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Math Help - Deriving Centripetal Acceleration?

  1. #1
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    Question 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
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  2. #2
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    Re: Deriving Centripetal Acceleration?

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


    \frac{\Delta v_y}{\Delta t}=\frac{v_y(t_2)-v_y(t_1)}{t_2-t_1}
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  3. #3
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    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...
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  4. #4
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    Re: Deriving Centripetal Acceleration?

    The magnitude of a two-dimensional vector is given by
    r=\sqrt{x^2+y^2}
    in which x is the x-component of the vector and 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 \theta. Let's measure \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:
    \theta=\arctan\left({\frac{y}{x}}\right)
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  5. #5
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    Re: Deriving Centripetal Acceleration?

    okay, thank you, this might help a lot!
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