# Vectors

• May 2nd 2006, 01:56 PM
Vectors
Hi again!

I was doing a vector question until I got stuck! I have done part a) of the question assuming its right I need help on part b) & c)

In b) what is it meant by x-y plane?

c) I remember doing this type of question before where you have to get rp = rq not sure if this is the case for this question.

Hope you guys can help.

Regards

• May 2nd 2006, 02:05 PM
ThePerfectHacker
Quote:

Hi again

In b) what is it meant by x-y plane?

That means the graph in terms of x and y.
Plane means flat and x-y means you made a graph as a function of x and y and you end up with a curve which is flat in this plane
• May 2nd 2006, 02:12 PM
ThePerfectHacker
You have,
$\displaystyle \bold{R}_p(t)=a\cos \omega t \bold{i}+a\sin \omega t \bold{j}$
This is the displacement vector. The velocity vector is its derivative thus,
$\displaystyle \bold{R}'_p(t)=-a\omega \sin \omega t \bold{i}+a\omega \cos \omega t\bold{j}$
To find the acceleration take derivative again,
$\displaystyle \bold{R}''_p(t)=-a\omega^2\cos \omega t \bold{i} -a\omega^2 \sin \omega t\bold{j}$
-------
For the other vector function ya got,
$\displaystyle \bold{R}_q(t)=bt\bold{i}+ct\bold{j}$
$\displaystyle \bold{R}'_q(t)=b\bold{i}+c\bold{j}$
$\displaystyle \bold{R}''_q(t)=\bold{0}$
• May 2nd 2006, 02:15 PM
ThePerfectHacker
The vector function,
$\displaystyle \bold{R}_p(t)=a\cos \omega t \bold{i}+a\sin \omega t \bold{j}$
Means, parametrically
$\displaystyle x=a\cos \omega t$
$\displaystyle y=a\sin \omega t$

Umm, I cannot sketch this you did not specify $\displaystyle a,\omega$

The same in vector function,
$\displaystyle \bold{R}_q(t)=bt\bold{i}+ct\bold{j}$
Means, parametrically
$\displaystyle x=bt$
$\displaystyle y=at$
This is a line, but again cannot graph it I need you to specify the values.
• May 2nd 2006, 02:19 PM
ThePerfectHacker
They collide when,
$\displaystyle a\cos \omega t \bold{i}+a\sin \omega t \bold{j}=bt\bold{i}+ct\bold{j}$
That happens when,
$\displaystyle a\cos \omega t=bt$
$\displaystyle a\sin \omega t=ct$

Divide equation 2 by equation 1 to get,
$\displaystyle \tan \omega t=\frac{c}{b}$
So the problem reduces to solving this equation.
• May 3rd 2006, 01:15 AM
re:
Hi thanks for that theperfecthacker! :)

How can I use: $\displaystyle a\cos \omega t=bt$ and $\displaystyle a\sin \omega t=ct$

To get it in terms of t.
• May 3rd 2006, 01:08 PM
re:
OR is it not possible to get it in terms of t?
• May 3rd 2006, 01:15 PM
ThePerfectHacker
Quote:

OR is it not possible to get it in terms of t?

Look at my previous post,
it says these equations reduce to,
$\displaystyle \tan \omega t= \frac{c}{b}$
Therefore use the arctangent function to get,
$\displaystyle \omega t=\tan^{-1} (c/b)+\pi k$
Thus,
$\displaystyle t=\frac{1}{\omega}\tan^{-1}(c/b)+\frac{\pi k}{\omega}$
• May 3rd 2006, 08:45 PM
hi thanks for that!

is it possible to be in terms of a, b and c?
• May 4th 2006, 12:37 PM
ThePerfectHacker
Quote:

hi thanks for that!

is it possible to be in terms of a, b and c?

Of corse you can write,
$\displaystyle t=\frac{a}{a\omega}\tan^{-1}(c/b)+\frac{a\pi k}{a\omega}$
:D
• May 4th 2006, 01:13 PM