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Ap Physics kinematics help!

  1. Sep 20, 2005 #1
    Question 1) Graphically determine the resultant of the following three vector displacements:

    a) 34m , 25 degree north of east
    b) 48 m, 33 degree east of north
    c) 22m, 56 degree west of south.

    Ax = 30.814
    Ay= 13.369

    Bx= 40.256
    By= 26.142

    Cx=x 12.30
    Cy= 18.238

    I added the x components and y components and used the pythagorean theorm to get 101 m as my resulant.. HOwever, the book says the answer is 58 m... need some help...

    thanks much
     
  2. jcsd
  3. Sep 20, 2005 #2

    Doc Al

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    Recheck this value.

    x and y are mixed up. (That's 33 degrees east of north.)

    Again, x and y are mixed up. And the sign is wrong.

    Pay more attention to the angles. Draw a diagram of each vector.
     
  4. Sep 20, 2005 #3
    Ax = 30.814
    Ay= 14.369

    Bx= 26.14 ( since I took theta as 90-33 = 57 degrees )
    By= 40.25 ( since I took theta as 90-33 = 57 degrees )

    Cx= -18.238 ( ( since I took theta as 90-56 = 34 degrees )
    Cy= -12.30 ( ( since I took theta as 90-56 = 34 degrees )

    is this correct?

    thanks much.
     
    Last edited: Sep 20, 2005
  5. Sep 20, 2005 #4

    Doc Al

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    Recheck the sign of the C components.
     
  6. Sep 20, 2005 #5
    how about now?
     
  7. Sep 20, 2005 #6

    Doc Al

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    Seems like you're just guessing. Instead, actually draw an arrow pointing south, then another pointing west of south. What are the signs of the x and y components of that arrow?

    (Note: North is +, South is -; East is +, West is -.)
     
  8. Sep 20, 2005 #7
    well I have a feeling that my diagram is wrong and hence I wasnt referring to it when I edite dthe answers..

    My diagram is setup like this:

    starting from the same origin, I have three lines: a, b , c ... A goes north of east at 25 degrees. B goes east of north at 33 degrees and C goes 56 degrees west of south ( and hence all the x and y values are negative ) .. Is this correct?

    EDIT: jsut realized while Iwas writing my diagram setup that both components of C will be negative since it is west of south and not west of north which is what i was reading it as..
     
  9. Sep 21, 2005 #8

    Doc Al

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    Now you've got it!
     
  10. Sep 21, 2005 #9
    Even though I did this and got the answer, I had a question.. Is using the components of 90-33= 57 degrees north of east correct? shouldnt it yield a different result than 33 degrees east of north? or is it one and the same thing?

    Thanks much for your help.
     
  11. Sep 21, 2005 #10

    Doc Al

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    57 degrees north of east is the same as 33 degrees east of north. To see this, draw yourself a picture.

    What you may not realize is that cos(33) = sin(57), and sin(33) = cos(57). So either way, you'll get the same answer. (As long as you don't make a mistake.)
     
  12. Sep 21, 2005 #11
    I see .. cool thanks..

    Another questoin:

    A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. If the plane is traveling horiszoontally with speed of 250 km/hr ( 69.4m/s) .

    A) how far in advance of the receipients must the goods be dropped?

    I found this to be 480 m which is correct.

    B) Suppose instead that the plane releases the supplies a horizontal distance of 425 m in advance of the mountain climbers. What vertical vleocity ( up or down) should the supplies be given so that they arrive precisely at the climbers positiron?

    c) With what speed do the supplies land in the latter case?

    HOw do i go about solving B first of all? Do i use X=Vox . t and try to find T and then plug it into the y=y0+v0t-.5gt^2 and find y... Then plug in y into y/t= V ?? I tried that but I dont get the right answer ( which text book says is 8.37 m/s

    Thanks much.
     
  13. Sep 21, 2005 #12

    Doc Al

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    Use [itex]x = v_x t[/itex] to find the time. Then use the vertical motion equation to solve for v0: you already know t, y0, and y.
     
  14. Sep 21, 2005 #13
    well I get the v0 as 16.20 m/s and not 8.37 s ....

    I solved the following equation: 0=235+(6.124)v0 - 4.9 ( 6.124)^2 to get v0 ..
     
  15. Sep 21, 2005 #14
    nevemrind.. I got 8.37 m/sec...

    For C) i used v^2= v0^2-2a(y-y0)
    v^2= (8.37)^2 - 19.6( 235)
    v= 67 m/sec and not 97m/sec

    what am I doing wrong?

    I apologize for askin you so many qustions.. got a test tomorrow so am trying to make sure I understand the homework probs..

    THanks much.
     
  16. Sep 21, 2005 #15

    Doc Al

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    y-y0 = -235
     
  17. Sep 21, 2005 #16
    the answer still remains 68 m/sec though unless im doing something wrong again.. doh..
     
  18. Sep 21, 2005 #17
    ahh I see.. there's probably an error in the text book then...

    Once again, thanks much :) .. appreciate your help..
     
  19. Sep 21, 2005 #18

    Doc Al

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    It's not an error in the book. 68 m/s is just the vertical component of the velocity. You have to also include the horizontal component to get the total speed.
     
  20. Sep 21, 2005 #19
    Ahh I see.. alright cool... thanks..
     
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