Hello, struck!

Point $\displaystyle O$ is the intersection of two roads which cross at right angles;

one road runs from north to south, the other from east to west.

Find the least distance apart of two motorbikes $\displaystyle A$ and $\displaystyle B$ which are initially approaching $\displaystyle O$

on different roads in the following cases:

(a) $\displaystyle A$ is 120 metres from $\displaystyle O$ travelling at 20 m/s.

. . .$\displaystyle B$ is 80 metres from $\displaystyle O$ travelling at 10 m/s. Code:

- Q *
: |
: 10t |
: |
80 B *
: | *
: | *
: 80-10t | * d
: | *
: | *
: | *
- O * - - - - - - * - - - - *
120-20t A 20t P
: - - - - 120 - - - - :

Motorbike $\displaystyle A$ starts at $\displaystyle P$: .$\displaystyle OP = 120$

In $\displaystyle t$ seconds, it moves $\displaystyle 20t$ m to point $\displaystyle A.$

. . Hence: .$\displaystyle OA \:=\:120-20t$

Motorbike $\displaystyle B$ starts at $\displaystyle Q$: .$\displaystyle OQ = 80$

In $\displaystyle t$ seconds, it moves $\displaystyle 10t$ m to point $\displaystyle B.$

. . Hence: .$\displaystyle OB \:=\:80-10t$

The distance between them is: .$\displaystyle d \:=\:AB$

And we wish to minimize $\displaystyle d.$

Note that, if we minimize $\displaystyle D = d^2$, we will minimize $\displaystyle d.$

In right triangle $\displaystyle AOB\!:\;\;D\:=\:d^2 \:=\:(120-20t)^2 + (80-20t)^2

$

This simplifies to: .$\displaystyle D \;=\;500t^2 - 6400t + 20,\!800$ .[1]

The graph of this function is an up-opening parabola.

. . Its minimum is at its *vertex.*

The vertex is at: .$\displaystyle t \:=\:\frac{\text{-}b}{2a} \:=\:\frac{\text{-}(\text{-}6400)}{2(500)} \:=\: 6.4$

Substitute into [1]: .$\displaystyle D \:=\:d^2 \:=\:500(6.4^2) - 6400(6.4) + 20,\!800 \:=\:320$

Therefore: .$\displaystyle d \;=\;\sqrt{320} \;=\;8\sqrt{5} \;\approx\;17.9\text{ m}$