A bird watcher meanders through the woods, walking 0.744 km due east, 0.925 km due south, and 3.97 km in a direction 18.3 ° north of west. The time required for this trip is 1.765 h. Determine the magnitudes of the bird watcher's (a) displacement and (b) average velocity.

Answer 1

a. D=0.744[0o] + 0.925[270o] + 3.97[161.7o] CCW.

X = 0.744+0.925*Cos270+3.97*Cos161.7 =
0.744 + 0 - 0.949 = - 0.205 km.

Y = 0 + 0.925*sin270 + 3.97*sin161.7 =
0 - 0.925 + 1.25 = 0.325 km.

Tan A = Y/X = 0.325/-0.205 = -1.58537.
A = -57.8o = 57.8o N. of W. = 122.2o CCW

D=Y/sinA=0.325/sin122.2=0.384 km[57.8o N
of W.] = Displacement.

b. V = (d1+d2+d3)/T = (0.744+0.925+3.97)/1.765=
3.19 km/h.