A plane flies in a direction NW at an airspeed of $141$ km/hr. If the wind at the plane's cruise altitude is blowing with a speed of $100$ km/hr directly from the north, what is the plane's actual speed and direction relative to the ground?
So I'm presuming you can do this with vectors. One of them pointing to the northwest, with a magnitude of 141 and one of them pointing to the south with a magnitude of 100. But how can I get the answer?
Seperate the North and West directed components of the vector directed to Northwest. They are $141*cos(45)$ and $141*sin(45)$. As $cos(45) = sin(45)= \sqrt2/2 \simeq 100/141 $, then those components are $100$ km/hr to North and $100$ km/hr to West. Then the wind and the speed to North direction deletes each other, and the plane flies to West with a speed of $100$ km/hr.
Make a drawing, at time $0$, with three points (one representing the plane, one representing an air molecule and one representing a pebble on the ground) all on top of eachother at the origin. Then make a drawing of the situation an hour later, still with the pebble at the origin. The air molecule will be at $\vec m = (0, -100)$, since it's been moving steadily southward, and the airplane $\vec p$ will be $141$ km straight northwest from the air molecule. Where in the coordinate system would that be? Is there some easy way to manipulate the vectors you've been given to arrive at this answer?
