plane 1 flies at 6 km/min
plane 2 flies at 8 km/min
so the distance z at time t minutes after plane 1 passes overhead is
z^2 = (6t)^2 + (8(t-10))^2 = 100t^2 - 1280t + 6400
2z dz/dt = 200t - 1280
dz/dt = 0 at t=6.4 min
since t < 10, what does that tell you?
An airplane flying east at 360 km/h passes over Pearson airport 10 minutes before a second airplane flying south at 480 km/h passes over the same point. Assuming that both airplanes are at the same attitude, at what time is the distance between them a minimum? Round the final answer to the nearest minute.
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1 answer