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+i6.17-43.73)/(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.
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?
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1 answer