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Homework Help: Time at given speed in freefall

  1. Jan 24, 2004 #1
    Here is a problem from my book:

    If a car rolls gently (initial velocity = 0) off a vertical cliff, how long does it take to reach 90km/h?

    It seems easy, and I have been breezing through all teh other ones like it's my job, but for some reason this one gets me. I knwo it's possible, but I jsut don't know how. The answer is supposed to be 2.6 seconds, but I somehow always wind up dividing 90 by 9.8. Any help?
     
  2. jcsd
  3. Jan 24, 2004 #2

    chroot

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    Are you familiar with the expression:

    [tex]v(t) = v_0 + \frac{1}{2} a t^2[/tex]?

    Just solve it for t.

    - Warren
     
  4. Jan 24, 2004 #3
    Alright, thanks...but I still seem to get the wrong answer consistently. I do:

    90=0.5(9.8)t^2
    90=4.9t^2
    18.37=t^2
    t=4.29

    what is going wrong?
     
  5. Jan 24, 2004 #4

    chroot

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    Convert 90 km/hr into m/s first, so that your units will agree.

    - Warren
     
  6. Jan 24, 2004 #5
    Thanks, I don't know how I skipped that. That's what happened on the quiz, too. Did all of them perfectly, then on one forgot to convert.

    Also, is it jsut me, or should I really end up with about 2.26 seconds?
     
    Last edited: Jan 24, 2004
  7. Jan 25, 2004 #6
    I get 2.258769757 on my calculator using g = 9.8 m/s^2.

    Does the problem explicitly say we can ignore air resistance?
     
  8. Jan 25, 2004 #7

    HallsofIvy

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    Chroot: I must confess that during the several days I spent in calculus (and physics) class, I never saw an equation that said
    "[tex]v(t) = v_0 + \frac{1}{2} a t^2[/tex]"

    I do recall seeing one that said "[tex]v(t)= v_0+ at[/tex]" and one that said "[tex]d(t)= d_0+ v_0t+ \frac{1}{2}at^2[/tex]".

    Assuming no air resistance and that a= -9.8 m/s2, with v0= 0, v(t)= -9.8 t m/s. Since there are 1000m in a km and 60*60= 3600 sec in and hour, -90 km/hr= -90 *1000/3600= -25 m/s.
    -25= -9.8 t so t= 25/9.8= 2.55 seconds.

    Looks pretty easy to me (as long as you don't listen to Chroot!).

    (C'mon, who is it that's using Chroot's good name?)
     
  9. Jan 25, 2004 #8
    I'm sure it was just an oversight. :smile:.
     
    Last edited: Jan 25, 2004
  10. Jan 25, 2004 #9

    chroot

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    GAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAH!

    *slinks into a corner and hides*

    - Warren
     
  11. Jan 25, 2004 #10
    Or you can use

    v^2 = v0^2 + 2aΔx.

    Solve for Δx and then time is just Δx/v.
     
  12. Jan 25, 2004 #11
    Optimus, you can't do that because once you've solved for the distance its fallen, dividing by v means you are assuming that the velocity is constant, which it obviously isn't.
     
  13. Jan 27, 2004 #12

    GAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAH!


    I blindly assumed chroot's equation was correct.

    Give me back the two minutes of my life I wasted!
     
  14. Jan 27, 2004 #13
    Besides, x is not given.
     
  15. Jan 27, 2004 #14
    Actually, under a constant acceleration, you can compute the time using the average velocity.

    Given the final velocity and displacement, to solve for t:

    [tex]t = \frac{2 d}{V_f}[/tex] or

    [tex]\frac{V_f}{2} = \frac{d}{t}[/tex]

    Vf/2 when Vi=0 and acceleration is constant gives the average velocity.
     
    Last edited: Jan 27, 2004
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