Question

A motorist drives south at 40.0 mi/h for 6.00 min, then turns west and travels at 50.0 mi/h for 9.00 min, and finally travels northwest at 60.0 mi/h for 5.00 min. For this trip, find the magnitude of the displacement vector, in miles. (Assume east is the positive x direction, and north is the positive y direction.) Answer: Check Determine the average speed, in m/s. Answer: Check

Answer 1

d_1 = (40 mi/h) (6/60 h) = 4 mi d_2 = (50 mi/h) (9/60 h) = 7.5 mi d_3 = (60 mi/h) (5/60 h) = 5 mi d_x = d_2 + d_3 cos 45 degree, west = 11.04 mi d_y = d_1 - d_3 sin 45 degree, south = 0.464 mi Vertical-bar d Vertical-bar = squareroot d^2_x + d^2_y = 11.05 mi Total distance = 4 + 7.5 + 5 = 16.5 mi: t = 6 + 9 + 5 = 20 mim = 0.33 h Therefore Avg. speed = Total distance/total time = 16.5/0.33 = 49.5 mi/h = 49.5 times 0.447 m/s = 22.1 m/s