Changing acceleration

  • Thread starter Dalau
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Main Question or Discussion Point

I am designing an air pressure cannon, that relies on a tank of compressed air to drive a projectile through, and eventually out of, a tube. The initial acceleration (when the projectile is at the start of the tube) will be greater than the final acceleration (when the projectile has reached the outlet end of the tube). How can I calculate the velocity of the projectile, when the acceleration is changing? This is a problem where the acceleration changes, based on distance covered by the projctile, not based on change in time. Thanks!
-Dan
 

Answers and Replies

ZapperZ
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Use the chain rule.

Since

[tex]a = \frac{dv}{dt}[/tex]

You can apply the chan rule to write this as

[tex]a = \frac{dv}{dx} \frac{dx}{dt}[/tex]

Since dx/dt = v, this means that

[tex]a = v \frac{dv}{dx}[/tex]

If you integrate both sides with respect to x, you then get

[tex] \int{a dx} = \int{v dv}[/tex]

You should be able to find the change in v here with respect to the displacement.

Zz.
 
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I tried that method, but I don't see how it works. The acceleration is changing, but when I integrate the last equation, I get ax = (v^2)/2. Did I do something wrong?
 
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you might be able to use conservation of energy.
 
ZapperZ
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I tried that method, but I don't see how it works. The acceleration is changing, but when I integrate the last equation, I get ax = (v^2)/2. Did I do something wrong?
Did you forget that your integration has a constant, or has limits? I didn't think I had to show that since this depends on the problem that you have. If I did that, I would have done the whole thing for you. All I had left for you was the last step in figuring out the limits to your problem. If you have an acceleration that is a function of position a(x), the way you had described it, then write down that expression in the left hand side, and do the integration!

You DID say that you had an acceleration that is a function of x, didn't you?

Zz.
 
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you might be able to use conservation of energy.
Yeah. I wikied some stuff about air pressure, and found that:
1 pascal (Pa) ≡ 1 N·m−2 ≡ 1 J·m−3 ≡ 1 kg·m−1·s−2

so does this mean that it takes 1 joule to increase the pressure of one cubic meter by one pascal?




For an example problem I set up, I'm just wondering if my math is correct:
Tank of air with a tube sticking out of the tank. When the projectile is at the start of the tube, the pressure in the tank is Pinit, and when the projectile is at the outlet end of the tube, the pressure in the tank + tube is now Pfinal.
Is this part correct?:
(volume of tank)*(change in pascals) = (change in joules)
 
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Did you forget that your integration has a constant, or has limits? I didn't think I had to show that since this depends on the problem that you have. If I did that, I would have done the whole thing for you. All I had left for you was the last step in figuring out the limits to your problem. If you have an acceleration that is a function of position a(x), the way you had described it, then write down that expression in the left hand side, and do the integration!

You DID say that you had an acceleration that is a function of x, didn't you?

Zz.
oh. d'oh. I forgot about limits. I'll try it again
 

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