Question

A bullet is fired through a board, 14.0 cm thick, with its line of motion perpendicular to the face of the board. If it enters with a speed of 450 m/s and emerges with a speed of 220 m/s, what is the bullet's acceleration as it passes through the board, in units of km/s2 , assuming constant acceleration?

Answer 1

Answer:

$a=-550.35\ m/s^2.$

Explanation:

Given :

Thickness of board , $d=14 \ cm =\dfrac{14}{100000}\ km=1.4\times 10^{-4}\ km.$

Initial speed of bullet , $u=450\ m/s=\dfrac{450}{1000}\ km/s=0.45\ km/s.$

Final speed of bullet , $v=220\ m/s=\dfrac{220}{1000}\ km/s=0.22\ km/s.$

We know by equation of motion.

$v^2-u^2=2as$  ( all sign have their usual meaning )

Putting all given values we get :

$0.22^2-0.45^2=2\times a\times 1.4\times 10^{-4}\\\\\dfrac{0.22^2-0.45^2}{2\times 1.4\times 10^{-4} }=a$

$a=-550.35\ m/s^2.$

Hence, this is the required solution.