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Revision:Kinematics

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Kinematics

These notes are based on the requirements of the M3 A Level mathematics module.


1.1 - Acceleration depending on time (t)

\displaystyle a = \frac{dv}{dt}

\displaystyle v = \frac{dx}{dt}

\displaystyle a = \frac{d^2x}{dt^2}


\displaystyle a = f(t)

\displaystyle v = \int f(t) dt + c


\displaystyle v = g(t)

\displaystyle x = \int g(t) dt + k


1.2 - Acceleration depending on displacement (x)

\displaystyle a = F(x)

\displaystyle \frac{dv}{dt} = F(x)


By the chain rule,

\displaystyle \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}


Therefore;

\displaystyle a = \frac{dv}{dt} = \frac{dv}{dx} \times \frac{dx}{dt}


Since, \displaystyle \frac{dx}{dt} = v

\displaystyle a = v.\frac{dv}{dx}


If \displaystyle y = v^2 where \displaystyle v = f(x)

\displaystyle \frac{dy}{dx} = \frac{dy}{dv} \times \frac{dv}{dx} = 2v.\frac{dv}{dx}


OR rewrite \displaystyle y = v^2 as \displaystyle y = v.v

\displaystyle \frac{dy}{dx} = v.\frac{dv}{dx} + v.\frac{dv}{dx} = 2v.\frac{dv}{dx} (by the product rule)


Hence, \displaystyle \frac{d(v^2)}{dx} = 2v.\frac{dv}{dx}


The acceleration, therefore, may also be written;

\displaystyle a = \frac{d(\frac{1}{2}v^2)}{dx}


If \displaystyle v = f(x) = \frac{dx}{dt}

\displaystyle \frac{1}{v} = \frac{1}{f(x)} = \frac{1}{\frac{dx}{dt}} = \frac{dt}{dx}

\displaystyle \int \frac{1}{v} dx = \int 1 dt = t


Comments

Originally written by Widowmaker on TSR forums.

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