1. vectors velocity

an airplane is headed north with a constant velocity of 450km/h. the plane encounters a wind from the west at 100 km/h.

a) in 3h, how far will the plane travel?
b) in what direction will the plane travel?

2. Originally Posted by william
an airplane is headed north with a constant velocity of 450km/h. the plane encounters a wind from the west at 100 km/h.

a) in 3h, how far will the plane travel?
b) in what direction will the plane travel?
In one hour the plane will travel $\sqrt{450^2+100^2}$ therefore in three hours it will cover $3\sqrt{450^2+100^2}$ km

b) $arctan \left(\frac{450}{100}\right)$

3. Originally Posted by e^(i*pi)
In one hour the plane will travel $\sqrt{450^2+100^2}$ therefore in three hours it will cover $3\sqrt{450^2+100^2}$ km

b) $arctan \left(\frac{450}{100}\right)$
i understand a, it makes sense but can you elaborate a bit more on b? the book gives the answer at 12.5^o east of north

4. Originally Posted by william
i understand a, it makes sense but can you elaborate a bit more on b? the book gives the answer at 12.5^o east of north
I should point out that there I made a mistake, that is the angle the plane makes with the horizontal so the actual angle is $90 - \arctan(4.5)$.

When I drew my diagram I put the plane at the origin and drew a horizontal line right to represent the wind (100) and a vertical line up to represent the plane going north (450)

I then made a right-angled triangle from these lines with an angle \theta between them. Part a is asking for the hypotenuse which is Pythagoras' as you got.

Part b is one of the trig ratios, since the opposite and the adjacent are known directly I used $tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{450}{100} = 4.5$.

However, this is the angle is makes with East (as I forgot in my last post), to find the answer from North we need to take this angle from 90.

$90 - \arctan (4.5) = 12.5$