A train at a constant 41.0 km/h moves east for 38 min, then in a direction 44.0° east of due north for 25.0 min, and then west for 59.0 min. What are the (a) magnitude (in km/h) and (b) angle (relative to north, with east of north positive and west of north negative) of its average velocity during this trip?

Answer 1

Vavg. = V1 + V2 + V3,
Vavg.=41km/h @ 0Deg + 41km/h @ 46Deg. +
41km/h @ 180Deg.

X = 41*cos(0) + 41*cos46 + 41*cos180 =
28.5 km/h.
Y = 41*sin(0) + 41*sin46 + 41*sin180 =
29.5 km/h.

tanA = Y/x = 29.5 / 28.5 = 1.03484,
A = 46 Deg.,CCW=44 Deg. East of North.

V = X/cosA = 28.5 / cos46 = 41 km/h.

V = 41 km/h @ 44 Deg. East of North

Answer 2

CORRECTION:

See later post: Sun,2-9-14, 6:16 AM.