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

Motorbike $A$ starts at $P$: . $OP = 120$
In $t$ seconds, it moves $20t$ m to point $A.$
. . Hence: . $OA \:=\:120-20t$

Motorbike $B$ starts at $Q$: . $OQ = 80$
In $t$ seconds, it moves $10t$ m to point $B.$
. . Hence: . $OB \:=\:80-10t$

The distance between them is: . $d \:=\:AB$
And we wish to minimize $d.$

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

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

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

The graph of this function is an up-opening parabola.
. . Its minimum is at its vertex.

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

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

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

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 $\theta$ with due E, where $\tan \theta = \frac{20}{10} = 2$.

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

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

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

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