Asailor in a small sailboat encounters shifting winds. she sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. her final position is 5.80 km directly east of the starting point (fig. p1.72). find the magnitude and direction of the third leg of the journey. draw the vectoraddition diagram and show that it is in qualitative agreement with your numerical solution.

Third displacement = 2.81 m which is 61.70° north of east.

Explanation:

   Let east represents positive x- axis and north represent positive y - axis. Horizontal component is i and vertical component is j.

  She sails 2 km east, displacement = 2 i

  Then 3.50 km southeast, means 3.5 km 315⁰ to + ve X axis

  Displacement = 3.5 cos 315 i + 3.5 sin 315 = 2.47 i - 2.47 j

  Let third displacement be x i + y j

  We have final displacement = 5.80 km east = 5.80 i

  From summation we have total displacement = 2 i + 2.47 i - 2.47 j + x i + y j

                                                                           = (4.47+x) i + (y - 2.47) j

  Comparing both , we have 4.47+x = 5.80

                                                        x = 1.33

                                                 y-2.47=0

                                                         y = 2.47 j

  So third displacement = 1.33 i + 2.47 j

  Magnitude of third displacement =

  θ = tan⁻¹(2.47/1.33) = 61.70°

  So third displacement = 2.81 m which is 61.70° north of east.

i think if it were me i whould divide 300 dide 15

step-by-step explanation:

true tan speed = ang vel x radius