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Vector magnitude and displacement

  1. Sep 10, 2010 #1
    1. The problem statement, all variables and given/known data

    A car is driven east for a distance of 48 km, then north for 25 km, and then in a direction 25° east of north for 28 km. Draw the vector diagram and determine the total displacement of the car from its starting point.

    whats the magnitude and direction(counterclockwise from east)

    2. Relevant equations

    A^2=Ax^2 + Ay^2

    3. The attempt at a solution

    i plugged in 48km for Ax and 25km for Ax and got 82.103km for my answer
  2. jcsd
  3. Sep 10, 2010 #2
    you seem to be missing the contributions of the 28 km. Draw the diagram and attach it.
  4. Sep 10, 2010 #3
    i noticed that the 28km was left out, i drew it on my picture but im not sure what forumula to use to utilize it.
  5. Sep 10, 2010 #4
    Are you able to break down the 28 km into its east and north components? If you can do that, then add all the east components together, and then .... can you do it now?
  6. Sep 10, 2010 #5
    do you mean like:

    x= 28cos25
    y= 28cos65
  7. Sep 11, 2010 #6
    yes, except I think your angles should be interchanged. Then to get the overall displacement you have an x component of 48+28sin25 and a y component of .... Then use pythagoras on the two overall components. In future, try to not think of 'plugging into a formula' without thinking what it means.
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