final velocity to 0?
so, setting that up, it would be

v^{2}=v^{2}_{0}+2a\Deltax

0= (350 m/s)^{2} - 2(1.0 x 10^{6}) (\Deltax)

and then using that number, dividing it by the 8 cm known of the thickness of the post, ending up with the answer...

 

[Siracuse]

Yeah, your logic is right. You can always use the other formulas for accelerated motion, but that's the easy way out.