Motion in one - dimension question

[rcwha]

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

 

[Doc Al]

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.)

 

[kyle8921]

for your question and explanation of your approach. I can see that you have correctly calculated the average acceleration of the bullet through the board, but there may be some confusion with the time calculation.

To find the total time that the bullet is in contact with the board, we need to consider the fact that the bullet is 2 cm long. This means that the front tip of the bullet will be in contact with the board for a shorter amount of time than the back tip.

To calculate the time, we can use the formula d=Vint*t + 1/2*a*t^2, where d is the distance traveled (in this case 10 cm), Vint is the initial velocity (420 m/s), a is the acceleration (-49000 m/s^2), and t is the time.

Since the bullet is 2 cm long, the front tip will travel a distance of 2 cm while the back tip travels 10 cm. We can set up two equations:

2 = 420*t + 1/2*(-49000)*t^2
10 = 420*t + 1/2*(-49000)*t^2

Solving for t in both equations, we get t = 0.0002 s for the front tip and t = 0.0004 s for the back tip. The total time the bullet is in contact with the board is the sum of these two times, which is 0.0006 s.

I hope this helps clarify the concept and how to approach the problem. Remember to always consider all the given information and use appropriate equations to solve the problem.