Question

A bullet is fired from rest and leaves a rifle with a velocity of 452 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.93 m. Determine the acceleration of the bullet.

A) 0.93 m/s²

B) 483.87 m/s²

C) 452 m/s²

D) Not enough information provided

Answer 1

The acceleration of the bullet is approximately 109,946 m/s^2.

Step-by-step explanation:

To determine the acceleration of the bullet, we can use the equation:

vf^2 = vi^2 + 2ad

Where vf is the final velocity, vi is the initial velocity (which is 0 m/s since the bullet starts from rest), a is the acceleration, and d is the distance traveled.

Plugging in the given values, we have:

452^2 = 0^2 + 2 imes a imes 0.93

204,304 = 0 + 0

204,304 = 1.86a

a ≈ 109,946 m/s^2

Therefore, the acceleration of the bullet is approximately 109,946 m/s^2. The correct answer is option B) 483.87 m/s².