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Motion in a Plane question dealing with circular motion

  1. Nov 8, 2004 #1
    A police cruiser chasing a speeding motorist travelled 60 km south, then 35 km northeast, and finally 50km west.

    a) calculate the total displacement of the cruiser
  2. jcsd
  3. Nov 8, 2004 #2
    make a diagram of the cruiser's displacement, the answer should be pretty straight forward.
  4. Nov 8, 2004 #3
    LOL but I did make a diagram....I got the answer from the book just don't know how it was done.....the answer is 43km s36w
  5. Nov 8, 2004 #4
    so your diagram should give you a straight line down, then a line heading towards the first quadrant, which stops in the 4th quadrant and heads due west to end in the 3rd quadrant. you know the final and initial position. the direction is determined from the angle between the final position and the y axis.

    hopefully this helps.
  6. Nov 8, 2004 #5
    yes now its a little clearer thanks......just the last thing is that I'm not sure what the length of the sides are to get the displacement...i think i have to use pythagorean theorem and i got 29.... which is wrong.....somehow i got the angle to be 90. Thanks for your patience but this is maybe too hard for me.....i'm getting frustrated :)
  7. Nov 8, 2004 #6
    don't get frustrated, if you pursue it you'll start to think differently about these problems. one day it all clicks. i'm majoring in physics and i'm still getting there myself.

    draw out the diagram. once you've your final coordinate you've got yourself a vector. make a triangle out of it and use sin, cos, or tangent to figure out the angle. 90 degrees is a right angle. /| your triangle will look like that. you're looking at the top angle, not the bottom one. the bottom is 90, and the top is 36 from what you said your answer is. try working out the sin or cos of 36, and see what the other legs of the triangle have to be. do they match what you drew?
  8. Nov 8, 2004 #7
    I appreciate your encouraging words.....
    I got the picture you drew /|
    I will choose to use sin theta= Opposite/Hypotenuse
    but I don't know the values of O and H
  9. Nov 8, 2004 #8
    just wondering but is there any benefits for you helping me out? Or are you just kind enough to do so?
  10. Nov 8, 2004 #9
    well, if you know the hypotenuse (magnitude of the vector)

    sin theta = o/h

    o = h sin theta : simliarly,
    a = h cos theta
    you do know the sides of the triangle if you graph it out though. you'd be at a position (x,y) where x is the base and y is the heigth.

    you know the answer is supposed to be 36 deg, so try that out and see what the distance in the x direction and y direction is.

    so by using the inverse function you can determine the angle theta.
  11. Nov 8, 2004 #10
    k thanks i got the stupid answer but I had to get out my ruler and do it very precisely by the cm.........is there any other better way......i'm not sure if were actualy suppost to get the answer from using a ruler and protractor....i mean what if there are mistakes......
  12. Nov 8, 2004 #11
    The easy way's to make one big triangle out of the 3 vectors you're given.

    Define the positive y direction as North, the positive x direction as East, so South is negative y and West is negative x.

    60km South, so we got y = -60km and x = 0km.

    Then 35km Northeast, so now we gotta break this one up into the vertical and horizontal components using a triangle. We got a hypotenuse of 35km and an angle of 45 degrees, so the x component is 35*cos(45) and the y component is 35*sin(45). Notice that these are both positive, since North is positive y and East is positive x. So now we add these two to our previous values of x and y.

    y = -60km + 35km*sin(45)
    x = 0km + 35km*cos(45)

    Now add the 50km west, which is an x of -50km.

    y doesn't change, so it's still -60km + 35km*sin(45)
    x adds -50, so we get x = 35km*cos(45) - 50km

    So we have a triangle with y component 35km*sin(45) - 60km, and an x component of 35km*cos(45) - 50km. Note that these are both numbers, although they may be negative or positive. Be sure to keep track of the sign. This is enough to tell you everything about the triangle, including the angle and the hypotenuse.

    By the way, what's this got to do with circular motion?

  13. Nov 8, 2004 #12
    Thats awesome dude thank you
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