[Physics] Calculate average speed of projectile with air resistance

homework-and-exerciseskinematicsnewtonian-mechanicsprojectile

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.)

Best Answer

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