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Drag and time to slow down -__-

  1. Oct 26, 2004 #1
    drag and time to slow down... and apparent weight. -__-

    A 1.58 m wide, 1.44 m high, 1300 kg car hits a very slick patch of ice while going 20.0 m/s. Air resistance is not negligible.

    If friction is neglected, how long will it take until the car's speed drops to 18.0 m/s?

    Ok drag is equal to = 1/4 * A * v^2

    v^2 = 400 and A = 1.44*1.58

    I divided this drag by the mass to get acceleration in the x direction and then used this equation:

    [tex]v_f = v_i + a t[/tex]

    t here is delta t...

    final velocity is what I want = 18 and initial is 20 while acceleration is the one I got from analyzing the drag force....

    I get 11.4 seconds and it's wrong. :cry:

    I did it again and again to make sure there are no errors but nothing changed.


    btw does anybody know how to get the mass from a chart of apparent weight vs time? I know it's w/g but I only have apparent weight and not weight... I can't tell when it's equal to the real weight. (when a = 0 but I can't tell from the chart).

    Here's the chart:

    Last edited: Oct 26, 2004
  2. jcsd
  3. Oct 26, 2004 #2


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    The rate of deceleration is not constant as you assumed. Also, where did the factor of 1/4 come from?
  4. Oct 26, 2004 #3
    (1/4 * A * v^2 ) <--- equation in my text for drag...

    and if deacceleration is not constant... how can I calculate the time?
  5. Oct 26, 2004 #4


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    You'll have to integrate

    [tex]\frac {dv}{dt} = - \frac {A}{4M} v^2[/tex]
  6. Oct 26, 2004 #5
    ok so integrating that would result in the same drag equation but [tex]v^2[/tex] becomes [tex]\frac {v^3}{3}[/tex] but since acceleration is not constant I can't use the kinematic equations to solve for delta t...


    now that I think about it I'm supposed to get an equation for velocity in t right? after getting that equation I'll just plug the velocity 18 to get t... :confused:


    ok integrating the derivative of velocity (which is acceleration) gives me the original equation for velocity right? I just plugged in 18 in that equation I integrated and the time is still wrong... :(
    Last edited: Oct 26, 2004
  7. Oct 26, 2004 #6
    you integration is wrong; try to rearrange the equation tide gave you so that v terms is together with dv and dt is at the other side of the equation.
  8. Oct 26, 2004 #7
    I got several equations and all of them were wrong.. :cry:

    re-arranging that equation gives me this.. right?

    [tex]v = \frac {-4 m}{A t}[/tex]

    do I need to integrate that?

    [edit]integrate the the t?

    this sucks. Our course is supposed to be calculus-based but out of 10 HW problems only 1 has to do with calculus... and of course the text barely goes over problems that need calculus. And yes I'm trying to save (a little) face. :(
    Last edited: Oct 26, 2004
  9. Oct 26, 2004 #8
    [tex]\int_{20}^{v} -\frac{4m}{Av^2} dv = \int_{0}^{t} dt[/tex]
    Know how to integrate this ?
  10. Oct 26, 2004 #9
    no. :(

    Is there any other way to solve the problem? the course I'm taking uses VERY easy and simple calculus in the rare occasions it's needed.... so there must be another way. =\
  11. Oct 26, 2004 #10
    there is other way of course but it is more complicated . something like alternative to integration, numerical methods. ok, i will integrate for you . tide must not be very happy with this.
  12. Oct 26, 2004 #11
    the time I get is still wrong...

    bah forget about it and thanks a lot man. I'll ask my teacher next week. The guy hates me but I hope not enough to refuse. :biggrin:
  13. Oct 26, 2004 #12
    is it 12.7 s?
  14. Oct 26, 2004 #13
    I got 13.7....

    *types in 12.7*

    it's correct.


    [edit] my mistake was rounding/significant figures. and yes I'm not dead. >_<
    Last edited: Oct 26, 2004
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