# 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