A pilot maintains a heading due west with an air speed of 240 km/h. After flying for 30 min, he finds himself over a town that he knows is 150 km west and 40 km south of his starting point.
What is the wind velocity, in magnitude and direction?
What heading should he now maintain, with the same air speed, to follow a course due west from the town?
I don't know how I'm suppose draw this out, I understand that d for the west is 120km but I'm lost after that.
Best Answer
What you want to do is look at two components of velocity acting in two directions: south and west. Since the components acting in each direction are independent of each other, you can break the problem down into two parts, find the velocity of the wind in both directions, and then you can work out the magnitude and direction. (As a note, I'll use the subscript "w" to refer to the wind and "p" to refer to the plane.)
West Components: If the pilot covers a distance of $150$ km west in half an hour, then $v_{\text{west}} = 300$. But, $v_{\text{west}}$ is the combined velocity of the plane and the west component of the wind's velocity. The plane's velocity is $240$ km/hr, so the first component of the wind's velocity becomes $60$ km/hr west.
South Components: If the pilot covers a distance of $40$ km south in half an hour, then $v_{\text{south}} = 80$. Since only the wind contributes, then the second component of the wind's velocity becomes $80$ km/hr south.
Finding the magnitude is simple, since both components are perpendicular to each other. Using the Pythagorean Theorem, the magnitude $v_{\text{w}}$, can be found:
$$\left(v_\text{w}\right) = \left(v_{\text{w}_\text{west}}\right)^2 + \left(v_{\text{w}_\text{south}}\right)^2 \iff \left(v_\text{w}\right)^2 = 60^2 + 80^2 \iff v_\text{w} = 100$$
The magnitude of the wind's velocity is therefore $100$ km/hr. To find the direction (south of west), you just need to find the angle formed by the west component and the resultant, which gives $\theta = \arctan\left(\frac{v_{\text{w}_\text{south}}}{v_{\text{w}_\text{west}}}\right) = \arctan\left(\frac{80}{60}\right) = \arctan\left(\frac{4}{3}\right)$. Note that you don't have to use $\arctan$, any inverse trig function works, but when you want the angle given the components, that's the best choice. (You have the magnitude anyway, so it won't make any difference here.)
For the second part, the approach is similar. If the pilot wants to maintain a heading due west, the south component of velocity has to be eliminated. So, the plane's velocity has to be $80$ km/hr in the north direction. To keep the same airspeed, the total magnitude has to stay the same, so the plane's velocity in the west direction has to decrease. You need to use $\left(v_\text{p}\right)^2 = \left(v_{\text{p}_\text{west}}\right)^2 + \left(v_{\text{p}_\text{north}}\right)^2$, and you have $v_{\text{p}} = 240$ and $v_{\text{p}_\text{north}} = 80$, so you can work out the rest by finding $v_{\text{p}_\text{west}}$ and the direction (north of west).