How to work out minimum distance bewtween 2 objects travelling perpendicular from one another at diferent speeds.

Question

Can I get some help on this question please. Cyclist A starts 10 miles east of cyclist B. Cyclist A starts to travel 20mph west whilst cyclist B starts to cycle 15mhp north. I want to find the minimum distance between the cylists and at what time that distance occurs.

My initial idea is to use Pythagoras to find the distance and then differentiate that. Once I have first derivative set it equal to zero to work out time? Then plug in answer for time into original Pythagoras equation to get the distance? I don't know this is just an idea and im stuck on the method to start.

I know both will travel a certain distance d after some time so for A would the distance be (10-(10*t) and B (0+15*t) and then expand and put in Pythagoras equation ?

Answer 1

The position of $A$ at instant $t$ is $(10-20t,0)$, the position of $B$ is $(0,15t)$, and $t\ge0$.

The distance is minimum when its square is minimum. The square of distance is $$f(t)=(10-20t)^2+(15t)^2$$

Since $f$ is differentiable everywhere, the distance is minimum either at a boundary of the interval $[0,+\infty)$ or at a point $t$ such that $f'(t)=0$.

First,

$$f'(t)=-40(10-20t)+30\cdot15t=1250t-400$$

Hence $f'(t)=0$ iff $t=\frac{8}{25}$, and there $f(t)=36$, so the distance is $6$.

For $t\to+\infty$ the distance tends to infinity and is certainly not minimum. And for $t=0$, $f(t)=100>36$ (the distance at $t=0$ is $10$). Hence the minimum distance is indeed $6$, at $t=\frac{8}{25}$.

Note: it's possible to work directly with the distance. The computations are slightly more complicated as there is a square root, but you can check for yourself that it's really not insurmountable.