Atrain moving at a constant seeed of S0.0 km/h mewes east for 3s.0 min, thenin·d
ID: 1873780 • Letter: A
Question
Atrain moving at a constant seeed of S0.0 km/h mewes east for 3s.0 min, thenin·dre ton 4so. ea t of due north for 23e hat· average velocity of the train during this run? (uet the -xi point towards the east and the y-axis point sowards the north. magntude direction to caiculete an aversge velacity, divide the displacement by the tse taken. That is, findtanitude of the displacement vector peinting from start to neish and then dvide by the time (Den't divide the s is the vector thet to nish it is denermined s in chepter a: Pind the net disacement pereliel to the for 250 0 min, and then west for 75.0 min, w km/h counterclockwise from the axis etal distance by the sime taken. ) The dsplacement --tan-asto find the megstide of ne vector and an n enewsent tornd an angle. Hnport th" !Explanation / Answer
Solution) Given velocity v=50km/h east for 35min
41°east of north for 25 min
West for 75 min
Let us calculate displacement along x and y directions
Along x
Displacement x=(50/60)(35)+(50/60)(25)sin(41°)+(50/60)(-75)
X= -19.66m
Similarly along y direction
Y=(50/60)(25cos(41°))
Y=15.72m
Net displacement D=(X^2 + Y^2)^(1÷2)
D=25.17
Total time T=((35+25+75)/60)
T=2.25
Average velocity = Total displacement/total time
V=25.17/2.25
V=11.18km/h
Direction theeta=tan inverse(15.72/(-19.66))
Theeta=-38.64°
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