Thread: 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.

Hello, struck!

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

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

Code:

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

Motorbike starts at : .
In seconds, it moves m to point
. . Hence: .

Motorbike starts at : .
In seconds, it moves m to point
. . Hence: .

The distance between them is: .
And we wish to minimize

Note that, if we minimize , we will minimize


In right triangle

This simplifies to: . .[1]

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

The vertex is at: .

Substitute into [1]: .

Therefore: .

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 with due E, where .

So when this line is drawn on the displacement diagram, the distance .

So if AP is drawn perpendicular to BC, angle PAC = .

You can use a calculator to find , or you can get an exact value (if you know the trig formulae) which is

m.

Grandad