# Motion in 2d/Centripetal Acceleration

1. Oct 11, 2004

### HurricaneH

...sorry for doing this, but having trouble with some problems

1) A particle starts from rest at t=0 at the origin and moves in the xy plane with a constant acceleration of a= (2i + 4j) m/s2. After a time t has elaped, determine (a) the x and y components of velocity. (b) the coordinates of the particle, and (c) the speed of the particle.

I get a) Vx= (2t)m/s and Vy=(4t)m/s, but not the rest...

and...

2) A ball swings in a vertical circle at the end of a rope 1.5 M long. When it is 36.9* past the lowest point on its way up, the ball's total acceleration is (-22.5i + 20.2j) m/s2. For that instant, a) determine the magnitude of its centripetal acceleration, and b) determine the magnitude and direction of its velocity.

I know that v^2/R= Ac, but i don't know how to get the velocity:(

Any help will be appreciated

2. Oct 11, 2004

### Pyrrhus

Use uniform acceleration kinematic equations, or use calculus. Integrate Acceleration, then integrate Velocity.

$$\vec{a} = \frac{d \vec{v}}{dt}$$

$$\vec{v} = \frac{d \vec{r}}{dt}$$

Remember

$$A_{c} = \frac{v^2}{r}$$

Refers to Magnitude.

3. Oct 11, 2004

### HurricaneH

but how do i find out the velocity?

4. Oct 13, 2004

### HurricaneH

Well I got it. For 1 b) x=t^2m y=2t^2, for c) a^2 + b^2= c^2, c= velocity.

For 2. The angles of the directionals were 36.9 and 53.1, then cos36.9=20.2/x, solve for x, cos53.1= 22.5/x2, solve for x2, add both xs, and you get 62.8 m/s for the magnitude of its centripetal acceleration. For c, v^2/r= aC, we have the aC and r, so v^2/1.5=62.8, simple algebra, and v=9.7m/s.