8EkinematicDEs


 * Kinematics and Differential Equations**

Acceleration as Rate of Change

Using the chain rule: math . \qquad a=\dfrac{dv}{dt}=\dfrac{dv}{dx} \centerdot \dfrac{dx}{dt} math we can get different rules for acceleration to use in different circumstances.

math \\ \text{If } a=f(t) \\. \\ . \qquad \text{Use } a=\dfrac{dv}{dt} \text{ or } a=\dfrac{d^2x}{dt^2} math

math \\ \text{If } a=f(x) \\. \\ . \qquad \text{Use } a=\dfrac{d}{dx} \Big( \frac{1}{2}v^2 \Big) math

math \\ \text{If } a=f(v) \\. \\ . \qquad \text{Use } a=\dfrac{dv}{dt} \; \; \text{ if initial conditions are in terms of v and t} \\. \\ . \qquad \text{Use } a=v\dfrac{dv}{dx} \; \; \text{ if initial conditions are in terms of v and x} math

Example 1

A particle moves in a straight line so that velocity is given by: math . \qquad \dot{x}(t)=\dfrac{1}{2-x}, \; x>2 \; \; \text{ and } \; x=4 \text{ when } t=0. math Find the position, x, as a function of t.

math \\ . \qquad \dfrac{dx}{dt}=\dfrac{1}{2-x} \\. \\ . \qquad \dfrac{dt}{dx}=2-x \\. \\ . \qquad t=\int2-x \, dx \\. \\ . \qquad t=2x-\dfrac{x^2}{2}+c math

when t = 0, x = 4, so c = 0 math \\ . \qquad t=2x-\dfrac{x^2}{2} \\. \\ . \qquad 2t=4x-x^2 \\. \\ . \qquad x^2-4x=-2t math

Complete the square math \\ . \qquad x^2-4x+4=-2t+4 \\ .\\ . \qquad \big( x-2 \big)^2 =-2t+4 \\. \\ . \qquad x-2=\pm\sqrt{-2t+4} math

reject negative because x > 2

math . \qquad x=2+\sqrt{-2t+4} \quad t\in[0,2) math

Note that from question, x > 2 so surd > 0 so t not equal to 2.

Example 2

An object moves in a straight line so that the acceleration is given by math . \qquad \ddot{x}(t)=x+1 math Find a relation for v in terms of x, given that v = 3 when x = 2

math \text{Use } a=\dfrac{d}{dx} \Big( \frac{1}{2}v^2 \Big) \textit{ because } a=f(x) math

math \\ . \qquad \dfrac{d}{dx} \Big( \frac{1}{2}v^2 \Big)=x+1 \\. \\ . \qquad \frac{1}{2}v^2 = \int {x+1} \,dx \\. \\ . \qquad \frac{1}{2}v^2 = \frac{1}{2}x^2+x+c \\. \\ . \qquad v^2=x^2+2x+2c math

when v = 3, x = 2 so 2c = 1

math . \qquad v^2=x^2+2x+1 math

{Notice that this is a relation, not a function}

**See also:** www.mathsonline.com.au Y12Extension --> Calculus --> Physical Applications --> Lessons 3, 4

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