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Intro to DIFF EQ (help )

  1. Feb 11, 2006 #1
    intro to DIFF EQ (help!!)

    a drag chute must be designed to reduce the speed of a 3000-lb dragster from 220 mph to 50 mph in 4 sec. Assume that the drag force is proportional to the velocity.

    a) what value of the drag coefficient k is needed to accomplish this?
    b) how far will the dragster travel in the 4-sec interval?

    the way i reasonned is the following:

    m(dv/dt) = sum of all forces
    m(dv/dt) = -mg -kv (F=-kv drag force "opposite direction to the velocity).

    mv' +kv = -mg
    v' +(k/m)v = -g

    integrating factor = e^(k/m)t

    multiplying the integrating factor by the diff eq and integrating

    we get: v(t) = -(mg/k) +Ce^(-k/m)t

    initial conditions t=0 v=v(0)=220

    --> C = v(0) +(mg/k)

    thus v(t) = -(mg/k)(1 - e^(-k/m)t) +v(0)e^(-k/m)t

    now in order to find the value of k
    we have v(t)=50 mph
    v(0) = 220 mph
    g =32
    m =3000
    t = 4 sec

    but we have a k^-1 and a k in the exponential.
    how can we get the value of k?

    it's a sophomore level course.
    I don't need very complicated equations :).

  2. jcsd
  3. Feb 11, 2006 #2


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    Your dragster is moving horizontally, so you won't need an mg term - that acts in the vertical direction anyway, and m(dv/dt) and kv act in the horizontal.

    That should simplify it a bit.

    Also, remember to keep your units compatible. If you are using imperial measure, then your velocity shoud be in feet/sec, not mph.
  4. Feb 11, 2006 #3
    i dont know how to transform from mph to feet per sec :)
    can you help me? :)
  5. Feb 11, 2006 #4


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    There are 1760 yards in a mile and 3 ft in a yard
    Convert miles to feet and divide by 3600 (60*60) to bring it from mph to ft/sec
  6. Feb 11, 2006 #5
    i got for 50 mph = 73.3 ft/sec
    and for 220 mph = 322.6 ft/sec

    my solution was v(t) = v(0)e^(-k/m)t

    v(t) = 73.3
    v(0) = 322.6
    t = 4 sec
    m = 3000 lb

    i got a value of k =1111.1
    but in solutionary we have k = 0.2469

    i must have done a terrible mistake.
    can u tell me what is it?

  7. Feb 11, 2006 #6


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    I've gone over the problem a few times, and I can't justify the given solution, of k = 0.2964.
    I get the same result as yourself, k = 1111.2.
    It looks like a wrong answer in the book/solution set.
  8. Feb 11, 2006 #7
    well i'm sorry to make you work it few times.
    And in fact, 1111.3 is the right answer.

    I still have some problem in the second part of the problem

    mv' = -kv

    m (dv/dt) = -kv

    m (dv/dx)(dx/dt) = -k(dx/dt)

    m(dv/dx) = -k

    mdv = -kdx

    integating mv = -kx
    thus we end up with a negative value for x
    or is there another way to solve for x?

    thanks again
  9. Feb 11, 2006 #8


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    That's the correct way to solve for x.
    x is supposed to be negative so that you have a decreasing exponential function. One that starts at 220 mph and ends up at 50 mph.

    Oh, btw, You didn't have to convert the velocity from mph to ft/s. Not in this particular problem, that is. You would have got the same value for k if you had used mph, since it only depended on the ratio of the velocities.
  10. Feb 11, 2006 #9


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    Notice that 3000 pounds is not mass. It is weight which is the gravitational force on the car. Since F= ma so m= F/a= 3000/32.2 slugs (or poundals).
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