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Homework Help: Motion in Two or Three Dimensions

  1. Sep 5, 2007 #1
    1. The problem statement, all variables and given/known data
    At one instant a bicyclist is 26.0 m due east of a park's flagpole, going due south with a speed of 14.0 m/s. Then 35.0 s later, the cyclist is 26.0 m due north of the flagpole, going due east with a speed of 14.0 m/s. For the cyclist in this 35.0 s interval, what are the (a) magnitude and (b) direction of the displacement, the (c) magnitude and (d) direction of the average velocity, and the (e) magnitude and (f) direction of the average acceleration? (Give all directions as positive angles relative to due east, where positive is measured going counterclockwise.)

    3. The attempt at a solution

    I honestly don't know where to begin. Would the magnitude =sq rt. (26^2+26^2)?

    Any help would be greatly appreciated.
  2. jcsd
  3. Sep 5, 2007 #2


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    yes, but what is the reasoning.

    Try to develop the coordinate system, e.g. 26 m due E is what in a Cartesian system, and 26 m due N is what? Assume the flagpole is the center (origin) of the coordinate system.

    If one is E going due S, and N going due E, what trajectory does that describe if acceleration is constant?

    Hint -
  4. Sep 5, 2007 #3
    I'd say that 26m due E just (26,0) and 26m due N is (0,26). That I get, but being at (26,0) heading South and then 35sec later being at (0,26) heading East doesn't make a lot of sense to me. Trajectory would be SE?

    The reasoning behind the magnitude is that the position changed from (26,0) to (0,26). and the distance between them is the square root of their sums squared.
  5. Sep 5, 2007 #4


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    Think about traveling clockwise. If one is initially traveling south, how does one end up north traveling east?
  6. Sep 5, 2007 #5
    Traveling in a circle.
    (a) I get 36.8m
    (b) I thought would be 45 since the arctan of (26/26) = 45?
    (c)1.05 m/s
    (d) I'm not sure how to go about calculating it.
    (e) Divide avg velocity by 35 sec?
    (f) 45 degrees?
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