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?
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 =
To solve this problem, we need to break it down into different parts and use the given information to calculate the time for each leg of the trip.
Let's denote the time for the eastbound trip (from SFO to YUL) as "t" hours. Since the westbound trip (from YUL to SFO) takes 51 more minutes than the eastbound portion, the time for the westbound trip can be expressed as "t + 0.85" hours.
According to the problem, the total time for the round trip, including the layover, is 13.9 hours. So, we can set up the equation:
t + (t + 0.85) + 0.81 = 13.9
2t + 1.66 = 13.9
2t = 12.24
t = 6.12
Now that we have the time for the eastbound trip (SFO to YUL) as 6.12 hours, we can calculate the time for the westbound trip (YUL to SFO):
t + 0.85 = 6.12 + 0.85 = 6.97 hours
Therefore, the time for each leg of the trip is approximately 6.12 hours for the eastbound trip and 6.97 hours for the westbound trip.
To calculate the average speed of the overall trip, we need to divide the total distance traveled (4100 km) by the total time taken (13.9 hours):
Average speed = Total distance / Total time
Average speed = 4100 km / 13.9 hours ≈ 294.96 km/h
Now, to calculate the average speed without the layover, we need to subtract the time for the layover (0.81 hours) from the total time taken (13.9 hours) and then divide the total distance traveled by the updated total time:
Updated total time = Total time - Layover time = 13.9 - 0.81 = 13.09 hours
Average speed without layover = Total distance / Updated total time
Average speed without layover = 4100 km / 13.09 hours ≈ 313.03 km/h
Therefore, the average speed of the overall trip is approximately 294.96 km/h, and the average speed without the layover is approximately 313.03 km/h.