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Instantaneous velocity when non-constant acceleration

  1. Jan 2, 2015 #1
    Hello.
    I've seen on Wikipedia (http://en.wikipedia.org/wiki/Airsoft_pellets#pellet_ballistics) a formula to find the aerodynamic drag (velocity dependent negative acceleration) for a spherical body. Knowing body's mass, diameter and air density I simplified it to the form:
    A = vt2 * 0.04
    where A is the drag, t is a point in time, vt is the instantaneous velocity at t.
    The problem comes now: how can I get vt at arbitrary times or distances? A changes in instants of time so I can't do a simple
    vt = v0 + A * t
    Say the ball has an initial v0 = 120m/s; what's the velocity at t = 1s?
     
  2. jcsd
  3. Jan 2, 2015 #2

    ShayanJ

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    You should write Newton's second law: [itex] \vec F_{net}=m \vec a [/itex] where [itex] \vec F_{net} [/itex] is the vector sum of all forces on the mass m. Drag is just one term in this sum appearing as [itex] -0.04 m |v_t| \vec{v}_t [/itex]. I suppose in the situation of your interest, the only other force is gravity. So you have [itex] \vec F_{net}=-0.04 m |v_t| \vec{v}_t-mg\hat z=m \vec a [/itex]. This is a differential equation which you should solve to get the trajectory of the ball.
     
  4. Jan 3, 2015 #3
    Thanks for replying.
    I need to find both At (acceleration) and vt which, problematically, are interdependent: A depends on v as per given formula (At = vt2 * 0.04), v depends on A by Newton's laws. We ignore gravity. I don't know why you brought force into discussion.
    I tried something on the paper but it yielded rubbish:
    At v0 = 120m/s it corresponds A0 = 574m/s2 (deceleration)
    At a constant A it follows that after 0.1s it's supposed to have 62.6m/s. So now:
    at v0.1 = 62.6m/s A0.1 = 156m/s2.
    But it's wrong as A continuously decreases as a function of v. During 0.1s the drag (A) has been dropping due to continuous deceleration, so v0.1 is bigger than 62.6m/s.
     
    Last edited: Jan 3, 2015
  5. Jan 3, 2015 #4

    HallsofIvy

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    That's why you need Calculus. Instantaneous acceleration is the derivative of velocity. You need to solve the "differential equation" [itex]\frac{dv}{dt}= 0.04 v[/itex] which can be separated as [itex]\frac{1}{v} dv= 0.04 dt[/itex]. Integrate both sides.
     
  6. Jan 3, 2015 #5
    Thank you very much. It looks like what I'm asking for. Surprisingly as it may seem -- all the calculus I know is raw formulae for derivatives like (3x4)' = 12x3, sin(x)' = cos(x) etc. So this drastically outweighs my mathematical knowledge.
    If I'm not asking too much, could you please, write an example so I can learn seeing the procedure?
    You say some v0 and a t, and obtain At and vt. Please.
     
  7. Jan 3, 2015 #6

    vanhees71

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    Let's solve the equation for the case of motion in one direction (it's more complicated for the general case). Then you have the differential equation
    $$\dot{v}=-\alpha v^2,$$
    with ##\alpha## a constant. This you can solve by "separation of variables":
    $$\int_{v_0}^{v} \mathrm{d} v \frac{1}{v^2}=-\alpha \int_{0}^t \mathrm{d} t'=-\alpha t.$$
    Integrating the left-hand side gives
    $$-\alpha t=\frac{1}{v_0}-\frac{1}{v} \; \Rightarrow \; v(t)=\frac{1}{1/v_0+\alpha t}=\frac{v_0}{1+v_0 \alpha t}.$$
     
  8. Jan 5, 2015 #7
    Thank you very much for your time. Sorry for the late reply. Reading your post and "constant of integration" on Wikipedia raised some intriguing facts:
    - A can be decomposed into terms of v and t making the formula simpler.
    - v(t) can't be really determined as α isn't determined in the first place. But the problem, as it looks, must have an unique solution. It may be not determined mathematically? Maybe through practical measurements? And once I determine α from a measured v(t) I can calculate any v(t). But I can't measure, literally. And I can't think of any trivial cases to get some v(t) in order to determine α.
    I tried using your formula with α = 1 as a neutral factor but for v(0) = 120 it gives v(0.1) = 9.23 which is not "it" for sure.
     
  9. Jan 5, 2015 #8

    vanhees71

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    I don't understand what you mean; ##\alpha## is simply a constant, and I gave the full solution for the velocity.
     
  10. Jan 5, 2015 #9
    I'm a total rookie at calculus and this holds me back from getting to the bottom of many problems like this :( that's why I probably misunderstood your solution.
    But to me it looks like this: α is an unknown constant -- constant yes, but yet unknown, undetermined; as such the recipe misses one ingredient. How do I determine α? Usually, I know that when there is an extra unknown, an extra equation is needed and sufficient. But I don't know how could I apply this to get α.
    Am I missing something? I tried calculating v(0.1) and I couldn't.

    I'm mentioning the problem is not for "academical" purposes but empirical. I'm into airsoft and I want to be able to calculate different bb velocities for different guns.
     
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