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Homework Help: Really fast bicyclist in the park

  1. Sep 19, 2008 #1
    I'm stuck at part b on this problem. ive been working at it for an hour at least! lol

    1. The problem statement, all variables and given/known data

    At one instant a bicyclist is 50 m due east of a park's flagpole, going due south with a speed of 7 m/s. Then, 24 s later, the cyclist is 70 m due north of the flagpole, going due east with a speed of 20 m/s. For the cyclist in this 24 s interval, find each of the following.

    (a) displacement
    magnitude: ? m
    direction: ? mark° north of due west

    (b) average velocity
    magnitude: ? m/s
    direction: ? ° north of due west

    (c) average acceleration
    magnitude: ? m/s2
    direction: ? ° north of due east

    2. Relevant equations

    [tex]\vec{}r[/tex] = [tex]\sqrt{}((r\hat{i})^{2} + (r\hat{j})^{2})[/tex]

    3. The attempt at a solution

    alright so for part (a) i used 50\hat{i} and 70\hat{j} in the equation above to get 86.023m
    then inversetan (70/50) to get 54.462 degrees. i thought that 54.462 was the angle for norht of due east, but its the correct answer. It would be nice if someone could explain that to me.

    then with part (b) im completely lost, i tried plugging in the 7m/s and the 20m/s the same way as i did for the displacement but got 21.190 degrees. thats wrong and now iv been stuck forever it seems. any help would be much appreciated. thanks!
  2. jcsd
  3. Sep 19, 2008 #2


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    Try drawing a figure with 3 points:
    A point representing the flagpole.
    Point #1, about 5 inches to the right (east) of the flagpole.
    Point #2, about 7 inches above (north of) the flagpole.

    Then draw an arrow from point 1 to point 2, to represent the displacement. What direction does that arrow point?

    You can use the displacement, and the elapsed time, to get average velocity.
  4. Sep 19, 2008 #3
    I see the point 1 to point 2 thing now, arrow points northwest.
    the one i drew before posting here went in the opposite direction (northeast) my point 2 was 7 inches right above where my 5 inch (east) mark was, and i drew an arrow from the flagpole to that. oops. thanks for clarifying that.

    so displacement divided by t= 24s would be 3.584 m/s!! cool
    since the Vav vector is a scalar of the displacement vector, the angle is the same, ya!?

    So doing the same thing to find Aav as i did to find Vav gives me 0.883 m/s[tex]^{2}[/tex]
    but i dont understand how thats right. I i dont know how to represent it graphically.
  5. Sep 20, 2008 #4
    am i allowed to bump this? :)
  6. Sep 20, 2008 #5


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    Sounds right. Graphically, use the same procedure as before, except you use the velocities instead of the displacements: draw points representing 7 units south, then 20 units east.
  7. Sep 20, 2008 #6
    Thats where im confused. is it 7 units south and 20 east, both starting at the flagpole (origin)? or 7 south from the 50 meter mark, and 20 east from the 70m mark?
    the wording on this problem is pwning me. lol
  8. Sep 20, 2008 #7


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  9. Sep 21, 2008 #8
    cool. thanks for helping me redbelly!
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