Question

This is a 3 part 1 question each part on how to locate the plane and working out the recovery details. (Vectors) (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 pictures: FIRST LEG: -the plane said they were headed N- 60 degrees-E at a velocity of 550 miles per hour. -the wind was heading S-10degrees at 20 mph- the leg of the flight lasted 45 minutes. -> 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: -the wind maintained direction, but the speed increased to 25 miles per hour- As a response, the crew increased the velocity of the plane to 560 miles per hour and directed the plane to N- 50degrees-E -the trip lasted 30 minutes -> 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. 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 understandAnd FINALLY, Work Out the Recovery Details Now that you know the location of the plane, how far from take off it is, and the direction, it is time to plan to recover the cargo. On this flight, the plane was transporting a water turbine to a hydroelectric plant overseas. The turbine they were transporting weights 2 tons (4,000 pounds). They will send a ship to pull it out of the water. To do this, the ship lowers a ramp to the water level and will pull the turbine up that ramp, over rollers, with a winch. Before the ship can be sent out, we need to calculate how many pounds the winch will need to handle to pull the cargo up the ramp. (Due to the length restrictions on the ship, the ramp has to be steeper than usual. As a result, the ramp will be sitting at an angle of 33 degrees to the water.) 1st question: How many pounds will the winch need to pull with to move the cargo up the ramp? Show all of the work and give a small explanation. The 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

To find the displacement of the plane during the third leg (as a vector), multiply the velocity of the plane times the time that it travels at that velocity:

$\Delta\vec{r_3}=\vec{v}_3t$

The velocity of the plane with respect to the ground v_PG is equal to the velocity of the plane with respect to the wind v_PW, plus the velocity of the wind with respect to the ground v_WG:

$\vec{v_{}}_{PG}=\vec{v}_{PW}+\vec{v}_{WG}$

The direction of the wind is θ=280°, and its speed is 35mph:

$\begin{gathered} \vec{v}_{WG}=(35\text{mph}\cdot\cos (280),35\text{mph}\cdot\sin (280)) \\ =(6.078\text{mph,}-34.47\text{mph)} \end{gathered}$

The direction of the plane with respect to the wind is θ=70, and the magnitude of its velocity is the same as in the 2nd leg, which is 560mph. Then:

$\begin{gathered} \vec{v}_{PW}=(560\text{mph}\cdot\cos (70),560\text{mph}\cdot\sin (70)) \\ =(191.5\text{mph,}526.2\text{mph)} \end{gathered}$

Add both velocities to find the velocity of the plane with respect to the ground:

$\begin{gathered} \vec{v}_{PG}=(6.078mph,-34.47mph)+(191.5mph,526.2mph) \\ =(6.078mph+191.5mph,-34.47mph+526.2mph) \\ =(197.6mph,491.9mph) \end{gathered}$

Since 20 minutes is 1/3 of an hour, then the displacement during the third leg of the trip, is:

$\begin{gathered} \Delta\vec{r}_3=\frac{1h}{3}(197.6mph,491.9mph) \\ =(65.9\text{miles,}164\text{miles)} \end{gathered}$

1st question:

To find the position of the plane from the origin, add the displacement vector to the position before the 3rd leg started, which is:

$\vec{r_2}=(576.5miles,359.15miles)$

Then, the final position after the third leg, is:

$\begin{gathered} \vec{r_{}}=\vec{r}+\Delta\vec{r} \\ =(576.5miles,359.15miles)+(65.9\text{miles,}164\text{miles)} \\ =(576.5miles+65.9\text{miles,}359.15miles+164\text{miles)} \\ =(642\text{miles,}523\text{miles)} \end{gathered}$

Then, the final location of the plane, is:

$\vec{r}=(642\text{miles,}523\text{miles)}$

2nd question:

To find how far from the take-off point the plane is, calculate the magnitude of the vector r:

$\begin{gathered} |\vec{r}|=\sqrt[]{r_x+r_y} \\ =\sqrt[]{(642mi)^2+(523mi)^2} \\ =828\text{miles} \end{gathered}$

Therefore, the plane is 828 miles away from the take-off point.

3rd question:

To find the direction of the plane (the angle θ), remember that:

$\begin{gathered} \tan \theta=\frac{r_y}{r_x} \\ \Rightarrow\theta=\arctan (\frac{r_y}{r_x}) \\ =\arctan (\frac{523miles_{}}{642\text{miles}}) \\ =\arctan (\frac{523_{}}{642}) \\ =39^{\circ} \end{gathered}$

Therefore, the direction of the plane is 39° from the East toward the North.