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Most likely an easy Integration problem

  1. Mar 3, 2005 #1
    However, im stuck on this problem

    Let v(t) be the velocity, in feet per second, of a skydiver at time t seconds, t>=0. After her parachute opens, her velocity satisfies the differential equation dv/dt = -2v -32, with initial condition v(0) = -50.

    A) Find an expression for v in terms of t, where t is measured in seconds.

    well I have problems right from the start...
    dv/(-2v-32) = dt
    integrate both sides --> ln(-2v-32) = -2t
    I solve for v and i got v = e^(-2t)/-2 - 16 + C

    C = -33.5

    However, i could also have done it like this

    dv/(-2(v+16)) = dt
    bring the -2 over and integrate it to get --> ln(v+16) = -2t
    solve for v = e^(-2t) -16 + C

    C = -35

    Why are they different?!!!! i must be doing something wrong...

    b) Terminal velocity is defined as lim t->infinity v(t). Find the terminal velocity of the skydiver to the nearest foot per second.

    well for the first equation i got
    i get -49.5 feet per second, since the 1/infinity = 0

    while the second one is -51 feet per second.... which is most likely not correct.

    c) It is safe to land when her speed is 20 feet per second. At what time t does she reach this speed?

    well for both cases i got t = negative seconds. I really don't think this is right. Can some one please help me figure out my mistake? Im really lost thanks...
     
  2. jcsd
  3. Mar 3, 2005 #2

    StatusX

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    In the first part, the constant C has to be added right after the integration. It ends up in the exponent and becomes a constant factor.
     
  4. Mar 3, 2005 #3

    dextercioby

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    I think they meant the absolute value for the velocity...Negative time would be when he's still in the plane...

    Daniel.
     
  5. Mar 3, 2005 #4
    the negative velocity will work if the equation is like this

    dv/(-2v - 32) = dt

    integrate both sides --> ln(-2v-32)/-2 = t + c

    -2v - 32 = e^(-2t - 2c)

    v = e^(-2t - 2c) /-2 - 16


    However the velocity will not work if it is like this
    ln(v+16) = -2t + c
    v = e^(-2t + c) - 16

    Im really confused as to which one is the correct one... please help
     
  6. Mar 3, 2005 #5

    dextercioby

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    No,there's one way to do this problem:
    [tex] \int_{-50}^{v} \frac{dv}{-2v-32}=\int_{0}^{t}dt [/tex]

    Solve this and tell me what u get...

    Daniel.
     
  7. Mar 3, 2005 #6
    alright, i get ln(-2v-32)/-2 = t + c

    so (-2v-32) = e^(-2t - 2c)

    v = e^(-2t - 2c)/-2 -16

    is this correct? im just wondering, why is it that if you bring out the -2 before you integrate it, you get a different answer?
    I don't want to plug in 50 yet... i just want to find out why the equation is different as compared to my previous posts.
     
  8. Mar 3, 2005 #7

    dextercioby

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    You did't do what i said,right...?Suit yourself.As for this issue,i believe StatusX gave the explanation...

    Daniel.
     
  9. Mar 3, 2005 #8
    Well first of all, i don't understand the integration values completely yet (50 - 0 ?) We've only just started this as you see :tongue2: ... Well i did what Status said, but i still don't get the same answer... Can anyone explain this?
     
  10. Mar 3, 2005 #9

    dextercioby

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    That's integrating with "corresponding limits".At t=0,the v=-50 ([ms^{-1}]),at t=t,the v=v([ms^{-1}])...There's no big deal.It's much more simple than integrating indefinitely & then imposing conditions on the antiderivative...

    Daniel.
     
  11. Mar 3, 2005 #10
    Your different answers are just due to errors in your solutions of the d.e. The answer is the same either way.

    First way:
    [tex]\begin{align*}\frac{dv}{2v+32} &= -dt \\
    \frac{1}{2}ln|2v+32| &= -t + c \quad \text{note: dv is not the deriv. of 2v}\\
    ln|2v+32| &= -2t + c'\\
    2v+32 &= e^{-2t + c'}\\
    &= Ce^{-2t}\\
    2v &= Ce^{-2t} -32\\
    v &= \frac{1}{2}(Ce^{-2t} -32)\\
    -50 &= \frac{C}{2} -16 \\
    C &= -68 \\
    v &= \frac{1}{2}(-68e^{-2t} -32)\\
    v &= -34e^{-2t} -16 \end{align*}[/tex]


    Second way:
    [tex]\begin{align*}\frac{dv}{v+16} &= -2dt \\
    ln|v+16| &= -2t + c\\
    v+16 &= e^{-2t + c}\\
    &= Ce^{-2t}\\
    v &= Ce^{-2t} -16\\
    -50 &= C -16 \\
    C &= -34 \\
    v &= -34e^{-2t} -16 \end{align*}[/tex]
     
  12. Mar 4, 2005 #11

    dextercioby

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    I believe and hope it's all clear and u realize where you went wrong...

    Daniel.
     
  13. Mar 4, 2005 #12
    thank's gnome and dexter.... it makes sense now. I didn't know that e^c can be just used as C.
     
  14. Mar 4, 2005 #13

    dextercioby

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    That's a standard trick...Once you get used to this type of integration,it will become obvious...

    Daniel.
     
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