[sheila1983]

If an airjet has 80GJ of energy available for a 4000km trip, what is the shortest time that the trip can be made?
Not sure how to answer this, also given that the jet has a mass of 7500kg, a maximum thrust of 30,000 Newtons and a maximum velocity of 237 metres per second.

[ebaines]

This is going to depend very much on assumptions you make regarding air resistance. If the plane must use 30,000 N of thrust to maintain a velocity of 237 m/s then it cannot make it the entire 4,000 Km distance at full power. Therefore it must fly slower than 237 m/s, and now you need to make an assumption as to how airspeed varies with thrust. For example perhaps you assume that airspeed is directly proportional to thrust - in this case you can simply use W=Fd to determine the thrust you can maintain, then from that determine the airspeed, and finally time required. If gets a bit more complicated if you include the energy required to accelerate the plane to its cruising speed using KE = (1/2)mv^2, and to lift the plane to its cruising altitude using PE = mgh.

[sheila1983]

This is going to depend very much on assumptions you make regarding air resistance. If the plane must use 30,000 N of thrust to maintain a velocity of 237 m/s then it cannot make it the entire 4,000 Km distance at full power. Therefore it must fly slower than 237 m/s, and now you need to make an assumption as to how airspeed varies with thrust. For example perhaps you assume that airspeed is directly proportional to thrust - in this case you can simply use W=Fd to determine the thrust you can maintain, then from that determine the airspeed, and finally time required. If gets a bit more complicated if you include the energy required to accelerate the plane to its cruising speed using KE = (1/2)mv^2, and to lift the plane to its cruising altitude using PE = mgh.

OK, so I calculated the force on the plane using W = F d, so that (80 x 10^9 Joules = F x 4000,000m) which gives a force of 20,000 Newtons.
Then, as you suggested, assuming that the airspeed is directly proportional to thrust, if the maximum thrust of 30,000 Newtons enables the plane to reach a maximum velocity of 237 metres per second, then 20,000 Newtons would give a velocity of about 158 metres per second (237 x two thirds).
Then, using time = distance divided by speed, t = 4000,000 m / 158 ms-1 = approximately 7 hours for the shortest time to make the trip.
Not sure if my calculations are entirely correct, but this does make a lot more sense now. It would be easy to ignore the energy value given in the question and just calculate the shortest time using the maximum velocity value of 237 metres per second. Then, you would get t = 4000,000 m / 237 ms-1 = approximately 4.7 hours for the shortest time to make the trip. However, then you would be assuming that the force used is the maximum thrust for the entire journey, which would be 30,000 Newtons. However, you could see from the equation for W = Fd, that you would need 1.2 x 10^11 Joules of energy to do it in this time, since W = (30,000 N x 4000,000 m) = 1.2 x 10^11 Joules.
So the W = Fd makes sense of it all. Thanks.