How Do You Calculate Bullet Acceleration in a Rifle Barrel?

[tjh855]

Homework Statement

the barrel of a rifle has a length of 0.916m. a bullet leaves the muzzle of a rifle with a speed of 642 m/s. (note: assume the acceleration in this is constant) what is the acceleration of the bullet while in the barrel? answer is m/s^2

Homework Equations

uhm i assume since I am looking for A i got a=V(final)+V(initial) divided by t

The Attempt at a Solution

uhm what i have goten so far is V(i)= 0m/s? V(f) = 642 m/s and time as .0014267913 ?

i get a wacked number that i know isn't right haha. any help?

also my paper has 2-3 problems like the one I am fixing to explain that I am little confused on.
1. find the average acceleration during the time interval 3s to 6s. answer in m/s^2
so far i got that the velocity is 5m/s from 3 to 6? and i guess T= 3? I am getting it wrong though the site says when i enter my answer.

2.find the instantaneous acceleration at 4 seconds. answer in m/s^2 uhm like earlier the velocity is 5m/s from 3s to 6 s. and i have no clue how to do that with the Acceleration formula my teacher gave me. any help helps

 

[joeyar]

What information do we have? We're looking for a. We have Vf = 642. Vi = 0. x = 0.916.

What equation has these four variables in it?

Wouldn't it be better to use Vf^2 = Vi^2 + 2ax?

 

[tjh855]

im getting 224980.3493? heh? O_O

 

[morbidwork]

3 knows
vi, vf, and x. solve for a.

Using Newton's equations of motion:
vf^2=vi^2+2ax

a = 224,989.3493 m/s^2