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Math Help - speed, velocity and acceleration

  1. #1
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    Question speed, velocity and acceleration

    hey everyone

    i recently came across a question which asks for speed and in AB calculus the only thing my teacher taught me was velocity. Can anyone explain to me the difference between speed, velocity and acceleration in terms of calculus and their relationship to each other like when velocity is increasing what happens to the speed or acceleration etc.

    Thanks alot
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  2. #2
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    Re: speed, velocity and acceleration

    Speed and velocity are basically synonyms, but Wikipedia says that velocity is a vector and speed is its magnitude (or, absolute value). Velocity is the derivative of position with respect to time, and acceleration is the derivative of velocity, i.e., the second derivative of position.
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  3. #3
    MHF Contributor ebaines's Avatar
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    Re: speed, velocity and acceleration

    One nuance in the fact that velocity is a vector, and hence has both magnitude and direction, is that it's possible for the magnitude to be constant (ie constant speed) but the direction could be varying with time, and hence the acceleraton is non-zero. For example think about a wheel spinning at a constant rate of  \omega radiand/second - a point on the edge of the wheel has constant speed equal to  \omega R, and so you might think the acceleration of the point is zero. But it's not, because the direction of velocity changes as the point revolves about the axis. Formally what you have is:

     \vec a = \frac {d \vec v}{dt} = \frac {d (\omega R \hat { \theta})} {dt}

    Here  \omega and  R are constants, and  \hat {\theta} is the unit vector in the direction of motion, which is tangential to the edge of the wheel. That direction changes as the wheel rotates, and the rate of change of  \hat {\theta} is given by:

     \frac { d \hat {\theta}}{dt} = - \omega \hat r

    where  \hat r is the unit vector in the radial direction. Hence for the case of constant  R and \omega (and hence constant speed), the acceleration is :

     \vec a = \frac {d (\omega R \hat { \theta})} {dt} = -\omega ^2 R \hat r
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