Dear Uncle Colin,

I have a little problem. You see, there’s this bird, A, in its nest at time $t=0$ - the nest is at $(20, -17)$ - and it travels with a velocity of $-6\bi + 7\bj$ (in the appropriate units). But there’s *another* bird, B, whose nest is at $(-8,9)$ and who travels at $p\bi + 2p\bj$. Then when $t=4$, it turns out that B is (get this) *south-west* of bird A! For\ some reason, I have to figure out the speed of bird B. How would you do it?

- Help Evaluating Right Ornithology Numbers

Hi, HERON, and thanks for your message!

Ah, a classical two-bird problem with vectors. Luckily, we have quite a lot of information and it should fall apart fairly quickly.

### A picture

Before doing anything else, I’d draw a picture of what’s going on.

I did that. I also plotted it in Desmos:

### Where are the birds?

We can write the position vectors of the birds as $\mathbf{r_A} = (20 - 6t)\bi + (-17+7t)\bj$ and $\mathbf{r_B} = (-8+pt)\bi + (9 + 2pt)\bj$.

Better than that, we’re only really interested in the situation when $t=4$, so we can substitute that in: $\mathbf{r_A} = -4\bi + 11\bj$, which corresponds to the picture, and $\mathbf{r_B} = (-8+4p)\bi + (9+8p)\bj$.

### Vector $\vec{BA}$

If B is southwest of A, vector $\vec{BA}$ is in the northwesterly direction: it’s a multiple of $\bi + \bj$.

But we can work out $\vec{BA} = \mathbf{r_A} - \mathbf{r_B}$, which is $((-4) - (-8+4p))\bi + ( 11 - (9+8p))\bj$.

More clearly, that’s $(4 - 4p)\bi + (2-8p)\bj$, which needs to be in the form $k\bi + k\bj$ - or rather, the components must be equal.

So $4-4p = 2-8p$, giving $2=-4p$ and $p = -\frac{1}{2}$.

### Finally the speed

B’s speed is $\sqrt{p^2 + (2p^2)}$, or $|p|\sqrt{5}$, which is $\frac{1}{2}\sqrt{5}$ of those appropriate units.

Hope that helps!

- Uncle Colin

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.