Instantaneous velocity when non-constant acceleration

  • Thread starter mireazma
  • Start date
  • #1
16
0
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?
 

Answers and Replies

  • #2
2,792
594
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
16
0
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:
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
961
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
16
0
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
vanhees71
Science Advisor
Insights Author
Gold Member
16,990
8,141
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
16
0
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
vanhees71
Science Advisor
Insights Author
Gold Member
16,990
8,141
I don't understand what you mean; ##\alpha## is simply a constant, and I gave the full solution for the velocity.
 
  • #9
16
0
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.
 

Related Threads on Instantaneous velocity when non-constant acceleration

Replies
16
Views
7K
Replies
9
Views
3K
  • Last Post
Replies
8
Views
7K
  • Last Post
Replies
2
Views
5K
Replies
22
Views
6K
Replies
14
Views
3K
Replies
5
Views
743
  • Last Post
Replies
5
Views
3K
Top