A bullet is fired through a wooden board with a thickness of 10.0 cm. The bullet hits the board perpendicular to it, and with a speed of +390 m/s. The bullet then emerges on the other side of the board with a speed of +234 m/s. Assuming constant acceleration (rather, deceleration!) of the bullet while inside the wooden board, calculate the acceleration and Calculate also the total time the bullet is in contact with the board (in sec).

The average velocity in the board is

vavg=(vout+vin)/2 figure that.

then the time in the board is

thickness/velocty average

To find the acceleration and the total time the bullet is in contact with the board, we can apply the kinematic equations of motion.

1. Start by finding the deceleration of the bullet while inside the wooden board. We can use the equation:

v^2 = u^2 + 2as

Where:
v = final velocity of the bullet (+234 m/s)
u = initial velocity of the bullet (+390 m/s)
a = deceleration of the bullet (-a, as it is slowing down)
s = distance traveled (-0.1 m, as the thickness of the board is 10.0 cm)

Rearranging the equation, we have:

a = (v^2 - u^2) / (2s)
a = (234^2 - 390^2) / (2 * (-0.1))
a = (54756 - 152100) / (-0.2)
a = -97344 / (-0.2)
a = 486720 m/s^2

The acceleration of the bullet inside the wooden board is -486720 m/s^2.

2. Next, we can determine the time the bullet is in contact with the board using the equation:

v = u + at

Where:
v = final velocity of the bullet (+234 m/s)
u = initial velocity of the bullet (+390 m/s)
a = deceleration of the bullet (-486720 m/s^2)
t = time taken

Rearranging the equation, we have:

t = (v - u) / a
t = (234 - 390) / (-486720)
t = -156 / (-486720)
t ≈ 0.00032 s

The total time the bullet is in contact with the board is approximately 0.00032 seconds.