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Kinematics problem giving me trouble help please

  1. Sep 9, 2004 #1
    Encountered the following question. For (a), I repetitively found the answer 12 for magnitude of displacement and 131° for the direction; after being told i was incorrect....i need help: :confused:

    Path A is 8.0 km long heading 60.0° north of east. Path B is 6.0 km long in a direction due east. Path C is 3.0 km long heading 315° counterclockwise from east.

    (a) Graphically add the hiker's displacements in the order A, B, C.

    What is the magnitude of displacement (km)? What is the direction of displacement (°counterclockwise from East)?

    (b) Graphically add the hiker's displacements in the order C, B, A.

    What is the magnitude of displacement (km)? What is the direction of displacement (°counterclockwise from East)?

    (c) What can you conclude about the resulting displacements?


    thanks.
     
  2. jcsd
  3. Sep 9, 2004 #2
    Sine & Cosine Laws

    [tex]a=\sqrt{6^2+8^2-2*6*8*cos\ 120}[/tex]
    [tex]a=\sqrt{148}[/tex]

    [tex]\frac{sin\ x}{8}=\frac{sin\ 120}{\sqrt{148}}[/tex]
    [tex]x = 35^0[/tex]

    [tex]c=\sqrt{3^2+148-2*3*\sqrt{148}*cos\ 100}[/tex]
    [tex]c=13\ km[/tex]

    [tex]\frac{sin\ y}{3}=\frac{sin\ 100}{13}[/tex]
    [tex]y = 13^0[/tex]
    The direction is (35-13)=22 degree
     

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  4. Sep 9, 2004 #3
    i dont know what level physics that is...

    but im in 11th grade honors...

    its the right answer, but i have no explanation

    i'm used to the more adding/subtracting vectors, forming pictures and finding the resultants using trig functions...

    but thanks, you did help
     
  5. Sep 9, 2004 #4
    you dont know the sine and cosine law?
     
  6. Sep 9, 2004 #5
    i know this...but no i havent reached class with emphasis on trig

    sin(angle)=opposite/hypo

    cos(angle)=adjacent/hypo

    im guessin thats not enough???
     
  7. Sep 9, 2004 #6

    Pyrrhus

    User Avatar
    Homework Helper

    Using Cosine Law and Sine Law is a way.

    You can get the magnitude and direction of the resultant vector ([tex] \vec{R} [/tex]) by using the components.

    [tex] \vec{R} = (R_{x}i + R_{y}j) [/tex]
    [tex] \vec{A} = (8 , 60^o) [/tex]
    [tex] \vec{B} = (6, 0^o) [/tex]
    [tex] \vec{C} = (3, 315^o) [/tex]

    [tex] R_{x} = (A_{x} + B_{x} + C_{x})i [/tex]
    [tex] R_{y} = (A_{y} + B_{y} + C_{y})j [/tex]

    [tex] A_{x} = 8\cos(60^o) = 4[/tex]
    [tex] B_{x} = 6\cos(0^o) = 6[/tex]
    [tex] C_{x} = 3\cos(315^o)= 2.12 [/tex]

    [tex] A_{y} = 8\sin(60^o) = 6.93[/tex]
    [tex] B_{y} = 6\sin(0^o) = 0[/tex]
    [tex] C_{y} = 3\sin(315^o) = -2.12[/tex]

    [tex] \vec{R} = (12.12i + 4.81j) [/tex]

    [tex] |\vec{R}| = \sqrt{R_{x}^2+R_{y}^2}[/tex]
    [tex] |\vec{R}| = 13.04 [/tex]
    [tex]\theta_R = arctan(\frac{R_{y}}{R_{x}})[/tex]
    [tex]\theta_R = 21.65^o [/tex]

    -Cyclovenom
     
    Last edited: Sep 9, 2004
  8. Sep 9, 2004 #7
    thanks..

    makes much more sense
     
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