A bullet is fired through a board 6.0 cm thick in such a way that the bullet's line of motion is perpendicular to the face of the board. If the initial speed of the bullet is 450 m/s and it emerges from the other side of the board with a speed of 350 m/s. (a) Find the acceleration of the bullet as it passes through the board.

Question

anonymous · Answer

Vi = 450 m/s
Vf = 350 m/s
Xi = 0 m
Xf = 0.06 m
a = ?

There is a kinematic equation for this you should know:

Vf^2 = Vi^2 + 2a(Xf - Xi)
a = (Vf^2 - Vi^2) / 2(Xf - Xi)

a = -6.67 x 10^5 m/s^2