A computation with vectors

I'm going through these lecture notes, and I don't understand how one of the example problems was solved. Can anyone show me a step by step solution?

Question: Suppose a Canada goose is flying northwest at 30mph in a wind blowing from the south at 20mph. What is the gooses true course and ground speed?

Thanks

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I have retagged the question since it consists of vector addition and has no calculus. – Neal Mar 24 '12 at 5:08
@Neal I am not quite sure, although I tend to agree with you. The velocity vector v(t) is the derivative of the position -- WolframMathWorld Velocity Vector – Américo Tavares Mar 24 '12 at 13:14
If position is given as a function of time, as a path through space, then velocity is the time-derivative. However, this question has no time dependence and is taking place in a constant-velocity setting or in an instant of time, so it doesn't involve any actual calculus. – Neal Mar 24 '12 at 17:11

From the perspective of the goose, the goose is going northwest at $30$ mph. However, the goose is not aware of any motion in the air. From the perspective of an observer on the ground, the air in which the goose is embedded is also traveling at $20$ mph, to the north. So from the perspective of the observer on the ground, the goose's motion has two components: northwest at $30$ mph and north at $20$ mph. These add together to the goose's velocity vector, from the perspective of the ground.
To be a little more formal, let us choose coordinates so that $(0,1)$ indicates flying north at $1$ mph and $(1,0)$ indicates flying east at $1$ mph. Now, from the goose's perspective, she is flying northwest at $30$ mph. Northwest is in the direction of the vector $(-1,1)$, with magnitude $30$, so her velocity vector is $(-\frac{30}{\sqrt{2}},\frac{30}{\sqrt{2}})$. (Draw the triangle to check components.)
From the perspective of the ground, the air in which the goose is flying is also moving northward at $20$ mph, so its motion is represented by the vector $(0,20)$. To find the total velocity of the goose from the perspective of the ground, add the goose's velocity and the air's velocity to get $(-\frac{30}{\sqrt{2}},20+\frac{30}{\sqrt{2}})$. Its ground speed is therefore $$\sqrt{\frac{1}{2}(30^2 + (20\sqrt{2}+30)^2)}.$$
@Lexy Yang From WolframMathWorld Velocity Vector. The velocity vector $v(t)$ is the derivative of the position. – Américo Tavares Mar 24 '12 at 13:10
@LexyYang the velocity vector describes how fast an object is moving and in what direction it's going. So when you specify that the goose is flying northwest at $30$ mph, in wind pushing it north at $20$ mph, you are specifying its velocity. – Neal Mar 24 '12 at 17:13