Motion in one - dimension question

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The discussion revolves around calculating the average acceleration and total time a bullet spends in contact with a board. The average acceleration was correctly determined to be -49,000 m/s² using the kinematic equation. However, the time calculation for the bullet's contact with the board was incorrect; the initial calculation only accounted for the acceleration phase. To find the total time, one must also consider the bullet's length, as the back of the bullet continues moving through the board after the front tip exits. Understanding this additional time is crucial for an accurate solution.
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An indestructible bullet 2.00 cm long is fired straight through a board that is 10.0 cm thick. The bullet strikes the board with a speed of 420 m/s and emerges with a speed of 280 m/s. (To simplify, assume that the bullet accelerates only while the front tip is in contact with the wood.)

the question asks for a)avg. acceleration of the bullet through the board?
b) total time that the bullet is in contact with board?


i found the avg. acceleration by using the formula v^2=Vinit^2 + 2aX
with V=280 Vint=420 and X=.1m and found acceleration to be -49000 m/s^2... i checked with my teacher and this answer is correct..however my answer for b) is wrong

for b) i used the equation V= Vint + at
with V=280, Vint=420 and a=-4900 and found time to = 2.87e-4

my t is wrong can somebody please help me...i would also like to add that my teacher said that there are a few ways to interpret this question..(for this question I've ignored the fact that the bullet is 2cm long, but i don't know how to use this bit of info to solve for time)

thank you
 
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What you found is the time for the bullet to accelerate from initial to final speed. But when the bullet reaches its final speed, only the front tip has emerged. How much additional time is needed for the rest of the bullet to lose contact with the wood? (That's why they gave you the length of the bullet.)
 
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