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Previously, I was helped in solving a projectile motion equation to model the velocity of the projectile with respect to distance with drag taken into account by using differential equations (which I am pretty new to). This is the drag equation I used:

$F_d = \frac{1}{2} \rho v^2C_DA$

As a note, on the right side of the equation, all of the variables are constants except for v. After all my work (which I dont find relevant to the problem, but if needed I can include), I get this as a final answer:

$$v(x) = v(0)\exp{\left(-\frac{1}{2m} pC_DA x\right)}.$$

The equation here models the velocity with respect to distance (or x), but a genius kid at my school challenged me to instead solve the equation so it models velocity with respect to time instead of distance. I am not in AP physics to begin with and just solving for what I already have was challenging for me. I have no idea where to start on this problem and was wondering if anyone has a solution or guidance on how to do this. Let me know if any more information needs to be included.

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closed as off-topic by tpg2114, AccidentalFourierTransform, CuriousOne, ACuriousMind, David Z Apr 12 '16 at 10:25

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  • $\begingroup$ The drag equation can be written as $v'(t)=\alpha v(t)^2$ for some constant $\alpha$ (which you have to work out: it depends on $p,C_D,$etc.). The equation $v'(t)=\alpha v(t)^2$ is a first order differential equation which you have to solve to get the function $v(t)$. There are general methods on how to solve 1st order ODE's, but for this one in particular it may help to use the change of variables $v(t)=1/u(t)$. $\endgroup$ – AccidentalFourierTransform Apr 11 '16 at 19:18
  • $\begingroup$ What is the geometry of your problem? $\endgroup$ – L. Levrel Apr 11 '16 at 20:02
  • $\begingroup$ The idealized model for drag force is meant to model force for 1 dimensional flow, that is the projected area is always pointed in the direction of $v$. But you need to consider that $v= \sqrt{v_x^2 +v_y^2}$ so you get coupling of the drag force between the two axes. Drag force affects forward as well as rising and falling motion. Projectile motion is a 2 dimensional problem when considering flow. $\endgroup$ – docscience Apr 11 '16 at 20:04
  • $\begingroup$ @AFT: simpler than changing variables, separate variables: $\frac{dv}{dt}=αv^2\ →\ \frac{dv}{v^2}=d(-\frac 1v)=α\,dt$ $\endgroup$ – L. Levrel Apr 11 '16 at 20:05
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    $\begingroup$ I don't think there is an analytical solution to the projectile with air resistance, for the differential equation see yukterez.net/ballistik/#plot $\endgroup$ – Yukterez Apr 12 '16 at 0:53
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You haven't really tackled projectile motion with drag, because that is a 2D problem i.e. a projectile like a cannonball moves in a curve. In the absence of drag this curve is a parabola but when you include drag the equations of motion turn out to have no analytic solution (except for the special case of purely vertical motion).

What you've done is to consider the motion of a particle moving in a straight line and subject to no force except drag (so no gravity). In that case the equation of motion is:

$$ \frac{dv}{dt} = -k v^2 \tag{1} $$

This is just Newton's second law rewritten as $a = F/m$. The constant $k$ is in this case $k = \frac{1}{2} \rho C_DA/m$ but let's keep it as $k$ to avoid clutter.

To get the result you quote we use the chain rule:

$$ \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx} $$

And equation (1) becomes:

$$ \frac{dv}{dx} = -kv $$

which is just the equation for exponential decay, hence your result. To solve equation (1) directly we rewrite it as:

$$ \frac{dv}{v^2} = -k dt $$

and then integrate both sides to get:

$$ \frac{-1}{v} = -kt + C $$

And you then just need to work out the constant of integration $C$ and rearrange as an equation for $v(t)$.

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  • $\begingroup$ Is there any real application of this equation? I.e. linear motion with no gravity just drag? Another point is that unless the object is moving in a turbulent regime, the drag coefficient is velocity-dependent. $\endgroup$ – nluigi Apr 12 '16 at 6:51
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Well let us start with our four states of motion. Force,Acceleration,Velocity,Position Since drag is measured by force, it would only make since to add drag to the force component of our four states. Depending how you measure your Force or "thrust" is how we will measure our drag. Since Drag is the opponent to force, we will write it like so... (Thrust-Drag)

Time=Time+1

Mass=20
Thrust=30 'kg
Drag=*0.101*(0.5*Cd*Density*Velocity^2*Area) 'kg
Velocity=((Thrust-Drag)/Mass)Time
Position=Velocity*Time

Taking Drag from Velocity

Velocity-(*0.101*(0.5*Cd*Velocity^2*Area)/Mass*Time)
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