a=(V^2-Vo^2)/2•s
s=11.5+1.85=13.35 cm=0.1335 m
a =2.95•10^5 m/s
t=(V-Vo)/a =0.000390 s
s1 = V^2/2•a = 0.542 m=54.2 cm
An indestructible bullet 1.85 cm long is fired straight through a board that is 11.5 cm thick. The bullet strikes the board with a speed of 400 m/s and emerges with a speed of 285 m/s.
(a) What is the average acceleration of the bullet as it passes through the board?
Give your answer in in scientific notation rounded to three significant figures.
(b) What is the total time that the bullet is in contact with the board?
Round your answer to the nearest millionth, if necessary. Give your answer in seconds.
**This I have correctly calculated to be 0.000390seconds***
(c) What thickness of board (calculated to 0.1 cm) would it take to stop the bullet, assuming the bullet's acceleration through all boards is the same?
Give your answer in centimetres.
1 answer