Question

This is a 3 part 1 question each part on how to locate the plane and a small explanation report working out the recovery details (Vectors) YOU WILL ONLY GET THE 20 POINTS IF YOU ANSWER THE QUESTIONS (If we get through this entire section: I will write a very long letter of gratitude and give you the highest rating for helping me understand. I truly just want to finish) There are only 3 steps, so this should be easy, but I need help! (Total of questions: 4) we must include an explanation on how we receive these answers. I’ve worked with a tutor earlier and progressed far, can you help me with the rest? HERE IS WHAT I HAVE SO FAR: I’ll include picturesFIRST LEG:-> R= (359.84,191.48) , which is from the origin, the vector that represents the flight path of the plane during the first leg. SECOND LEG: -> R=(576.5,359.15) , which is from the origin, the vector that represents the direct route to the location of the plane after the second leg of the flight. I’ll include a picture of the recovery details which was the last question NOW that that’s out of the way, HERE IS WHAT I NEED MOVING FORWARD: (Please include a small explanation, so I can read it and understand what is going on.) 3rd leg :- the wind maintained direction, but increased to 35 mph -the crew changed their direction to N-20degrees-E and maintained speed. - the flight lasted 20 minutes. 1st question: from the origin, what is the vector that represents the direct route to the location of the plane after the third flight? This is the final location of the plane. (Now that you have the location of the plane, calculate how far away the plane is from take off, and in what direction. Calculate the magnitude of the vector and its direction) 2nd question : How far from the take - off point, the origin , is the plane? please show how you got this with an explanation and full work 3rd question: What is the direction from take -off , the origin, to the FINAL location of the plane? Please show me how you did this with an explanation and work, so I can understandThe LAST short QUESTION: We need to write a small report and It must include : •the explanation of the events •How it connects to the work given in the questions •A complete explanation of all of our work if we can connect that into the report. (If you made it this far, I would like to thank you for your service and patience with my tedious questions. Thank you for granting me the ability to see the understanding of these questions. You are greatly appreciated) 

Answer 1

Third leg.

The crew flies at a speed of 560 mi/h in direction N-20°-E.

The wind has a speed of 35 mi/h and a direction S-10°-E.

We then can draw this as:

We have to add the two vectors to find the actual speed and direction.

We will start by adding the x-coordinate (W-E axis):

$\begin{gathered} x=560\cdot\sin (20\degree)+35\cdot\sin (10\degree) \\ x\approx560\cdot0.342+35\cdot0.174 \\ x\approx191.53+6.08 \\ x\approx197.61 \end{gathered}$

and the y-coordinate (S-N axis) is:

$\begin{gathered} y=560\cdot\cos (20\degree)-35\cdot\cos (10\degree) \\ y\approx560\cdot0.940-35\cdot0.985 \\ y\approx526.23-34.47 \\ y\approx491.76 \end{gathered}$

Then, the actual speed vector is v3=(197.61, 491.76).

The starting location for the third leg is R2=(216.66, 167.67) [taken from the previous answer].

Then, we have to calculate the displacement in 20 minutes using the actual speed vector.

We can calculate the movement in each of the axis. For the x-axis:

$\begin{gathered} R_{3x}=R_{2x}+v_{3x}\cdot t \\ R_{3x}=216.66+197.61\cdot\frac{1}{3} \\ R_{3x}=216.66+65.87 \\ R_{3x}=282.53 \end{gathered}$

NOTE: 20 minutes represents 1/3 of an hour.

We can do the same with the y-coordinate:

$\begin{gathered} R_{3y}=R_{2y}+v_{3y}\cdot t \\ R_{3y}=167.67+491.76\cdot\frac{1}{3} \\ R_{3y}=167.67+163.92 \\ R_{3y}=331.59 \end{gathered}$

The final position is R3 = (282.53, 331.59).

To find the distance from the origin and direction, we transform the cartesian coordinates of R3 into polar coordinates:

The distance can be calculated as if it was a right triangle:

$\begin{gathered} d^2=x^2+y^2_{} \\ d^2=282.53^2+331.59^2 \\ d^2=79823.20+109951.93 \\ d^2=189775.13 \\ d=\sqrt[]{189775.13} \\ d\approx435.63 \end{gathered}$

The angle, from E to N, can be calculated as:

$\begin{gathered} \tan (\alpha)=\frac{y}{x} \\ \tan (\alpha)=\frac{331.59}{282.53} \\ \tan (\alpha)\approx1.1736 \\ \alpha=\arctan (1.1736) \\ \alpha=49.56\degree \end{gathered}$

If we want to express it from N to E, we substract the angle from 90°:

$\beta=90\degree-\alpha=90-49.56=40.44\degree$

Answer: the final location can be represented with the vector (282.53, 331.59).

1) The distance from the origin is 435.63 miles and

2) the direction is N-40°-E.