A bullet is fired into a tree trunk (Figure 1.12), striking it with an ini- tial velocity of 350 m/ s. If the bullet penetrates the tree trunk to a depth of 8.0 cm and comes to rest, what is the acceleration of the bullet?
To find the acceleration of the bullet, we need to use the kinematic equation:
v² = u² + 2as
Where:
v = final velocity (0 m/s since the bullet comes to rest)
u = initial velocity (350 m/s)
a = acceleration (unknown)
s = displacement (8.0 cm = 0.08 m)
Substituting the given values into the equation:
0² = (350 m/s)² + 2a(0.08 m)
0 = 122500 m²/s² + 0.16a
Rearranging the equation:
0.16a = -122500 m²/s²
Dividing both sides by 0.16:
a = -(122500 m²/s²) / 0.16
a ≈ -765,625 m²/s²
Therefore, the acceleration of the bullet is approximately -765,625 m²/s².
To find the acceleration of the bullet, we can use the equation for deceleration:
acceleration = (final velocity - initial velocity) / time
In this case, the final velocity is 0 m/s because the bullet comes to rest. The initial velocity is 350 m/s.
We need to find the time it takes for the bullet to come to rest. To do this, we can use the equation for displacement:
displacement = (initial velocity + final velocity) / 2 * time
In this case, the initial velocity is 350 m/s, the final velocity is 0 m/s, and the displacement is the penetration depth of the bullet, which is 8.0 cm or 0.08 m.
We can rearrange the equation to solve for time:
time = 2 * displacement / (initial velocity + final velocity)
Substituting the values we have:
time = 2 * 0.08 m / (350 m/s + 0 m/s)
time = 0.16 m / 350 m/s
time = 0.000457 s
Now we have the time it takes for the bullet to come to rest, we can substitute it back into the formula for acceleration:
acceleration = (final velocity - initial velocity) / time
acceleration = (0 m/s - 350 m/s) / 0.000457 s
acceleration = -762928.64 m/s^2
Therefore, the acceleration of the bullet is approximately -762928.64 m/s^2. The negative sign indicates that the bullet is decelerating.
To find the acceleration of the bullet, we can use the kinematic equation:
vf^2 = vi^2 + 2ad
Where vf is the final velocity, vi is the initial velocity, a is the acceleration, and d is the distance traveled.
In this case, the final velocity is 0 m/s because the bullet comes to rest. The initial velocity (vi) is given as 350 m/s. The distance traveled (d) is the penetration depth of the bullet, which is 8.0 cm or 0.08 m.
Putting these values into the equation, we have:
0 = (350 m/s)^2 + 2a * 0.08 m
Simplifying the equation, we get:
0 = 122500 m^2/s^2 + 0.16 a
To find the acceleration, we need to isolate the acceleration term:
0.16 a = -122500 m^2/s^2
Dividing both sides of the equation by 0.16, we get:
a = -122500 m^2/s^2 / 0.16
Calculating the acceleration, we have:
a = -765625 m^2/s^2
Therefore, the acceleration of the bullet is approximately -765625 m^2/s^2. The negative sign indicates that the direction of the acceleration is opposite to the initial velocity.