# Aircraft landing (drag -> variable acceleration?)

1. Dec 1, 2009

### meb09JW

1. An F-15 touches down with a velocity of 100m/s. The only decelerating force acting on it is the drag, given by-

How far does the aircraft move before it stops?

2. D = 1/2 x Cd x ρ x A x v^2 (add in your own values for Cd, ρ, and A)

a = (-D/m)

3. Since the D depends on the velocity (decreasing all the time), the acceleration (-D/m)(make up your own m) is variable. Does this mean I can't use the suvat equations?

Any ideas guys?

Cheers,
Josh

2. Dec 1, 2009

### HallsofIvy

Staff Emeritus
Isn't there something missing here?

Yes, what is the drag force? In particular is it proportional to v or v2?

3. Dec 1, 2009

### Hootenanny

Staff Emeritus
Welcome to Physics Forums.
Indeed, SUVAT equations are only valid when the acceleration is constant. What you really have is a first order differential equation for velocity. However, since you're after the stopping distance what you really really have is a second order differential equation in displacement. Both are non-linear equations, but luckily for you they have solutions in terms of elementary functions (which isn't generally the case for non-linear DEs). So, the equation you need to solve is

$$x^{\prime\prime}\left(t\right) = - \alpha \left[x^\prime\left(t\right)\right]^2$$

where I have grouped all your constants together in $\alpha$. What do you suppose our next step is?

Edit: Halls was quick on the draw there!

4. Dec 1, 2009

### meb09JW

So,

a = -(D/m) = -(0.5CdAp/m)v^2

a = -kv^2

d2s/dt2 = -k(ds/dt)^2

Do I need to start integrating here? Somehow move the v^2 over to the left?

5. Dec 1, 2009

### Hootenanny

Staff Emeritus
I'd actually start with finding the velocity, since it is a much simpler ODE to solve. Using your notation, we simply have

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

Do you recognise this type of first order ode?

6. Dec 1, 2009

### meb09JW

So,

dv/dt = -kv^2

(v^-2)dv = (-k)dt

integrating

-(1/v) = -kt + c

v = 1/kt +c

So do I now have the velocity of the plane at any certain point t?
What now?

7. Dec 1, 2009

### Hootenanny

Staff Emeritus
Up to here is good. However, the next line is not so good.

8. Dec 1, 2009

### meb09JW

Hmm,

v = 1/(kt +c)

is that any better, or am i totally off?

Last edited: Dec 1, 2009
9. Dec 1, 2009

### Hootenanny

Staff Emeritus
Looks good to me. Now, as I said earlier what you really want to find is the displacement. So, can you now write this equation in terms of the displacement?

10. Dec 1, 2009

### meb09JW

Right,

v = 1/(kt +c)

ds/dt = 1/(kt +c)

ds = (1/(kt+c))dt

(int)ds = (int)(1/(kt+c))dt

s = ln(kt + c) + x

Not so sure about that integration...
Also, is a second constant needed?

Thanks.

11. Dec 1, 2009

### Hootenanny

Staff Emeritus
You can always check your anti-derivatives by differentiating them to see if they match the original equation. In this case, I would re-check your working. Where did the x come from and what is it?

12. Dec 1, 2009

### SystemTheory

Although this is a good exercise in the solution of the DE, I think it is a trick question. I will explain my logic (as to why the plane mathematically never stops) if it would not interfere with the flow of the thread.

13. Dec 1, 2009

### meb09JW

Ok.

(int)ds = (int)(1/(kt+c))dt

s = (1/k)ln(kt + c)

14. Dec 1, 2009

### Hootenanny

Staff Emeritus
Don't forget the additional constant. You can determine the value of c knowing the plane's initial speed, whilst the additional constant can be determined by setting s=0 at t=0.

15. Dec 1, 2009

### meb09JW

s = (1/k)ln[kt+c] + x

---Finding c

-(1/v) = -kt + c

c = kt - (1/v)

at t=0, v=100, so c = -0.01

----Finding x

s = (1/k)ln[kt+c] + x

x = s - (1/k)ln[kt+c]

at t=0, s=0, so-

x = (1/k)ln[-0.01]

x = 4.61/k

So the overall model for s becomes-

s = (1/k)ln[kt-0.01] + 4.61/k

------------------------------
Is this correct?
If so, how do I go about finding the distance the plane takes to stop? I can't put v=0 into the 'Finding c' part since it is 1/v.

Thank you.

16. Dec 1, 2009

### meb09JW

Yep, I've just realised this too. Of course, in reality, wheel braking force will also slow the plane down, and this isn't (easily) related to the velocity (like drag is).

17. Dec 1, 2009

### SystemTheory

Schaum's Outline: Theoretical Mechanics, Spiegel (1967):

A particle is projected at initial velocity vI. The only force is negative in proportion to velocity squared.

Although the speed of the particle continually decreases it never comes to rest.

In engineering the exponential (linear function of velocity) or quadratic (square function of velocity) are considered "zero" when within about 2% of the asymptotic limiting velocity.

Vertical trajectory with quadratic drag on hyperphysics (time to reach terminal velocity is estimated based on characteristic system time):

http://hyperphysics.phy-astr.gsu.edu/hbase/airfri3.html#c1

Last edited: Dec 1, 2009
18. Dec 1, 2009

### meb09JW

How could I estimate the distance travelled then?

The distance until the velocity drops to 10m/s?

19. Dec 1, 2009

### SystemTheory

Plug in a small velocity, see if the distance makes sense (it will be far), and explain why the plane does not stop. I did not have time to check your math btw.

20. Dec 1, 2009

### Hootenanny

Staff Emeritus
This is almost correct, except for the minus sign. C should be positive. This error will also follow through for the remainder of the problem.

Regarding when the aircraft stops, the answer as you have already found is never. That's it, no approximation, the answer is simply that it never stops.