The barrel of a rifle has a length of 0.886 m. a bullet leaves the muzzle of a rifle with a speed of 623 m/s.
what is the acceleration of the bullet while in the barrel? a bullet in a rifle barrel does not have constant acceleration, but constant acceleration is to be assumed for this problem

First off we need to find what time is equal to. In order to do that we need to derive an equation that will work here. 

Let's start with the base equation for Velocity. 

ΔV = ΔX/ΔT

ΔV, is the change in Velocity... so the Initial Velocity minus the Final Velocity.

ΔX, is the change in Distance... so the Initial Distance minus the Final Distance

ΔT, is the chagne in Time... so the Initial Time Minus the Final Time. 

Let's see what information we can get from the problem before we proceed. 

ΔV = V_f - V_o 
The Initial Velocity is whatever the bullet was before we hit the trigger. We can assume that it was 0 (inside the barrel). 
The Final Velocity is given to us. 623 m/s. 

So! ΔV = 623 - 0 = 623m/s 

ΔX = X_f - X_o 
The Initial Distance is how much distance has been covered already... so 0 in this case. The Final Distance is given to us in the problem, 0.886m. 

So! ΔX = 0.886m - 0 = 0.866m

As for ΔT, that's what we're looking for. So we'll be solving for it!
Now back to making that equation work for us!

ΔV = ΔX/ΔT

Multiply ΔT on both sides.

ΔV*ΔT = ΔX

Get ΔT alone. 

ΔT = ΔX/ΔV

Let's plug in our numbers into that equation to get ΔT. 
Remember to round off to three significant figures! (I do it, but I thought I'd mention that) 
0.886/623 = 1.42*10^-3 seconds

Next we will use a kinematic equation to solve for Acceleration. 

Δx = V_o*Δt+(1/2)a*(Δt)^2

Plug in our information, and get to solving it! (Remember V_o = 0, and anything multiplied by 0 is 0)

0.886m = 0 + (1/2)a*(1.42x10^-3)^2

Seeing as all the numbers are multiplying one term (a) we can divide them all out of the right side and of course divide them on the left side as well. 

0.886/((1/2)(1.42x10^-3)^2 = 878,000m/s^2

Yup no bull frogs here man