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The Song Remains the Same

  1. Sep 29, 2004 #1
    Help!!!

    A motorboat traveling at a speed of 2.4 m/s shuts off its engines at t = 0. How far does it travel before coming to rest if it is noted that after 3.0 s its speed has dropped to half its original value? Assume that the drag force of the water is proportional to v.
     
    Last edited: Sep 29, 2004
  2. jcsd
  3. Oct 1, 2004 #2
    ....anyone?
     
  4. Oct 1, 2004 #3

    Tide

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    The equation of motion would be
    [tex]\frac {dv}{dt} = -k v[/tex]
    Can you integrate that?
     
  5. Oct 1, 2004 #4
    Do I get lnv=-kt ? What do I need to do after I find what k is?
     
    Last edited: Oct 1, 2004
  6. Oct 1, 2004 #5
    Whats the answer given in your reference book?
    -Cheers.
     
  7. Oct 1, 2004 #6
    The answer is 10m, and i have no idea how they got it
     
  8. Oct 1, 2004 #7
    When you solve the differential eqn you will have 2 variables(1 from integration & other 'k')Use the initial conditions given to find them.
    At t=0,v=? and one more.
    Else,if you did definite integration ,you have to figure out k by the 2nd condition given.
     
  9. Oct 1, 2004 #8

    Tide

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    [tex]v = v_0 e^{-kt}[/tex]
     
  10. Oct 1, 2004 #9

    HallsofIvy

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    Since ln v=-kt+ C (you forgot to add the constant), v= Ce-kt which has two unknown parameters, C and k. Now use the information you were given: "traveling at a speed of 2.4 m/s shuts off its engines at t = 0". Okay, when t=0, v= Ce-k(0)= C= 2.4. "it is noted that after 3.0 s its speed has dropped to half its original value" Okay, when t= 0, v= "half its original value" which is 2.4/2= 1.2 m/s. v= 2.4e-k(3)= 1.2 . Solve that for k.
     
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