
Question on trigo?
Hi I really have no idea which catergory this question falls into, especially so when I don't even know how to start off the question..
If Person A is 20Km north of Person B at 5pm. Person A walks south at a rate of 7Km/h and Person B walks west at a rate of 9Km/h. What time do these two peopole stop approaching each other and instead, drift further apart?
How do I start and what formula should I use? I'm really clueless. Is it under bearings..?
Thank you so much, really!

Re: Question on trigo?
let person B start at the origin, $\displaystyle (0,0)$
person A's position as a function of time, $\displaystyle t$, in hrs ... $\displaystyle (0, 207t)$
person B's position ... $\displaystyle (9t , 0)$
distance, $\displaystyle r$ ,between the two at any time $\displaystyle t$ in hrs ...
$\displaystyle r = \sqrt{[0  (9t)]^2 + [(207t)  0]^2}$
simplify, then determine when $\displaystyle \frac{dr}{dt}$ changes sign from negative to positive.

Re: Question on trigo?
This is how I would set it up:
Let north be the positive yaxis and west be the positive xaxis. Distances are in km and time is in hrs.
Person A's position at time t is:
$\displaystyle A(t)=(0,207t)$
Person B's position at time t is:
$\displaystyle B(t)=(9t,0)$
Let $\displaystyle D(t)$ be the distance separating the two. Hence:
$\displaystyle D^2(t)=(207t)^2+(9t)^2=130t^2280t+400$
We see this is a parabola opening upward, so all you need to do is find the axis of symmetry to find the time when their distance is a minimum.
Recall the axis of symmetry for the parabola $\displaystyle y=ax^2+bx+c$ is $\displaystyle x=\frac{b}{2a}$.