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Homework Help: Relative Motion Problem been trying to figure out for hours

  1. Sep 9, 2015 #1
    1. The problem statement, all variables and given/known data
    In hot pursuit, Agent Logan of the FBI must get directly across a 1200-m-wide river in minimum time. The river's current is 0.80 m/s, he can row a boat at 1.60 m/s, and he can run 3.00 m/s. Describe the path he should take(rowing plus running along the shore) for the minimum crossing time, and determine the minimum time.

    2. Relevant equations
    x = vt
    x = x + (vi)t - (1/2)gt^2
    and other kinematic equations

    3. The attempt at a solution
    So I have been trying to figure out this problem for quite a long time and this is what I have come up with so far:

    https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/71/71483-dee2797c012ada0b48bcaee676e9239a.jpg [Broken] The last equation should be y = +3t2

    My first step is to take equations 2 and 3 and equate them. Next I replace the t1 value from the second equation with the first equation.

    when I simplify, I end up getting t2 = (750(1.6sinΘ -.8))/(3cosθ)

    from there I am not really sure what to do. If anyone could help me step-by-step I would really appreciate it.

    The solution is "Row at an angle of 24.9° upstream and run 104m along the bank in a total time of 862 seconds."

    Attached Files:

    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Sep 9, 2015 #2


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    You need to type out all your working, it's hard to tell what exactly it is you've done. include all units.

    This is a optimisation problem.
    you need an expression for the total time (t1+2) in terms of theta, then find the derivative, set to zero to locate the critical point (s).
  4. Sep 9, 2015 #3

    Ray Vickson

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    Homework Helper

    Your attachment will not open for me, so I cannot comment on what you have done. Please type your work, rather than attaching images. (If you read the PF requirements, you will find that to be stated policy, although sometimes it is relaxed.)
  5. Sep 9, 2015 #4
    Why would I find the derivative and set it equal to zero? I get the t1 + t2 in terms of theta to be:

    t1 + t2 = (750/cosΘ) + (250(1.6sinΘ - 0.8))/(cosΘ)

    edit: I uploaded my work
    Last edited: Sep 9, 2015
  6. Sep 9, 2015 #5


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    To find the critical point where the value of theta corresponds to the smallest possible value of t1+2.
    You have one equation and two unknowns, t1+2 and theta, it can't be solved with algebra alone.

    Have you done an introductory calculus course? You may want to go over your notes on optimisation, or here's some resources:


  7. Sep 9, 2015 #6

    At the optimum angle the change in the total time taken for a small change in angle will be zero, so dt/dΘ = 0
    Last edited: Sep 9, 2015
  8. Sep 9, 2015 #7
    Alright I understand that now how would I go about taking the derivative dt/dΘ and then computing what Θ in that equation is = to 0?
  9. Sep 9, 2015 #8


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    Based on your posts so far I'm going to guess the answer to this is a no.
    In light of that, I think your time would be better spent working on physics problems that don't require calculus.
    As you said, you've spent hours on this problem already, it might be best to cut your losses and return to it after a calc course.
  10. Sep 9, 2015 #9
    ...I have taken calc just this problem is kind of difficult in its own way. No need to be condescending my peers couldn't figure this out either who have also taken calc...all I wanted was some help in case I was missing out on a detail or something.
  11. Sep 10, 2015 #10
    So you now have

    t1 + t2 = (750/cosΘ) + (250(1.6sinΘ - 0.8))/(cosΘ)

    To find the derivative of time with respect to θ, or dt/dθ you need to apply the rules of derivation (sometimes called differentiation rules). Its easier now to call the total time t, it just makes things look clearer. So you have

    t = (750/cosΘ) + (250(1.6sinΘ - 0.8))/(cosΘ)

    To start off you will need to use the sum rule, which put simply is:


    t = a + b


    dt/dΘ = da/dΘ + db/dΘ

    But then to find da/dΘ which for this specific situation is d(750/cosΘ)/dΘ , you are going to have to apply other rules such as the quotient and the trigonometry differentiation rules.

    Similarly to find db/dΘ or d((250(1.6sinΘ - 0.8))/(cosΘ)) /dΘ you need to use the rules of derivation again.

    Once you have done all this and got an equation for dt/dΘ which you know equals zero for the optimum angle, you can then set this equation to equal zero and solve to find Θ by using trigonometric identities.
    Last edited: Sep 10, 2015
  12. Sep 10, 2015 #11
    Ok when I take the derivative, the equation comes out to be (750sinθ + 200)/(cos^2θ) = 0,

    That would mean θ is equal to 15.5° which isn't the correct answer
  13. Sep 10, 2015 #12
    You need to check back for mistakes.

    You have been given that the answer is Θ = 24.9 degrees. Have you tried plugging this into your equation

    t1 + t2 = (750/cosΘ) + (250(1.6sinΘ - 0.8))/(cosΘ)

    to see if this gives the answer 862 seconds
  14. Sep 10, 2015 #13
    Ah ok, the original equation should have been t1 + t2 = (750/cosθ) - (400sinθ - 200)/(cosθ). Then when I took the derivative, I forgot to distribute a sinθ. It should have been (950sinθ - 400)/(cos^2θ) = 0. This gives me θ to equal 24.9 degrees. Finally done with this problem. Thank you so much for your help.
  15. Sep 10, 2015 #14
    You're welcome !
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