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Parametric equations motion problem

  1. Apr 7, 2012 #1
    The question states:
    Two towns A and B are located directly opposite each other on a river 8km wide which flows at a speed 4km/h. A person from town A wants to travel to a town C located 6km up-stream from and on the same side as B. The person travels in a boat with maximum speed 10km/h and wishes to reach C in the shortest possible time. Let x(t) be the distance travelled upstream and y(t) be the distance travelled across the river in t hours. The person heads out at angle theta.

    a) Show that x(t)=10tcos(theta)-4t and y(t)=10tsin(theta)
    b) What is the angle theta and how long would the trip take?

    Relevant equations:
    So far I have used v=d/t along with some vector diagrams.

    My attempt:
    I have proven a) already by using v=d/t. The net velocity for x was equal to 10cos(theta)-4 and I just rearranged for x. I did the same to find y.

    I then found the angle theta by saying that sin(theta)=8/10, therefore theta=arcsin(4/5). Also, I found the theta in terms of arccos which was theta=arccos(3/5). I found these by using a distance triangle with adjacent=6, opposite=8 and hypotenuse=10.

    I then equated x(t)=6 ==> 10tcos(theta)-4t=6
    10tcos(arccos(3/5))-4t=6
    10t(3/5)-4t=6
    6t-4t=6
    t=3
    And equated y(t)=8 ==> 10tsin(theta)=8
    10tsin(arcsin(4/5))=8
    10t(4/5)=8
    8t=8
    t=1
    This is where I'm having problems. Shouldn't the time value be equal? If anyone could please help me out I would greatly appreciate it.
     
  2. jcsd
  3. Apr 7, 2012 #2

    tiny-tim

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    welcome to pf!

    hi cjh12398! welcome to pf! :smile:
    no, the question means that the boat is heading at angle θ relative to the water, not to the land

    use your two equations in a) to find θ (by eliminating t)​
     
  4. Apr 7, 2012 #3
    Thank you, I continued working out the times for both x and y after using theta by eliminating the t, but I had to use a graphics calculator to solve for theta.
    The equation I got to solve theta was:
    sin(theta)-(4/3)cos(theta)+(8/15)=0.
    In my course it is calculator free, so is possible to solve this by hand? I've tried for about 15 mins....
     
  5. Apr 7, 2012 #4

    tiny-tim

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    try to re-arrange it so that it's of the form sinθcosψ - cosθsinψ = A,

    then use one of the standard trigonometric identities :wink:
     
  6. Apr 7, 2012 #5

    tms

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    You left out the [itex]t[/itex]. Your equation was [itex]y(t) = 10\; t \sin\theta[/itex].

    It would help if you first solved this (and all other) problems symbolically, and only plugged in numbers as the very last step. It would also help if you put spaces in your equations; they don't cost anything, and they let the various terms and factors stand out more clearly.
     
  7. Apr 11, 2012 #6
    I'm also stuck with this one.

    I hit the same brick wall by myself with the trig identities, been trying for about 2 hours now to find a way to do this question :(

    Treating it as a vector addition from a point before the river flow is considered is something I will try.
     
    Last edited: Apr 11, 2012
  8. Apr 13, 2012 #7
    I'm sorry I can't get it at all near that form. Using the Asin(x) + Bcos(x), i can solve for theta with a correct answer.
     
    Last edited: Apr 13, 2012
  9. Apr 13, 2012 #8

    tiny-tim

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    that is that form (with B/A = -tanψ) :wink:
    i don't think they'd expect you to do it by hand

    change it to arcsin(-4/5), and use sin tables or your calculator :smile:
     
  10. Apr 13, 2012 #9
    Aye, I've worked through it all and trying to forget about it now ;). It didn't occur to me originally, but eventually I realised to just leave it as the difference of some inverse trig functions. Messy but it does appear to be the best I can do... substituting theta back in even more messy :P

    Thanks for your help. I'd not seen that trig equation before.
     
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