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NSW Y12 Maths - Advanced Motion and Rates Integrating With Respect to Time

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Integrating With Respect to Time Theory

Given the acceleration function \(a=f(t)\) for \(t=\) time, Then the velocity function \(v(t)=\displaystyle \int f(t)\, d t\).\\ The displacement function \(x(t)=\displaystyle \int v(t)\, d t\)\\ \textbf{Example 1}\\ A particle is moving in such a way that after \(t\) seconds its acceleration \(a \mathrm{~cm} / \mathrm{sec}^2\) is given by \(a=12 t-6\) Initially the particle is at the origin moving with a velocity \(-12 \mathrm{~cm} / \mathrm{sec}\).\\ Find:\\ (i) \(v\) as a function of \(t\)\\ (ii) \(x\) as a function of \(t\)\\ (iii) the velocity at \(t=1\)\\ (iv) when the particle is at rest.\\ \textbf{Example 1 solution}\\ $\begin{aligned} \text{(i) } \quad a=12 t-6 . \quad t=0, &x=0, v=-12 \mathrm{~cm} / \mathrm{sec}\\ v(t) & =\int 12 t-6 \,d t \\ & =6 t^2-6 t+c \\ -12 & =0-0+c \\ \therefore v(t) & =6 t^2-6 t-12 \end{aligned}$\\ $\begin{aligned} \text { (ii) }\quad x(t)&= \int 6 t^2-6 t-12 \,d t \\ &= 2 t^3-3 t^2-12 t+k \\ 0&= 0-0+0+k \\ x(t)&= 2 t^3-3 t^2-12 t \\ \\ \text { (iii) } \quad v(1)&= 6-6-12 \\ &=-12 \mathrm{~cm} / \mathrm{s} . \\ \\ \text { (iv) } \quad v(t)&= 6 t^2-6 t-12=0 \\ & t^2-t-2=0 \\ &(t-2)(t+1)=0 \\ & \therefore t=2 \text { or } t=-1 \quad(t \neq-1) \\ & \therefore \, t=2 \end{aligned}$\\

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