Kinematics proble

1. Nov 5, 2008

devanlevin

a particle move according to the vector
$$\vec{r}$$=3cos$$^{2}$$(6t)$$\hat{x}$$-5sin($$\frac{PI}{6}$$t)$$\hat{y}$$

what is the formula for the route it takes, y(x)? when will the particle stop and how many times will it stop in an hour??

i said, the vectors in the direction x and y
x(t)=3cos$$^{2}$$6t=3(1-sin$$^{2}$$6t)
y(t)=-2sin6t
-------------
sin6t=$$\frac{-y}{2}$$

x=3(1-($$\frac{-y}{2}$$)$$^{2}$$)
x(y)=3-$$\frac{3}{4}$$y$$^{2}$$

this course is a parabula lying on its side, (minus infinity) on its x axis with its maximum at Max(3,0)

to find out when the particle stops i say
the particle will stop when the parabula peaks-- $$\frac{dx}{dy}$$=0
x(y)=3-0.75y$$^{2}$$
$$\frac{dx}{dy}$$=-1.5y
y=0
the particle will stop every time y=0
y(t)=-2sin(6t)=0
sin(6t)=0
6t=$$PI$$*K
t=$$\frac{PI}{6}$$*K (K being a positive whole number)

does this mean that the particle will stop every $$\frac{PI}{6}$$ seconds??

how can this be, since the route the particle moves on is a parabula and doesnt peak more than once, i realise that where i have misunderstood something is in the transition from the trigonometric equations x(t) and x(y) which repeat themselves to the parabula x(y).

2. Nov 5, 2008

Hootenanny

Staff Emeritus
Careful

3. Nov 5, 2008

devanlevin

sorry, the -5sin...was a typo, , meant to be -2sin(6t)y like i wrote in all the calculations

4. Nov 5, 2008

Hootenanny

Staff Emeritus
What about the $\pi/6$ in the question? Was that a typo as well?

5. Nov 5, 2008

devanlevin

correct equation is r=3cos^2(6t)x-2sin(6t)y
but that doesnt really make a diffeerence to what im asking, its more of a technical question, the end result as a number isnt of that much importance,,,
after reading over what i wrote i now think that i was wrong, and that the particle will stop when |v|=0 meaning when cos(6t)=0
6t=PI/2 +PI*K
t=PI/12 + (PI/6)*K

from this how do i know how many times it will stop in an hour??

6. Nov 5, 2008

Hootenanny

Staff Emeritus
The particle doesn't stop when

$$\frac{dy}{dx}=0$$

Rather, it stops when

$$\frac{d\bold{r}}{dt}=0$$

7. Nov 5, 2008

devanlevin

thats essentially what i did by saying Vx=0 and Vy=0

but then i get to t=PI/12 + (PI/6)*K how do i translate that into how many times it will stop in an hour