How to Combine SUVAT and Calculus for Falling Objects with Air Resistance?

[fereak]

Hey,
Not sure if this is in the right place, move if necessary.

So, I know about SUVATs, and non-constant acceleration, using Calculus.
I also (basically) understand how to calculate the force of drag on an object.
How could I combine the 2, calculate the final velocity/acceleration of a falling object, taking into account air resistance?

Thanks, my knowledge is limited to AS Level Pure Maths and GCSE Physics, for now. :)

 

[tiny-tim]

Welcome to PF!

Hey fereak ! Welcome to PF!

How could I combine the 2, calculate the final velocity/acceleration of a falling object, taking into account air resistance?

You'd get an equation something like dv/dt = -kv + c, and just integrate it.

Or do you have something more complicated in mind?

 

[fereak]

Hey fereak ! Welcome to PF!


You'd get an equation something like dv/dt = -kv + c, and just integrate it.

Or do you have something more complicated in mind?

Where did you get that from? What is k? and c?

 

[tiny-tim]

Where did you get that from? What is k? and c?

(k and c are constants)

No … you first … what did you get?

 

[fereak]

Well I know the SUVATs:

and the drag equation:

But, for the drag equation, the velocity must be given. We can calculate that using SUVATs/Calculus but it would surely be affected by the drag, but to work that out I need velocity! And once I know the drag force, how can I apply it to the velocity to see how its affected?
Or am I overcomplicating? Could I just substitute them together?

 

[tiny-tim]

ah … that drag equation has force proportional to v², so the differential equation would be of the form dv/dt = -kv² + c …

again, just integrate

 

[fereak]

Sorry, but integrate what?

 

[tiny-tim]

Sorry, but integrate what?

Integrate dv/(-kv² + c) = dt

 

[fereak]

Integrate dv/(-kv² + c) = dt

Wha? Why??
I'm deeply confused.

 

[HallsofIvy]

What is confusing? Do you not know how to integrate? If you know how to integrate, then integrate:
\int \frac{dv}{-kv^2+ c}= \int dt

 

[fereak]

Yeah, of course I can integrate, by why dv/(-kv2 + c)? Where is that from?

 

[Mentallic]

a=\frac{dv}{dt}=-kv^2+c

but you can't differentiate -kv²+c with respect to time because there is no time in the equation.

So what you can instead do is use the rule \frac{1}{\frac{dv}{dt}}=\frac{dt}{dv} and now we have \frac{dt}{dv}=\frac{1}{-kv^2+c}

and now we can integrate with respect to velocity.

 

[fereak]

OK! But where has -kv2+c come from? Seriously, wtf? How was that expression calculated? How is it relevant?

 

[Mentallic]

Ahh so that's what you didn't understand. Well, at least you made it more clear this time

the a=-kv²+c is basically saying that as your velocity increases in any direction, the drag force of the air will increase at a quadratic rate (as expressed by the v²). Think of it logically for a second. Air resistance is fairly negligible at low speeds, i.e. you barely feel still air when you're walking through it, but as your speed increases to a run, it quickly becomes noticeable. When you pop your head out of the car it's very distinct, and even painful. When a space shuttle enters the atmosphere, the friction is profound enough to cause massive heating and creates very large G-forces on the shuttle and astronauts.
Ok, so as you can see, doubling your velocity doesn't necessarily mean twice as much drag force, it seems to be more than that. Now, for some reason (if anyone has deeper knowledge on the subject, please let us know) some had come to the conclusion - probably through experimental data - that the relationship between velocity and drag is a quadratic one.

a=-kv²+c
k>0, some number with a value probably depending on the atmospheric pressure (correct me if I'm wrong, this is a wild guess)
c=gravity, the only force acting when v=0 is gravity, i.e. the drag is negligible.

I hope this makes more sense.

 

[JanClaesen]

Ahh so that's what you didn't understand. Well, at least you made it more clear this time

the a=-kv²+c is basically saying that as your velocity increases in any direction, the drag force of the air will increase at a quadratic rate (as expressed by the v²). Think of it logically for a second. Air resistance is fairly negligible at low speeds, i.e. you barely feel still air when you're walking through it, but as your speed increases to a run, it quickly becomes noticeable. When you pop your head out of the car it's very distinct, and even painful. When a space shuttle enters the atmosphere, the friction is profound enough to cause massive heating and creates very large G-forces on the shuttle and astronauts.
Ok, so as you can see, doubling your velocity doesn't necessarily mean twice as much drag force, it seems to be more than that. Now, for some reason (if anyone has deeper knowledge on the subject, please let us know) some had come to the conclusion - probably through experimental data - that the relationship between velocity and drag is a quadratic one.

a=-kv²+c
k>0, some number with a value probably depending on the atmospheric pressure (correct me if I'm wrong, this is a wild guess)
c=gravity, the only force acting when v=0 is gravity, i.e. the drag is negligible.

I hope this makes more sense.

If c = g, does this mean that this formula is used for calculating the drag on for example rockets?
Since drag force and gravity have the same direction (in this case), is c < 0?
By the way, when do you use this formula, since I always learned that the drag force is -λ*v(t)

 

[tiny-tim]

reality? don't talk to me about reality …

By the way, when do you use this formula, since I always learned that the drag force is -λ*v(t)

Hi JanClaesen!

In my experience, exam questions are nearly always about drag proportional to v, not v² …

whether that represents reality, I've no idea!

 

[Mentallic]

If c = g, does this mean that this formula is used for calculating the drag on for example rockets?
Since drag force and gravity have the same direction (in this case), is c < 0?
By the way, when do you use this formula, since I always learned that the drag force is -λ*v(t)

Of course it would be -λ*v(t) not -kv² because I highly doubt the rate of drag is proportional to v² precisely, I mean, this is reality we are talking about, do you really think it would be that simple?

Also, think about terminal velocity (the highest speed achieved by an object in free fall, due to force of gravity and drag equalizing). If a dense spherical mass is left to free-fall, at the moment it is dropped, the force acting is only g. As its velocity increases, the drag force becomes more significant and opposes g. The drag force would never be larger than the gravity force so basically it is a limiting value of acceleration being 0⁺ (downwards being positive) as time in free-fall gets large.

If c = g, does this mean that this formula is used for calculating the drag on for example rockets?
Since drag force and gravity have the same direction (in this case), is c < 0?
By the way, when do you use this formula, since I always learned that the drag force is -λ*v(t)

When a ball drops down, the gravity is in the direction of the fall, so g>0 and the drag force is in the upward direction (it is always in the opposite direction to the moving object) so drag<0. For a rocket, gravity is in the opposing direction, so g<0, but again drag<0 so basically these forces are now additive rather than opposing.


Just if you're curious, to try and find an acceleration equation that is more closely related to reality, or like the a=-λ*v(t), I would try something like
a=g.e^{-kv^2}
g=gravity, k some constant >0