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

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88elephants
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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[tex]^{6}[/tex] N while going through the post.

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


Homework Equations



v[tex]^{2}[/tex]=v[tex]^{2}_{0}[/tex]+2a[tex]\Delta[/tex]x
F=ma
collision equations, maybe?

The Attempt at a Solution



I took the equation F=ma and plugged in (3.6x10[tex]^{6}[/tex])=(.03 kg)a and then got a=1.0x10[tex]^{8}[/tex] m/s[tex]^{2}[/tex].
I then plugged that into this equation, v[tex]^{2}[/tex]=v[tex]^{2}_{0}[/tex]+2a[tex]\Delta[/tex]x and got
v[tex]^{2}[/tex]= (350 m/s)[tex]^{2}[/tex] + 2(1.0x10[tex]^{8}[/tex] m/s[tex]^{2}[/tex])(.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 :)
 
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88elephants said:

Homework Statement


It is retarded by an average Force is 3.6 x 10[tex]^{6}[/tex] 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.
 
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[tex]^{2}[/tex]=v[tex]^{2}_{0}[/tex]+2a[tex]\Delta[/tex]x

0= (350 m/s)[tex]^{2}[/tex] - 2(1.0 x 10[tex]^{6}[/tex]) ([tex]\Delta[/tex]x)

and then using that number, dividing it by the 8 cm known of the thickness of the post, ending up with the answer...
 
Yeah, your logic is right. You can always use the other formulas for accelerated motion, but that's the easy way out.