Instantaneous velocity when non-constant acceleration

In summary, In order to calculate the aerodynamic drag (velocity dependent negative acceleration) for a spherical body, you need to solve a differential equation.
  • #1
I've seen on Wikipedia ( 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?
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  • #2
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.
  • #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.
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  • #4
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.
  • #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.
  • #6
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
$$-\alpha t=\frac{1}{v_0}-\frac{1}{v} \; \Rightarrow \; v(t)=\frac{1}{1/v_0+\alpha t}=\frac{v_0}{1+v_0 \alpha t}.$$
  • #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.
  • #8
I don't understand what you mean; ##\alpha## is simply a constant, and I gave the full solution for the velocity.
  • #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.

1. What is instantaneous velocity when non-constant acceleration?

Instantaneous velocity when non-constant acceleration refers to the rate of change of an object's displacement at a specific moment in time, when the acceleration of the object is not constant. It takes into account both the magnitude and direction of the object's velocity at that particular instant.

2. How is instantaneous velocity when non-constant acceleration calculated?

To calculate instantaneous velocity when non-constant acceleration, you can use the formula v= u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time interval. Alternatively, you can also use the formula v = ds/dt, where s is the displacement and t is the time interval.

3. How is instantaneous velocity different from average velocity?

Instantaneous velocity is the velocity of an object at a specific moment in time, whereas average velocity is the total displacement of an object divided by the total time taken. Instantaneous velocity can vary greatly throughout the motion of an object, while average velocity gives an overall picture of the object's motion.

4. What factors can affect the instantaneous velocity of an object?

The instantaneous velocity of an object can be affected by several factors, including the object's initial velocity, acceleration, and any other external forces acting on it. Gravity, friction, and air resistance are also common factors that can affect the instantaneous velocity of an object.

5. Why is it important to consider instantaneous velocity when studying non-constant acceleration?

Instantaneous velocity is crucial for understanding the motion of an object with non-constant acceleration. It gives us a more detailed and accurate picture of the object's motion at a specific moment in time, rather than just the average motion over a period of time. This information can be useful in analyzing and predicting the behavior of moving objects in real-world scenarios.

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