# Aircraft landing (drag -> variable acceleration?)

• meb09JW
In summary, an F-15 touches down with a velocity of 100m/s. The only decelerating force acting on it is the drag, given by--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)
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

meb09JW said:
1. An F-15 touches down with a velocity of 100m/s. The only decelerating force acting on it is the drag, given by-
Isn't there something missing here?

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

Welcome to Physics Forums.
meb09JW said:
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
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!

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?

meb09JW said:
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?
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?

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?

meb09JW said:
So,

dv/dt = -kv^2

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

integrating

-(1/v) = -kt + c
Up to here is good. However, the next line is not so good.
meb09JW said:
v = 1/kt +c

Hmm,

v = 1/(kt +c)

is that any better, or am i totally off?

Last edited:
meb09JW said:
Hmm,

v = 1/(kt +c)

is that any better, or am i totally off?
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?

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.

meb09JW said:
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.
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?

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.

Ok.

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

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

meb09JW said:
Ok.

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

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

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 xs = (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.

SystemTheory said:
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.

Yep, I've just realized 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).

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:
How could I estimate the distance traveled then?

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

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.

meb09JW said:
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
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.

Thanks you all so much.

## 1. What is drag and how does it affect aircraft landing?

Drag is the force that opposes an aircraft's motion through the air. During landing, drag plays a crucial role in slowing down the aircraft and reducing its speed to a safe level for touchdown. It is caused by air resistance and is affected by various factors such as the shape and size of the aircraft, wind conditions, and air density.

## 2. How does variable acceleration impact an aircraft's landing?

Variable acceleration refers to the change in an aircraft's speed during the landing process. This can be due to various factors such as wind gusts, engine power changes, and pilot inputs. Variable acceleration can affect the aircraft's stability and control, and pilots must make adjustments to compensate for it during landing.

## 3. How do pilots account for drag during landing?

Pilots account for drag during landing by using various techniques such as adjusting the flaps and landing gear, using spoilers to increase drag, and making adjustments to the aircraft's angle of attack. They also use their knowledge and experience to anticipate and compensate for drag factors such as wind conditions and air density.

## 4. How does the weight of an aircraft affect its drag and landing?

The weight of an aircraft can significantly impact its drag and landing. A heavier aircraft will experience more drag due to its larger mass, which requires more force to slow down. This can affect the aircraft's speed and landing distance, and pilots must take this into account when planning and executing a landing.

## 5. Are there any safety concerns related to drag during aircraft landing?

There are potential safety concerns related to drag during aircraft landing, as it can affect the aircraft's stability and control. Strong gusts of wind or unexpected changes in air density can increase drag and make it challenging for pilots to maintain control of the aircraft. It is crucial for pilots to stay alert and make necessary adjustments to compensate for any drag factors during landing to ensure a safe and smooth landing.

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