Recent content by lexie47

But the velocity of the plane WRT the ground is due E and V of the wind WRT the ground is due S; since those two vectors from a right triangle, that leaves V of the plane WRT the wind as the hypotenuse, so the vectors have to be subtracted since we're solving for V of the plane WRT the ground.

with numbers plugged in its V_pg = √(70)^2 - (20)^2 = 67 km/hr

That's what my teacher posted as the answer though, unless she's wrong? I'm getting Vpg= √((V_pw )^2-(V_wg)^2)

I got it, thank you, its 67 km/hr if using the theorem.

Can't find another way to attach, sorry for being a pain, I'm not familiar with forums. Thanks for looking anyway.

It just uploaded that way I'm sorry, trying to attach another way.

Sorry, this is my first post, will stick to the template next time! I attached a picture of my breaking the vectors into components, which lead me nowhere because I ended up with V_pg = V_pg.

Q: The pilot of an aircraft flies due east relative to the ground in a wind blowing 20 km/h toward the south. If the speed of the aircraft in the absence of wind is 70 km/h, what is the speed of the aircraft relative to the ground? I set up the vector of wind WRT ground pointing due south...