1. ## Word problem using quadratic equations..

I have to use quadratics to solve this:

Point 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 A and B which are initially approaching O on different roads in the following cases:

(a) A is 120 metres from O travelling at 20 m/s, B is 80 metres from O travelling at 10 m / s.

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

3. ## Relative velocity: closest approach

Hello struck
Originally Posted by struck
I have to use quadratics to solve this:

Point 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 A and B which are initially approaching O on different roads in the following cases:

(a) A is 120 metres from O travelling at 20 m/s, B is 80 metres from O travelling at 10 m / s.
I'm not sure why you think this is a Quadratics question. In fact, it's all about relative velocity, and the least distance between two moving bodies.

In the attached diagram, the Displacement Diagram shows the initial positions of A and B. I've assumed that A is moving due N and B due E.

The technique is this:

• Draw a triangle of velocities to find the direction of the velocity of B relative to A.
• On the displacement diagram, superimpose the direction you've just found to find the displacement of B relative to A.
• Draw a perpendicular from A onto this line. The length of this perpendicular is the shortest distance between A and B.

So, in the Triangle of Velocities, you'll see that the velocity of B relative to A is the velocity of B minus the velocity of A.

This line makes an angle $\displaystyle \theta$ with due E, where $\displaystyle \tan \theta = \frac{20}{10} = 2$.

So when this line is drawn on the displacement diagram, the distance $\displaystyle OC = 80\tan\theta = 160 \Rightarrow AC = 40$.

So if AP is drawn perpendicular to BC, angle PAC = $\displaystyle \theta \Rightarrow AP = 40 \cos\theta$.

You can use a calculator to find $\displaystyle \cos\theta$, or you can get an exact value (if you know the trig formulae) which is $\displaystyle \cos\theta = \frac{1}{\sqrt{5}}$

$\displaystyle \Rightarrow AP = \frac{40}{\sqrt{5}}= 8\sqrt{5}$ m.