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I have tryed this problem 40 times and cant solve the last 2 parts

  1. Sep 29, 2008 #1
    At one instant a bicyclist is 27.0 m due east of a park's flagpole, going due south with a speed of 14.0 m/s. Then 31.0 s later, the cyclist is 27.0 m due north of the flagpole, going due east with a speed of 14.0 m/s. For the cyclist in this 31.0 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.)

    (a)Number 38.1 Units m corrrect
    (b)Number 135 Units ° (degrees) correct
    (c)number 1.23 Units m/s correct
    (d)Number 135 Units ° (degrees) correct
    (e) Number Units__________ m/s^2 ___ need help
    (f) Number Units____________° (degrees) need help
  2. jcsd
  3. Sep 29, 2008 #2


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    Homework Helper

    Welcome to PF.

    Think about what Acceleration is.

    Isn't it the Change in Velocity over time? In this case it is the change in the Velocity vector divided by time. You have the final Velocity vector of 14 East and you subtract the initial Velocity vector 14 South. The negative of the initial is 14 North, so your vector addition should be adding together the 14 East Vector with a 14 North vector. You find the magnitude (and divide by the time of course) as you did in part c) and direction as you did in part d)
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