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2-D relative motion Problem

  1. Sep 17, 2006 #1
    Suppose the river is moving east at 4.80 km/h and the boat is traveling Theta = 43.0° south of east with respect to the Earth. You will have to solve two equations and 2 unknowns.

    (a) Find the speed of the boat with respect to the Earth. kh/m

    (b) Find the speed of the boat with respect to the river if the boat's heading in the water is 60° south of east.


    I am having trouble on where to start and what to do. I tried sqrt((4.80(cos43*))^2 + (4.80(sin43*))^2) but still give a wrong answer.
  2. jcsd
  3. Sep 17, 2006 #2
    OK, let's look at part a.
    I think:
    Basic concept is [itex]v_{BE}=v_{BR}+v_{RE}[/itex]
    where V_BE= velocity of boat with respect to earth
    v_BR is velocity of boat with respect to river
    v_RE is velocity of river with respect to earth

    Solve the above equation separately for x and y components, then use [itex]v_{BE}=\sqrt{(v_{BEx})^2+(v_{BEy})^2}[/itex]

    But my brain is fried. See if it makes sense, see if it works--good luck!

    EDIT: never mind, the problem doesn't give you enough info to do all that. I guess you already know that, though... Sorry.
    Last edited: Sep 17, 2006
  4. Sep 17, 2006 #3
    Yea, I tried doing what you said above but I did not have enough info to do it. I tried randomly doing things and closest I got the the correct answer was 14.8 but not sure how I did that.
  5. Sep 17, 2006 #4
    Anyone? I been stuck on this question for the last 2 hour.
  6. Sep 17, 2006 #5
    OK, take 2.

    Those two angles are not two separate situations. Its the same exact boat at the same exact time doing the same exact thing. the 48 degree angle is with respect to the shore, and the 60 degree angle is with respect to... to... ah, hell, I don't know. And notice there's no handy right triangle to play with, you might have to draw another line to get one...
  7. Sep 17, 2006 #6
    Okay, I tried using cos and sin in anyway I can but it seem to be no use
    since that way i wont really learn anything from it.
  8. Sep 17, 2006 #7
    Well, I'm a little bum out right now not able to get anywhere with this problem so maybe someone can shine some light here and help me out with this instead:

    A daredevil decides to jump a canyon. Its walls are equally high and 12 m apart. He takes off by driving a motorcycle up a short ramp sloped at an angle of 14°. What minimum speed must he have in order to clear the canyon? m/s

    I am not sure where to start with this problem. Any pointers would be great.
  9. Sep 18, 2006 #8
    okay for the above problem this is what I did and still got the wrong answer.

    V_oy = V_osin14*
    V_ox = V_ocos14*

    V_osin14* = (9.8)(1/2V_ocos14*)
    2V_o^2(sin14)(cos14) = 9.8
    V_o^2 = 9.8/(2(sin14)(cos14))
    V_o = 4.54

    which is incorrect when i plug in V_ox = 4.54cos14*

    then i tried using this formula when I got deltay (which was probably wrong)
    V_o = (sqrt(2G(Deltay))/(sin14) and i got 101 m/s which is also incorrect.
    Last edited: Sep 18, 2006
  10. Sep 18, 2006 #9
    you're off witht he right foot in breaking the velocity up, but after that i am not sure what you are trying to do. it looks like you have a key idea here that the x and y parts are independent of one another, but that one will let you solve for the other. trying using the position function to solve this one.

    first, we know the V_x=(V_0)cos14. so you have the velocity function and you need the position function. how would you go about getting that? once you have the position function (with respect to time), what would the position function be for the y direction and how would you relate it to the x direction?
  11. Sep 18, 2006 #10
    you give no velocity for the boat itself, just the rivers.
  12. Sep 18, 2006 #11


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    Homework Helper

    Okay, first let's write what we already know, so [tex]\vec{v}_{BE}=\vec{v}_{BR}+\vec{v}_{RE}[/tex]. [tex]\vec{v}_{RE}=4.8\vec{i}, \vec{v}_{BR}=v_{BR}cos60\vec{i}+v_{BR}sin60\vec{j}, \vec{v}_{BE}=v_{BE}cos43\vec{i}+v_{BE}sin43\vec{j}[/tex]. Now all we have to do is plug these equations into the equation [tex]\vec{v}_{BE}=\vec{v}_{BR}+\vec{v}_{RE}[/tex], from which we can easily retrieve two scalar equations with two unknowns (one for the i direction, and one for the j direction).
  13. Sep 18, 2006 #12
    Okay, I am completely lost. It seem that the teacher did not go over this over the lecture last week. I tried looking up in my notes and pluging the equations into the equation but do not get what to do. I seriously need a new teacher....
  14. Sep 18, 2006 #13
    let's see if this helps.

    we know that the x component of the velocity is V_x=V_0cos14.
    if we integrate this equation with respect to time we get X=V_0cos14(t).

    now, what would be the velocity function equation for the y direction? what would the position function be?

    once you have that, how could you relate the y-position function to the x-position function?

    if you work with me on this you end up getting some very helpful equation for projectile motion.
  15. Sep 18, 2006 #14
    if X = V_0cos14(t) i would plug that in Deltay = V_0y(t) - 1/2(g)(t)^2

    which is t = x/(V_0cos14) which is
    deltay = V_0y(x/(V_0cos14) - 1/2(-9.8)((x/(V_0cos14)^2

    Sorry but physics is a really hard subject for me to understand but i know with help I will eventually get it.
  16. Sep 18, 2006 #15
    you're doing perfect!
    now, what is V_0y equal to? when you place that into the equal and simplfy, what do you get?
  17. Sep 18, 2006 #16
    is V_0y = 0?

    deltay = - 1/2(-9.8)((x/V_0cos14)^2 does x = 12?
    deltay = -1/2(-9.8)(12.37/V_0^2)

    im not sure if i should go further since i think i made a mistake somewhere
  18. Sep 18, 2006 #17
    well, v_0y != 0 though. what is the vector component of the y velocity? you're getting really close to the answer though.
  19. Sep 18, 2006 #18
    deltay = 60.61/V_0^2

    V_0^2deltay = 60.61

    deltay = 60.61V_0^2 if that is it do i sqrt to get rid of the V_0^2 or did i made a mistake.
  20. Sep 18, 2006 #19
    we are not at the point for solving anything just yet; we are still doing equation manipulation here.

    you said
    which is right and exactly what direction you need to be heading in.
    if V_0=V, then what are the vector components (V_0y and V_0x)?

    how can you use them in y=V_0y(x/(V_0cos14) - 1/2(-9.8)((x/(V_0cos14)^2?
  21. Sep 18, 2006 #20
    okay V_0y would be 0 since there is no initial velocity for it.

    so would y = V_0sin14

    making it V_0sin14 = -1/2(-9.8)(x/(V_0cos14)^2
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