1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differential Equation of a moving car with drag

  1. Oct 18, 2011 #1
    1. The problem statement, all variables and given/known data

    So the question I've been asked is, that there is a moving car, at time t = 0, moving with a speed of 50 km/h. At t = 0, the car stops accelerating, and thereby slowing down due to two forces:

    The air resistance drag force

    A friction which is created by some magnets in the motor (description of this in section 2).

    Write a differential equation for the system, as a function of the cars decceleration.


    2. Relevant equations

    The air resistance is given by: Fd = -(1/2)pCdAv2. Where p, Cd and A are constants

    The friction in this motor is given by: Fe = -k*v. Where k is a konstant. This is basicly just a description of a spring using Hooke's law.

    3. The attempt at a solution

    Writing a differential equation for this systems seems fairly easy. All the forces that acts on the car is equal to Newton's 2nd law F = ma. The constants in the air resistance is given the letter C = -(1/2)pCdA.

    ma = C*v2-k*v

    Writing this as a differential equation gives:

    m (dv/dt) = C v2 - k v

    However solving it seems much more difficult. This is what I've come up with so far, but I simply just can't get on:

    k v + m (dv/dt) = C v2 (+ k v on both sides)

    (k v)/(c v2)+(m/(c v2)) dv = dt (divided by c v2 on both sides, and multiplied by dt).

    (k/c) (1/v) + (m/c) (1/v2) dv = dt (Rewriting the fractions).

    I the integrate both sides, where k/c and m/c are both constants, so I only focus on 1/v and 1/v2, which gives:

    (k/c) ln|v| + (m/c) ln|v2| = t + K (Where k is the constant.)

    Then I move both of the constants to they other side.

    ln|v| + ln|v2| = (t + k) (c/k) (c/m)

    then I take the exponential of both sides to get v.

    v + v2 = e(t+c)*(c2/(k m))

    However this just seems completely wrong, because solving the remaining eqution for v just wont make sense. There must be an easier way to do this, but I can't figure it out.
     
  2. jcsd
  3. Oct 18, 2011 #2
    Try separating the variables. For instance:

    dy/dx = y

    This implies Integral of dy/y = Integral of dx
     
  4. Oct 18, 2011 #3
    EDIT: If "Write a differential equation for the system, as a function of the cars decceleration." that's the question, then you don't even have to solve the differential equation... anyhow:

    From what I can see you're having problems solving the actual differential equation for a function of velocity.

    "(k v)/(c v2)+(m/(c v2)) dv = dt (divided by c v2 on both sides, and multiplied by dt)."

    In this step you correctly divided by c.v^2, but when you multiplied the equation by dt you skipped the (k.v)/(c.v^2) expression, you also integrated (1/v^2) incorrectly, it's (-1/v) not (ln|v^2|).

    That's the wrong I see in your solution... although I would recommend solving the differential equation with a substitution (u = y^-1) or (u = 1/y). This form of differential equation is actually called a Bernoulli ODE. You can find more information by googling "Bernoulli ODE". However I'll try to give a quick run-over of how to solve it.

    You have:

    rewrite 1 as:

    then define z and (dz/dt):

    then substitute (5) into (2)

    Now if you divide (6) by ((v^2)/-1) you'll have

    Remember that (3): (z = v^-1) ... substitute (3) into (7), and you have a linear ODE you can solve normally!


    To solve this, you must define the integrating factor (the shorthand method, for linear equations is as follows):

    then the solution (I -> I(t))

    in this case (1/I) is just

    therfore

    Solve it (a simple integration), then find v(t)..

    Then you can use the initial condition (v(0)=50) to solve for the constant c (small c, not the big one), and you should have the final equation.


    ***Disclaimer, I think I'm right, although I'm not 100% sure. I certainly didn't check the physics of how you obtained expression (1), as I haven't done this sort of physics in a looong time. Thought I'd chip in with the mathematics tho ;). If anyone spots any faults with my solution please PM me, or let me know otherwise. I have exams to prepare for, and any major flaws in my understanding could use pointing out :D. Thanks!

    ***PS, I just edited to use quotes so that the equations are more readable.

    ***PS,2, I had a minor mistake in equation (1), I copied it incorrectly :S, I revised the steps - fixed another -.-, and another. < well I've totally botched this up, typing it on PC is confusing :/, but you got the general idea. Research bernoulli ODEs.
     
    Last edited: Oct 18, 2011
  5. Oct 18, 2011 #4
    While I have not gone through your solution, what happens to it for large t? It is always good to test the function at its limits to see if it all makes sense.
     
  6. Oct 18, 2011 #5
    I haven't gone through the solution either, but I suppose as t -> infinity, v(t) -> 0. That's how it should be afaik.

    EDIT: had a quick look at the second term: c(e^kt/m) in z(t). I didn't solve the integral for z(t), but as t->infinity, then c(e^kt/m) also goes to infinity, unless the constant is 0 (which it's not, I'm sure). As c(e^kt/m)->infinity, z(t) -> infinity, therefore v(t)->0 (1/infinity). Which looks right... providing the integral on the first term is valid(which I'm sure it is).
     
    Last edited: Oct 18, 2011
  7. Oct 18, 2011 #6
    wow thanks for the quick answers:)

    Ahmendbasil, thanks for the great go through for the equation. However I just have a few questions.

    The thing, with making another equation that is z = 1/v, was that just a good idea you got. And also. How does the derivative of z become -1*v^-2 dv/dt ? I do get that dv/dz is the derivative of z, however where does dv/dt fit it in the picture?

    Also, isn't it easier to solve it as a seperable differential equation? Just thinking of rewriting it a few times etc :). Because I did ask my teacher today if I could solve it as a linear equation and, the way he answered that question, it looked like that would not be a possible option. It should definitely be solved as a seperately equation.

    Anyway I'm still trying to learn this, so I'm most likely wrong:) Just nice to know how to get the smart ideas, and also learn how to solve them the easiest way
     
  8. Oct 18, 2011 #7
    No problem at all.

    Regarding "my idea" of substituting z=1/v, that's not my idea/method :P, it's a textbook method.



    Well, regarding your question "how the derivative of z wrt t becomes 1*v^-2 dv/dt". It's simple.

    You have [tex]z = v^{-1}[/tex], where v is a function of t. As we are differentiating w.r.t to t we must use the chain rule:

    [tex]\frac{dz}{dt} = \frac{dz}{dv} \frac{dv}{dt}[/tex]

    In this case [tex]\frac{dz}{dv} = \frac{-1}{v^{-2}}[/tex]multiplied by[tex]\frac{dv}{dt}[/tex] gives [tex]\frac{dv}{dt} = \frac{-1}{v^{-2}} \frac{dv}{dt}[/tex]



    As to your second inquiry, it can be solved using separation of variables, and you would end up with:

    [tex](k v - C v^2) dv = \frac{1}{m} dt[/tex]

    -- then --

    [tex]\frac{k v^2}{2} - \frac{C v^3}{3} = \frac{t}{m}[/tex]

    ^ good luck making it in the form v=f(t)

    IMO it's more convenient to solve the equation as a linear ODE as opposed to the above. Mainly because as a linear equation you get the equations in the form of v(t) = expression. Solving as a separable equation involves overhead work to get it in the form v(t) = expression. Sometimes it becomes very difficult. The substitution method looks long and tedious, but it's really simple. You don't have to prove it every time you use it. When I actually use it I use the general formula for an ODE:

    [tex]\frac{dy}{dx} + p y = q y^n[/tex]

    [tex]\frac{dz}{dx} + (1 - n)pz = (1 - n)q[/tex]

    where [tex]z = y^{1 - n}[/tex] and p and q are functions of x.


    It doesn't take time or effort to change it to a linear form then solve and re-sub y.
     
    Last edited: Oct 18, 2011
  9. Oct 19, 2011 #8
    Ah okay it makes sence now :)

    I actually spoke to my teacher today and he said that's it way easier to do it by using the Bernoulli ODE, however since we havn't learned to use it yet, he'd rather have we solve it as a seperately equation.

    But been working on it all day and finally got a solution in both ways. But no doubt that the Bernoully ODE was way faster. Solving the equation as seperable equation was truely a pain in the ***.

    But once again thanks for the help :) Was a great help to solve this assignment :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Differential Equation of a moving car with drag
  1. Two cars drag racing (Replies: 3)

  2. Drag Equation (Replies: 4)

Loading...