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!

Homework Help: 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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Science Advisor

    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).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook