A bullet is fired with a speed of 450m/s through a board 20 cm thick. If the bullet is subject to a large deceleration of magnitude 106m/s2, will the bullet be able to emerge from the other side of the board ?
The distance X required to stop the bullet when decelerating from Vo = 450 m/s at that rate (a) is given by
Vo = sqrt(2 a X), which can be rewritten
X = Vo^2/(2a)
It is not clear whether your deceleration rate is 106 m/s^2 or 10^6 m/s^2. It makes a difference in your answer.
To determine whether the bullet will be able to emerge from the other side of the board, we need to calculate the time it takes for the bullet to travel through the board and compare it with the time it takes for the bullet to come to a stop.
First, let's convert the thickness of the board from centimeters to meters: 20 cm = 0.2 m.
The bullet's initial speed is given as 450 m/s, and the deceleration is given as 106 m/s².
We can use the kinematic equation:
vf = vi + at
where:
- vf is the final velocity (which will be zero when the bullet comes to a stop)
- vi is the initial velocity (450 m/s)
- a is the acceleration (deceleration, -106 m/s²)
- t is the time it takes for the bullet to stop
Rearranging the equation to solve for time (t):
t = (vf - vi) / a
t = (0 - 450) / -106
t = 4.245 seconds (rounded to 3 decimal places)
Next, we can calculate the distance traveled by the bullet as it decelerates:
d = vi * t + (1/2) * a * t²
d = 450 * 4.245 + (1/2) * -106 * (4.245)²
d = 1910.25 meters (rounded to 2 decimal places)
Comparing the distance traveled by the bullet (1910.25 meters) with the thickness of the board (0.2 meters), we can see that the bullet will have emerged from the other side of the board since the distance traveled is significantly greater than the thickness of the board.