Asked by Alyssa
A jet takes off from SFO (San Francisco, CA) and flies to YUL (Montréal, Quebec). The distance between the airports is 4100 km. After a 0.81-h layover, the jet returns to San Francisco. The total time for the round-trip (including the layover) is 13.9 h. If the westbound trip (from YUL to SFO) takes 51 more minutes than the eastbound portion, calculate the time for each leg of the trip. What is the average speed of the overall trip? What is the average speed without the layover?
Answers
Answered by
Henry
13.9h - 0.81h = 13.09 h. = Travel time.
EB trip = T Hours.
WB trip = (T+51/60) Hours.
T + (T+0.85) = 13.09 h
T + T+0.85 = 13.09.
2T = 12.24.
T = 6.12 h. = EB trip.
T+0.85 = 6.97 h. = WB trip.
a. Vavg=Dt/Tt=(2*4100)/13.9 = 590 km/h.
b. Vavg = (2*4100)/13.09 =
EB trip = T Hours.
WB trip = (T+51/60) Hours.
T + (T+0.85) = 13.09 h
T + T+0.85 = 13.09.
2T = 12.24.
T = 6.12 h. = EB trip.
T+0.85 = 6.97 h. = WB trip.
a. Vavg=Dt/Tt=(2*4100)/13.9 = 590 km/h.
b. Vavg = (2*4100)/13.09 =
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