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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.)

Community
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R2800
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3 Answers3

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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 $$ \left<v\left(t\right)\right>_{12} = \frac{1}{t_2-t_1} \int_{t_1}^{t_2} dt \ v\left(t\right) $$

Eric Angle
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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)

Gremlin
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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

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

John Alexiou
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