Understanding Kinematic Relative Motion: Solving Problem 1

[leprofece]

HElp With Problem 1 View attachment 2701see image

Ahora View attachment 2702

In THIS problem Part 1 without wind I understand it perfectly But where are the 10 km and 20 km come from?
y b) Why do arcsin ( 40/220)?

 

[leprofece]

MODIFIED

An airplane pilot sets a compass course due west and main¬tains an air speed of 220km/h. After flying for 0.0005 h is over a town 120 km west and 20 km south of the turning Point. (a) Find the wind velody (magnitude and direction). (b) If the wind vekcity is 40 kmih due west, in what direction should the pilot set her course to travel due west ? Use the same airspeed of 220 km/h.

above you have my doubts

 

[leprofece]

ANSWERED IN TAHOO ANSWERS

So in 1/2 hr., she has flown an extra 10km. west than she would have at 220kph., so wind west component must be 20kph.
In 1/2 hour, she has been blown 20km. south, so the south component must be 40kph.
Sqrt.(20^2 + 40^2) = 44.72kph. wind.
Direction = arctan (40/20) = 63.43 degrees south of west.

b) Something's wrong again, as if the wind velocity is 40kph. due west and that's the direction she needs to fly, she must point the aircraft west. the answer cannot be 10.5 degrees south of west.
She will have a groundspeed (220 + 40 ) = 260kph. directly west.
Yesterday I pointed this out also, but didn't give a groundspeed, just said she would arrive at her destination earlier than otherwise.

 

[MarkFL]

First, let's take the time to properly and in a readable manner, present the question:

3.81 An airplane pilot sets a compass course due west and maintains an airspeed of 220 km/h. After flying for 0.500 h, she finds herself over a town 120 km west and 20 km south of her starting point. (a) Find the wind velocity (magnitude and direction). (b) If the wind velocity is 40 km/h due south, in what direction should the pilot set her course to travel due west? Use the same airspeed of 220 km/h.

(a) Let's orient our coordinate axes such that her starting point is at the origin, and her ending point is (-120,-20). If we let $w_x$ be the horizontal (east/west) component of the wind's velocity vector, and $w_y$ be the vertical (north/south) component, then using vector addition, we have:

$$\left\langle -220,0 \right\rangle+\left\langle w_x,w_y \right\rangle=\left\langle -240,-40 \right\rangle$$

Adding like components, we obtain:

$$-220+w_x=-240\implies w_x=-20$$

$$0+w_y=-40\implies w_y=-40$$

Hence, for the wind's velocity we find:

Magnitude:

$$|w|=\sqrt{w_x^2+w_y^2}=\sqrt{(-20)^2+(-40)^2}=20\sqrt{5}\text{ km/h}$$

Direction:

Since both components are negative, we know the angle is in the third quadrant, and so we may use:

$$\theta=\text{W}\,\frac{180}{\pi}\tan^{-1}\left(2\right)\,\text{S}\approx\text{W}\,63.435^{\circ}\,\text{S}$$

(b) Let's let $a_x$ and $a_y$ be the horizontal and vertical components of the pilot's airspeed vector. We may then write:

$$\left\langle a_x,a_y \right\rangle+\left\langle 0,-40 \right\rangle=\left\langle x,0 \right\rangle$$

And this implies:

$$a_x+0=x\implies a_x=x$$

$$a_y-40=0\implies a_y=40$$

We also know:

$$220^2=a_x^2+a_y^2=x^2\implies x=-220$$

And so her direction must be:

$$\theta=\text{W}\,\frac{180}{\pi}\tan^{-1}\left(\frac{2}{11}\right)\,\text{N}\approx\text{W}\,10.305^{\circ}\,\text{N}$$