Quadratics - long questions 2.(a)

IFoundWonderland

Watttt?

Reply 1

thefatone

Original post by IFoundWonderland

Watttt?

LOL they just didn't take out a common factor that's all xD

Reply 2

Zacken

Original post by IFoundWonderland

Watttt?

A diagram is always useful!

So, the green car starts at (0, -30) another way of saying this is that its initial position vector is

\begin{pmatrix}0 \\ -30\end{pmatrix}

.

It has a speed of 15 km per hour northwards - that is, it's speed is given by

\begin{pmatrix}0 \\ 15 \end{pmatrix}

- that is, the 15 is positive because it's moving upwards and the 0 in the i component is 0 because it doesn't move left and right, just up.

So it's position at any general time

t

is given by

\begin{pmatrix}0 \\ -30\end{pmatrix} + t\begin{pmatrix}0 \\ 15\end{pmatrix}

. We can rewrite this as the (green) car having coordinates

(0, -30 + 15t)

at some general time t.

Can you see why the above is true? In the first hour, it's travelled an 15 kilometres so the new position vector is

\begin{pmatrix}0 \\ -30\end{pmatrix} + \begin{pmatrix}0 \\ 15\end{pmatrix}

.

In the second hour it's moved 30 kilometres upwards and it's position vector is given by

\begin{pmatrix}0 \\ -30\end{pmatrix} +2\begin{pmatrix}0 \\ 15\end{pmatrix}

, etc...

That is, in general for any time

t

in hours the position vector is

\begin{pmatrix}0 \\ -30\end{pmatrix} + t\begin{pmatrix}0 \\ 15\end{pmatrix}

Can you do the same for the other car? Write their coordinates as a function of

t

as I have above and then use use the distance formula

d^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2

.

Let me know if there's anything I can clarify! A big part of this sorta question is figuring out a way to represent the given information mathematically, the actual maths is easy, but getting it from words to equations and vectors or coordinates is the part you really need to think about.

Hope you don't mind that I've given the cars colours, it makes it easier to talk about them instead of "car 1" and "car 2". :lol:

Edit:

Spoiler

(edited 7 years ago)

Reply 3

IFoundWonderland

@Zacken

Thank you :h:

I got the right answer with your help:

Thr only issue is that I've never actually come across the distance formula you said to use before :s. Thid is also in the first section of the book (quadratics) and so I'm surprised that they require prior knowledge of vectors, which is a much later chapter. I'll make a note of the formula now, but it's weird that they're asking a question that requires something I've never been taught ffs.

Reply 4

Zacken

Original post by IFoundWonderland

The "distance formula" isn't really anything more than Pythagoras.

You have a triangle with height

(-30 + 15t)

and base

(-50 + 20t)

so the hypotenuse squared is...?

Also - the distance formula seems to be in your formula booklet under "prior learning" (I've had a gander at the HL copy here but presumably your SL one will have the same?)

It's not really a vector question and it doesn't require vectors per se, it's just knowing how the coordinates change w.r.t to time and I thought that exhibiting that in colum vector form would be the most conducive; you don't actually need to know anything about vectors.