How Does Wind Affect the Flight Path and Duration of an Aircraft?

[sunshinegirl]

I am doing a homework assignment for my General Engineering class and am really frustrated because I can't get it, and I remember doing this sort of problem in high school, but forget how to do it. Here it is

A plane flies (in still air) at 250 mi/hr. On Monday morning the plane leaves point A and flies due East for one hour and then flies due south for one hour to arrive at B. Throughout the wind velocity is 50 mi/hr to the east
a) What is the distance from B to A?
b) What is the heading (the direction the plane must point in) that the plane must fly to return to B?
c) How long would it take for the return flight?

I drew a nice little picture and figured out when it is flying east it will be flying 300 mi/hr with the wind, but I forget what to do when he is flying south and the wind is east. Any help would be greatly appreciated!

 

[enigma]

Add the two components.

If it changes direction to the south, and the wind blows it to the East, then the path it took isn't directly to the south.

All you need to do is add up the E-W component, and the N-S component, and then do Pythagoras to get the distance.

 

[wais]

First of all, don't worry, it's completely normal to forget certain concepts or formulas from high school. It's great that you're reaching out for help, that shows determination and a desire to learn.

To solve this problem, we need to use vector addition. Think of the plane's velocity as a vector, with the magnitude being its speed and the direction being its heading. In this case, the plane's velocity in still air is 250 mi/hr due east. Now, we need to add the wind velocity vector, which is 50 mi/hr due east.

To find the resultant velocity, we need to use the Pythagorean theorem. The resultant velocity will be the hypotenuse of a right triangle, with the two legs being the plane's velocity in still air and the wind velocity. So, the resultant velocity will be √(250^2 + 50^2) = √(62500 + 2500) = √65000 = 254.95 mi/hr.

Now, to solve for the distance from B to A, we can use the formula d = vt, where d is distance, v is velocity, and t is time. Since the plane flew for 2 hours (1 hour east + 1 hour south), the distance from B to A will be 254.95 mi/hr * 2 hours = 509.9 miles.

To find the heading the plane must fly to return to B, we need to use trigonometry. We know the opposite and adjacent sides of the triangle (50 mi/hr and 250 mi/hr respectively), so we can use the tangent function to find the angle. tan θ = opposite/adjacent = 50/250 = 0.2. Taking the inverse tangent of 0.2, we get θ = 11.31 degrees. So, the plane must fly at a heading of 11.31 degrees to return to B.

Finally, to find the time it would take for the return flight, we can use the same formula d = vt. Since we know the distance (509.9 miles) and the velocity (254.95 mi/hr), we can solve for t. So, t = d/v = 509.9 miles/254.95 mi/hr = 2 hours. Therefore, the return flight would take 2 hours.

I hope this explanation helps you understand the problem better