Posted by Robin on Sunday, February 9, 2014 at 6:16am.
A train at a constant 41.0 km/h moves east for 28 min, then in a direction 53.0° east of due north for 15.0 min, and then west for 64.0 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?

Physics  Henry, Tuesday, February 11, 2014 at 10:13pm
41km/h[0o], 28min.
41km/h[37o], 15min.
41km/h[180o], 64min
a. d1=41km/h[0o] * (28/60)h=19.13km[0o]
d2=41km/h[37o] * (15/60)h=10.25km[37o]=
8.19 + i6.17
d3=41km/h[180o] * (64/60)h=43.73km[180o]
V=(d1+d2+d3)/(t1+t2+t3)=(19.13+8.19+i6.1743.73)/(28+15+64) =
(16.41 + i6.17)/107=17.53[159.4o]/107 =
0.16384km/min = 9.83 km/h.
b. Tan Ar = Y/X = 6.17/16.41 = 0.37599
Ar = 20.61 = Reference angle.
A = 20.6 + 180 = 159.4o, CCW. = 69.4o
West of North.
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