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Homework Help: What is the magnitude of the runner's total displacement?

  1. Sep 16, 2005 #1
    A football player runs directly down the field for 35 m before turning to the right at an angle of 25 degrees from his original direction and running an additional 15 m before getting tackled. What is the magnitude of the runner's total displacement?

    I got 38.08 but it is not right can someone tell me how to do it?
  2. jcsd
  3. Sep 16, 2005 #2
    One vector pointing straight north has magnitude 35m, then another vector pointing 25 degrees to the right of north (65 north of east) has magnitude 15m, find the vector sum.
  4. Sep 16, 2005 #3
    Is the vector sum 50 meters?How do you get it?
    Last edited: Sep 16, 2005
  5. Sep 16, 2005 #4


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    Science Advisor

    Perhaps drawing the situation might help.

    Then find the over all change in position that occured as if he ran in a straight line the entire distance.
  6. Sep 16, 2005 #5
    You might want to break down components, your vectors looks like this

    [tex] \vec{x_1} = <35,0> [/tex]

    [tex] \vec{x_2} = <15\cos(25), 15\sin(25)> [/tex]
  7. Sep 16, 2005 #6
    Thank you I just got the answer right (49)
  8. Sep 17, 2005 #7


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    Science Advisor

    That can't be the correct answer, as you phrased the problem. "Displacement" is a vector, not a number so no number could be the correct answer. 49 m (not just "49") might well be the magnitude of the displacement.
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