# Average Bullet Velocity Inside the barrel

1. Feb 12, 2010

### waterwalker10

1. The problem statement, all variables and given/known data

The M61 Vulcan cannon is a Gatling Gun-type weapon deployed on an aircraft. It fires a 100g, 20mm diameter projectile at a rate of up to 6,000 shots per minute. Barrel length is 170cm, and muzzle velocity (speed of the projectile leaving the barrel) is 1030 m/s.

-Assuming constant acceleration, what is the mean (average) velocity of the projectile in the barrel?

(It has been 8 years since a physics class...not looking for an answer but an arrow in the correct direction. Thank you)

2. Relevant equations
V=D/T

$$\overline{V}$$ = $$\Delta$$ Distance/Time

$$V^{2}$$=$$V^{2}_{0}$$ + 2as

3. The attempt at a solution

Where do I start...I began with the second equation. V = 170cm / T Well if it shoots 6000rpm then it would be broken down to 100rps. Given that I would say that fires a bullet every .01 seconds. Plug the variables into the equation.

V = 170 cm / .01 sec or 17000cm or 170m/s

If you see the muzzle velocity of 1030m/s then you would know right off the bat that 170m/s is wrong.

So I used the second equation...and came up with some crazy answer that I know is definitely wrong. See below...

10302 = $$1030^{2}_{0}$$ + 2a170

1060900 = 0 (Initial Bullet Velocity is zero) + 2a170 (Solve for A)

a = V2/2*170cm

a = 1060900/340cm or 3120.29cm or 312,029 meters

Plug back into the equation

1060900 = 0 + 2*312,029*1.7

And they magically equal. Some how I think I might be using the wrong equation but I need someone to help me.

2. Feb 13, 2010

### Anti-Meson

To achieve this answer you need to know how much time the bullet spends in the barrel.

$$V^{2} = V^{2}_{0} + 2as$$ [1]
$$V = V_{0} + at$$ [2]

Divide 1 by 2 to eliminate a, as we dont know it. V0 is 0

$$V = \frac{2s}{t}$$
$$t = \frac{2s}{V}$$

which is t = 2 * 1.7 m/ 1030 m/s = 0.0033 seconds.

See where you can go from there...

Last edited: Feb 13, 2010
3. Feb 13, 2010

### waterwalker10

So it would be fair to say that the bullet travels the length of the barrel in .0033 seconds.

Solve for Avg Velocity...

$$\bar{V}$$ = $$\Delta$$d / $$\Delta$$t

$$\bar{V}$$ = 170cm/.0033 secs = 51515.15cm/s or 515.15m/s

Would you have to change the time to correlate with cm and not meters?

$$\bar{V}$$ = 170cm/.000033 secs = 5151515.15cm/s or 51515m/s

Wouldn't the bullet be more close to it's terminal velocity? Or because this is the "average" velocity, you wouldn't have such a large number...???

I'm trying to learn via a few MIT/Stanford University lec. videos...

4. Feb 13, 2010

### raymo39

the average velocity should be the middle point, of your velocities, because acceleration is constant, therefore it accelerates from 0 to 1030 in the length of the barrel.
(1030+0)/2 = 515m/s is your average velocity.
i give the answer because you have already stumbled upon it, but hopefully my view of things allowed your mind to take an easier path to the answer :)
hit back with any more questions and i can go into further detail

5. Feb 13, 2010

### Anti-Meson

Not quite sure what you mean here. 51500 cm/s is exactly the same as 515 m/s

Get this idea out of your head quickly. This is confusing and illogical.

This is correct but I prefer my reasoning as the mathematical derivation from the equation provides a mathematical basis, in addition to physical intuition (which can sometimes be flawed).

6. Feb 13, 2010

### waterwalker10

Well good to know I was on the right track. I understand mean as average and only average. If I wanted the middle I would take the length of the barrel and divide it by half to find the velocity at that point.

When I was in my algebra class I was taught to simplify. I guess it isn't wrong but is not fully relative to the muzzle velocity which is in m/s.

Oh and totally forgotten the seconds issue. The answer was in seconds not m/s or cm/s...

THANK YOU!! I'm starting to understand now!

7. Feb 13, 2010

### waterwalker10

Then to find the acceleration of the bullet inside the barrel I would use the following formula.
a = Δv/Δt= (vfinal - vinitial) / (tfinal - tinitial)

Initial thought was to use 1030m/s which is the muzzle velocity given above. And .0033 is the time it takes the bullet to travel the length of the barrel.

a = (1030-0) / (.0033-0) = 312121 m/s

For some reason I believe that this one is correct below. $$\downarrow$$

For Vfinal I calculated 170cm/.0033 = 51515cm

a = (515.15-0)/(.0033-0) = 156106 m/s

Correct?

8. Feb 13, 2010

### willem2

This one is indeed correct. The unit of acceleration is $ms^{-2}$
You already nearly computed this correctly from $v_f^2 = v_0^2 + 2as$
in your first post (except for the units)

The difference in the numbers is round-off error. If you round off the numbers to 3
significant figures, you'll get 3.12 * $10^4$ $ms^{-2}$ from both calculations.

Here you use the average speed instead of the final speed.

9. Feb 13, 2010

### waterwalker10

So is this
the correct way to write acceleration?

10. Feb 13, 2010

### willem2

Yes. Oops, it's $3.12 * 10^5$ $ms^{-2}$

11. Feb 13, 2010

### waterwalker10

Excellent. I've been watching a few more lecture videos so they seem to be helping too. Thank you! Mark this one as solved!