The_Postmaster wrote:
"Long flights at midlatitudes in the Northern Hemisphere encounter the jet stream, an eastward airflow that can affect a plane's speed relative to Earth's surface. If a pilot maintains a certain speed relative to the air (the plane's airspeed), the speed relative to the surface (the plane's ground speed) is more when the flight is in the direction of the jet stream and less when the flight is opposite the jet stream. Suppose a round-trip flight is scheduled between two cities separated by 4700 km, with the outgoing flight in the direction of the jet stream and the return flight opposite it. The airline computer advises an airspeed of 1010 km/h, for which the difference in flight times for the outgoing and return flights is 68 min. What jet-stream speed (in km/h) is the computer using?"
This is a problem from the assignment my physics teacher gave us, and I haven't been able to figure it out. Could anyone explain how to find the answer?
This is an rt=d type of problem where you have two trips of the same distance. In one direction, the rate is v
airplane-v
jetstream while it is v
airplane+v
jetstream in the other with v
airplane=1010 km/hr.
Furthermore, it will take t hours traveling with the jet stream and t+68/60 hours traveling against the jet stream.
Thus, you have two equations
(1010 km/hour+v
jetstream)*t=4700 km
and
(1010 km/hour-v
jetstream)*(t+68/60 hours)=4700 km
From the first, you can get
t=(4700 km)/(1010 km/hour+v
jetstream).
Substituting that into the second equation (and removing the units), you should have
(1010-v
jetstream)*(68/60+4700/(1010+v
jetstream))=4700
Solve for v
jetstream. It's a messy calculation, so be careful.
Make sure you check your results. For the number you get for v
jetstream, plug that into the first equation to get t. Then plug both v
jetstream and t into the second equation to make sure that the distance is 4700 km.
14 Oct 2012, 11:51 pm
