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Any other ways to solve?

  1. Jan 9, 2005 #1
    Joe is swimming in a river 50 metres away from the bank. He would like to get back to his parents who are 100m along the river bank from where he is. If he can swim at 3 m/sec and run at 7m/sec, how far along the bank should he land in order to get back as quickly as possible?

    So far, I've figured out how to do this:

    Let [tex]x[/tex] be the distance along the bank that he should land in order to get back as quickly as possible.

    Since he's 50 m away from the bank, he has to swim a distance of [tex]\sqrt{x^2 + 50^2}[/tex] metres, and then run a distance of [tex]100-x[/tex] metres.

    The total amount of time taken to do this is [tex]\frac{7}{100-x} + \frac{3}{\sqrt{x^2 + 50^2}}[/tex].

    Now, I can tabulate everything on an Excel spreadsheet, and figure out the answer from there, which is: he should land roughly 23.7 metres along the bank from where he is. Is there a more elegant way of getting around this?
     
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  3. Jan 10, 2005 #2

    dextercioby

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    I'm sorry,but the formula is wrong...U divided speed by distance to find time...It is wrong.It's the other way around.
    The total time is
    [tex] t(x)=\frac{100-x}{7}+\frac{\sqrt{2500+x^{2}}}{3} [/tex]

    Do you know calculus??

    Daniel.
     
  4. Jan 10, 2005 #3

    learningphysics

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    Your answer is correct. But in your formulas above you've put in speed/distance instead of distance/speed for time.

    Do you know calculus? If you do, then the way I did it was to calculate the derivative of the time function, and solve for 0.
     
  5. Jan 10, 2005 #4
    Thanks a lot, you two. I just thought that there might be a more elegant way of solving this problem ,rather than using brute force through calculus or tabulating. I'm not sure if I'm allowed to solve this problem through calculus.
     
  6. Jan 10, 2005 #5
    Edit: Recon, basically graph it and find its vertex, lowest point. Good enough answer...
     
    Last edited: Jan 10, 2005
  7. Jan 10, 2005 #6

    dextercioby

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    Calculus yields the exact answer to
    [tex] x_{extrema}=\pm \frac{75}{\sqrt{10}} [/tex]

    U can check that the physically acceptable solution (with the plus) corresponds to a maximum,plugging the value in the second derivative.The answer should be <0.

    Daniel.
     
  8. Jan 10, 2005 #7
    Isnt the second derivative a constant?
     
  9. Jan 10, 2005 #8

    dextercioby

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    Nope,the second derivative is a pretty ugly (sic) function.

    Differentiate 2 times the original functions and convince yourself.

    Daniel.
     
  10. Jan 10, 2005 #9
    dextercioby, to answer a question you asked earlier, I know only a little bit of calculus.

    I tried calculus and came out with this nasty looking equation:

    [tex]\frac{\frac{21}{49}^2 \times 2500}{1 - \frac{21}{49}^2} = x^2[/tex]

    Solving for x yields the correct answer. However, I would like to know how you managed to simplify x to [tex]\frac{75}{\sqrt{10}}[/tex]
     
  11. Jan 10, 2005 #10

    dextercioby

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    The derivative of the function is
    [tex] t'(x)=-\frac{1}{7}+\frac{x}{3\sqrt{2500+x^{2}}} [/tex]

    Equate it to zero and square the resulting equation.

    Daniel.
     
  12. Jan 10, 2005 #11
    Oh ok i see. I get [3sqrt(2500+x^2)-((x^2)/3sqrt(2500+x^2))]/[9(2500+x^2)] Is that right for the second derivative?
     
  13. Jan 10, 2005 #12

    dextercioby

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    It looks okay...It might have helped by tired eyes if u had used LaTex.

    Daniel.
     
  14. Jan 10, 2005 #13
    1/49=(x^2)/(22500+9x^2)

    9(2500+x^2)/49 = x^2 do I just use the quadratic equation? That gets messy... how did you get the nice numbers?

    How do i use latex?
     
  15. Jan 10, 2005 #14

    dextercioby

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    You're on the right track.
    [tex] 22500+9x^{2}=49x^{2} [/tex]
    Solve for "x" and chose the "+" solution

    Daniel.
     
  16. Jan 10, 2005 #15
    Oh duh... lol 40x^2 - 22500 ahaha

    +/- sqrt(-4*40*(-22500)) / 80 woohoo... lets see....

    2*2*150*sqrt(10) / 80

    7.5sqrt(10) * sqrt(10)/sqrt(10) = 75/sqrt(10) yay!!! Is this how you did it? or is there an easier methoid, you got it fast.
     
  17. Jan 10, 2005 #16

    dextercioby

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    You chased your tail a little bit,but it's okay... :tongue2: As to answer your question,i don't know,maybe i'm fast...????? :confused:

    Daniel.
     
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