Question

Please help and explain I have no idea it is math homework

Answer 1

First, find the net speed of the plane in the return flight.

Use the distance of the first half of the trip (510 mi) and the speed during the first half (255 mi/h) to calculate the time that it takes for the plane to travel the first 510 miles:

`\begin{gathered} v=(d)/(t) \\ \\ \Rightarrow t=(d)/(v)=\frac{510\text{ mi}}{255(mi)/(h)}=2h \end{gathered}`

It takes two hours for the plane to complete the first half of the trip. Since the round-trip took 3.9 hours to complete, then, the time that it took for the return trip to complete was:

`3.9h-2h=1.9h`

Find the net speed of the plane during the return trip if it travels 510 miles in 1.9 hours:

`v=(d)/(t)=(510mi)/(1.9h)=268.42...(mi)/(h)`

Since the speed is greater than the speed of the plane in still air, the air is blowing in the same direction that the plane is moving (it is a tailwind). Subtract the speed of the plane in still air from the net speed of the plane to find the speed of the wind:

`v_(wind)=268.42(mi)/(h)-255(mi)/(h)=13.42(mi)/(h)`

Therefore, the answers are:

a) To the nearest tenth, the air in the return flight is 13.4 miles per hour.

b) The wind on the return flight is a tailwind.