d1 = 61km/h[0o] * (35/60)h = 35.6km[0o].
d2=61km/h[40o] * (10/60)h=10.2km[40o].
d3=61km/h[180o] * (49/60)h=49.8km[180o].
T = (35+10+49)min * 1h/60min = 1.57 h.
X=35.6+10.2*cos40+49.8*cos180=-6.39km.
Y = 10.2*sin40 + 49.8*sin180 = 6.56 km.
tan Ar = Y/X = 6.56/-6.39 = -1.02605
Ar = -45.7o = Reference angle.
A = -45.7 + 180=134.3o, CCW.=Direction
d = Y/sin A = 6.56/sin134.3 = 9.17 km.
V = d/t = 9.17/1.57 = 5.84 km/h.
A train moving at a constant speed of 61.0 km/h moves east for 35.0 min, then in a direction 50.0° east of due north for 10.0 min, and then west for 49.0 min. What is the average velocity of the train during this run?
What is the magnitude of that velocity?
What is the direction (counterclockwise from east)?
1 answer
1 answer