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Old Feb 2nd 2017, 03:03 PM   #1
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Question Jet making round trip

Problem: A jet takes off from airport A and flies to airport B. The distance between the airports is 571.1 miles(903). After a 1.00hr layover, the jet returns to airport A. The total time for the round trip (including layover) is 3.50hr. The southbound trip from airport B to airport A takes 20.0min more than the northbound trip from airport A to airport B.

A) calculate the time required for each leg of the trip and the average velocity for each leg of the trip.

B) calculate average speed for overall round trip

c) calculate average speed for overall round trip without layover.

I'm pretty lost with this problem. I'm not sure how I'm supposed to find the time for each leg of the trip when I don't have the velocity or acceleration. I just don't even know how to start this. Please help, thank you!
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Old Feb 2nd 2017, 03:22 PM   #2
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Originally Posted by SeanChrisXIV View Post
Problem: A jet takes off from airport A and flies to airport B. The distance between the airports is 571.1 miles(903). After a 1.00hr layover, the jet returns to airport A. The total time for the round trip (including layover) is 3.50hr. The southbound trip from airport B to airport A takes 20.0min more than the northbound trip from airport A to airport B.

A) calculate the time required for each leg of the trip and the average velocity for each leg of the trip.
Let $\displaystyle t_{AB}$ be the time to fly from A to B, and let $\displaystyle t_{BA}$ be the time to fly from B to A. Calling T the total trip time we have:
$\displaystyle T = 3.50 = t_{AB} + 1 + t_{BA}$

We also know that $\displaystyle t_{BA} = t_{AB} + 1/3$. (The 1/3 is 20 minutes in terms of hours.)

We now have two equations in two variables. Putting the equation for $\displaystyle t_{BA}$ (the second equation) into the first equation gives $\displaystyle 3.5 = t_{AB} + 1 + (t_{AB} + 1/3)$.

Can you go from there? If you can then see if you can finish the whole thing.

-Dan
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Old Feb 7th 2017, 04:03 PM   #3
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Right on, thank you for the help Topsquark!
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