# Thread: Deriving Centripetal Acceleration?

1. ## Deriving Centripetal Acceleration?

ok i have no idea how to do this, can you guys please help me out with this for physics?

"For no apparent reason, a poodle is running counter-clockwise at a constant speed of 3.90 m/s in a circle with radius 2.3 m. Let v_1 be the velocity vector at time t_1, and let v_2 be the velocity vector at time t_2. Recall that average acceleration = delta_v/delta_t."

i really need help on finding the magnitude and direction of the average acceleration at a given delta_t, for example, delta_t=0.5

i've used the formulas v_x=-w sin wt and v_y = w cos wt, but i'm not sure if that's right. anyways, thanks in advance to anyone who helps

2. ## Re: Deriving Centripetal Acceleration?

$\displaystyle \frac{\Delta v_x}{\Delta t}=\frac{v_x(t_2)-v_x(t_1)}{t_2-t_1}$

$\displaystyle \frac{\Delta v_y}{\Delta t}=\frac{v_y(t_2)-v_y(t_1)}{t_2-t_1}$

3. ## Re: Deriving Centripetal Acceleration?

cool, thanks for adding fancy versions of the equations, but that doesn't really answer my question, which was for the magnitude and direction...

4. ## Re: Deriving Centripetal Acceleration?

The magnitude of a two-dimensional vector is given by
$\displaystyle r=\sqrt{x^2+y^2}$
in which $\displaystyle x$ is the x-component of the vector and $\displaystyle y$ is the y-component. This is an application of Pythagoras theorem.

As for the direction, that depends on which coordinate system you want to use to express the direction. A possibility is to express the direction as angle $\displaystyle \theta$. Let's measure $\displaystyle \theta$ as the angle between x-direction and the vector, being positive in counter-clockwise direction (this is a common convention). The angle is then given by:
$\displaystyle \theta=\arctan\left({\frac{y}{x}}\right)$

5. ## Re: Deriving Centripetal Acceleration?

okay, thank you, this might help a lot!