Draw a diagram, where the train starts at (0,0). The speed makes it easy to convert from minutes to km. The final location is
(40,0)+(10√2,10√2)+(-50,0) = (10√2-10,10√2)
So, just convert that location to distance and bearing from (0,0).
A train with a constant speed of 60km/h goes east for 40 minutes. Then it goes
45◦ north-east for 20 minutes. And finally it goes west for 50 minutes. What
is the average velocity of the train?
2 answers
Displacement = V*T1 + V*T2 + V*T3.
Disp. = 60*(40/60) + 60*(20/60)[45o] + (-60(50/60),
Disp. = 40 + (14.14+14.14i) + (-50) = 4.14 + 14.14i = 14.73km[73.7o].
T = T1+T2+T3 = 40 + 20 + 50 = 110 Min. = 1.833 Hours.
V = Disp./T = 14.73km[73.7o]/1.833 = 8.03km/h[73.7o].
Disp. = 60*(40/60) + 60*(20/60)[45o] + (-60(50/60),
Disp. = 40 + (14.14+14.14i) + (-50) = 4.14 + 14.14i = 14.73km[73.7o].
T = T1+T2+T3 = 40 + 20 + 50 = 110 Min. = 1.833 Hours.
V = Disp./T = 14.73km[73.7o]/1.833 = 8.03km/h[73.7o].
2 answers