1. ## 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

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

3. 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,
$\displaystyle \bold{R}_q(t)=bt\bold{i}+ct\bold{j}$
$\displaystyle \bold{R}'_q(t)=b\bold{i}+c\bold{j}$
$\displaystyle \bold{R}''_q(t)=\bold{0}$

4. 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.

5. 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.

6. ## 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.

7. ## re:

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

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,
$\displaystyle t=\frac{1}{\omega}\tan^{-1}(c/b)+\frac{\pi k}{\omega}$

9. hi thanks for that!

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

$\displaystyle t=\frac{a}{a\omega}\tan^{-1}(c/b)+\frac{a\pi k}{a\omega}$