There's a project that I'm working on that been bugging me for awhile.

The project involves this senario:

A projectile is released at 1000 feet per second and moves in a straight line infinitely with only wind resistance acting on it. Of course air resistance is going to start slowing down the projectile. I can calculate the force of wind resistance on the object since I'm given a value for drag coefficient.

I want to find the average speed over a certain time interval from the point the object is released.

I can calculate the acceleration (deceleration) for an infinitely small moment in time just as the projectile is released. a=f/m. As the object slows, the force decreases by the square of the velocity, and the rate of acceleration will decrease along with it.

I can't really figure out how I can model this with time as the x axis and acceleration (or force) as the y axis.

If I could do this, I would be able to get an average acceleration over a time interval by breaking up the x axis into small intervals and summing the y values of those intervals together, then dividing by the number of divisions. I think in calculus, there's a name for this, but i haven't gotten that far in my book yet---- :-)

One suspicion that I have is that the graph will resemble a=1/sqrt(x)

I can already calculate the path of a projectile with no air resistance with theta as the x axis, and range as the y axis, but this requires that you have an average speed, (which I could find given a constant acceleration.)

  • $\begingroup$ What you are looking for is called a differential equation. In this case it would be something as simple as y"(t)=-cy'(t)^2. One can solve some of the simple ones without calculus, but it's much easier if you had the tools that calculus gives you. If you didn't have calculus, yet, mathematics software can tell you the solutions. wolframalpha.com/input/?i=y%27%27%28x%29%3D-cy%27%28x%29*y%27%28x%29%2C+y%280%29%3D1000 is an example of how this works. Having said that, unless you learn calculus, the software is of limited use. $\endgroup$
    – CuriousOne
    Oct 3, 2014 at 10:45
  • $\begingroup$ Use physics.stackexchange.com/a/15620/392 to find speed when acceleration is a function of speed. Then a time integral would yield the average speed. $\endgroup$ Nov 4, 2014 at 16:04

3 Answers 3


You said this is a one dimensional problem, and the drag force is proportional to the square of the velocity $v$, so $$ F = - b v^2 = m \frac{dv}{dt} \ \ \ \Rightarrow \ \ \ \int_{v\left(0\right)}^{v\left(t\right)}\frac{dv}{v^2} = - \frac{b}{m} \int_0^t dt' $$ The above integral (I'll let you do it, or see this) gives $v\left(t\right)$. The average velocity (or speed, since $v\left(t\right)>0$) from $t=t_1$ to $t=t_2$ is then $$ \left<v\left(t\right)\right>_{12} = \frac{1}{t_2-t_1} \int_{t_1}^{t_2} dt \ v\left(t\right) $$


You need to use a differential equation. We can write Newton's second law as

$$ F = m a = m \frac{dv}{dt} = m \frac{d^2x}{dt^2} $$

Your wind resitance will look like $$ F = - c v = - c \frac{dx}{dt} $$

so you can write

$$ m \frac{d^2x}{dt^2} = - c \frac{dx}{dt} $$

And it's possible to solve for x.

Once you have x, the average speed is just x(t)/t

(Hint, x(t) won't look like a sqrt, more like exponential)


If the acceleration at any time is $a(v) = -\beta v^2$ then the time to go from the initial velocity $v_1$ to some other speed $v$ is

$$ \left. t = \int \limits_{v_1}^v \frac{1}{-\beta v^2}\,{\rm d}v = \frac{1}{\beta} \left(\frac{1}{v} - \frac{1}{v_1} \right) \right\} v(t) = \frac{v_1}{1+\beta v_1 t} $$

Now the average is

$$ \left<v\right> = \frac{1 }{T} \int \limits_0^T v(t)\,{\rm d}t = \frac{ \ln \left(1+\beta v_1 T\right)}{\beta T} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.