Question

The speed of a bullet is 300. m/s as it emerges from the 1.00-meter-long barrel of a rifle. Assume that the acceleration of the bullet is constant along the entire length of the barrel, and that the bullet started from rest. What is the acceleration, a, for the bullet while it is in the barrel? (Hint: Time doesn’t matter in this part. Think about which kinematics equation to use.

Answer 1

Answer: 45000 m/s^2

Step-by-step explanation:

First, we know:

Bullet's Initial Velocity (Vo) = 0 m/s

Distance Travelled (D) = 1 m

Final Velocity (Vf) = 300 m/s

We need to know the bullets acceleration to reach the final velocity on that distance. Because the bullet stars from rest, his Initial Velocity is equal to zero.

We dont know on how much time it takes for the bullet, but we know the acceleration on the barrel is constant.

We can use the following equation for this

`Vf^(2) = Vo^(2)+2ax`

Because Vo is zero, the equation simplify, so we have.

`Vf^(2) = 2ax`

Searching for the accelaration we obtain

`a = ((Vf)^(2) )/(2x)`

Changing values

`a = ((300 m/s)^(2) )/(2 * 1m)`

`a = ((300 m/s)^(2) )/(2 * 1m)`

`a = 45000 m/s^(2)`

This is the acceleration needed to reach the velocity at the end of the barrel.