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A plane flies 457 km east from city A to city B

  1. Jun 15, 2008 #1
    1. The problem statement, all variables and given/known data

    A plane flies 457 km east from city A to city B in 41.0 min and then 949 km south from city B to city C in 1.30 h. For the total trip, what are the (a) magnitude (in km) and (b) direction of the plane's displacement, the (c) magnitude (in km/h) and (d) direction of its average velocity, and (e) its average speed (in km/h)? Give your angles as positive or negative values of magnitude less than 180 degrees, measured from the +x direction (east).

    3. The attempt at a solution

    Part A

    457cos(0) + 949sin(270)

    = 457i + -949j
    lRl=sqrt[(457)^2+(-949)^2]
    R=1053.3 km

    Part B

    arctan(-949/457) = -64.29 degrees

    I am not sure if either of those are correct and as for the rest of the problem I need help on how to solve for those things.
     
  2. jcsd
  3. Jun 15, 2008 #2
    Never mind everyone I finally got it

    (c) Average velocity = Total displacement/total time
    = 1053 km/(41+78)/60 = 530.9 km/h
    (d) Direction is the direction of displacement; - 64.3 degrees

    (e) Average speed = total distance/total time
    = (457 + 949) km/1.9833 h = 709 km/h
     
  4. Jun 15, 2008 #3
    You have the distance traveled and the time for each direction can't you calculate the velocity of each vector?
     
  5. Jun 15, 2008 #4
    yea I got it now thanks for your help
     
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