Average Bullet Velocity Inside the barrel

[waterwalker10]

Homework Statement

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)

Homework Equations

V=D/T

\overline{V} = \Delta Distance/Time

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

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.

 

[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 don't 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...

 

[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} = \Deltad / \Deltat

\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...

 

[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

 

[Anti-Meson]

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} = \Deltad / \Deltat

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

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

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

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.

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).

 

[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.

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

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!

 

[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?

 

[willem2]

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

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.

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

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

 

[waterwalker10]

So is this

3.12 * 10^{4}ms^{-2}

the correct way to write acceleration?

 

[willem2]

So is this the correct way to write acceleration?

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

 

[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!