Abullet fired straight through a board  0.10  m  thick strikes the board with a speed of  480  m/s  and emerges with a speed of  310  m  .
part a
what is the  x  component of the average acceleration of the bullet while in the board? the positive  x  direction is the initial direction of motion of the bullet.
express your answer with the appropriate units.

Explanation:

givens

vf = 310 m/s

vi = 480 m/s

d = 0.1 m

formula

vf^2 = vi^2 + 2*a*d

Solution

310^2 (m/s)^2 = 480^2 (m/s)^2 + 2 * a * 0.1 meters

96100  = 230400  + 2*a * 0.1

96100 - 230400 = 2*a*0.1

-134300 = 2*0.1*a

-134300 = 0.2 *a

-134300/0.2 = a

a = -671500 m/s^2

The important thing to notice is that the acceleration is minus which means the bullet is slowing down.

50 inches should be the resultant vector