# Motions involving Acceleration, Velocity and Time

\begin{align} a &= \displaystyle \frac{d}{dx} \left( \frac{1}{2} v^2 \right) \\ v &= \frac{dx}{dt} \end{align}

## Example

A particle is moving so that the acceleration $a = 32x^3 + 48 x^2 + 16x$. Initially $x=1$ and $v=-8$.

## Part 1

Show that $\displaystyle a = \frac{d}{dx} \left( \frac{1}{2}v^2 \right)$.

\require{cancel} \displaystyle \begin{align} a &= \frac{dv}{dt} \\ &= \frac{\color{red}{dx}}{dt} \times \frac{dv}{\color{red}{dx}} \\ &= v \times \frac{dv}{dx} \\ &= \frac{d}{dv} \left( \frac{1}{2} v^2 \right) \times \frac{dv}{dx} \\ &= \frac{d}{\color{red}{\cancel{dv}}} \left( \frac{1}{2} v^2 \right) \times \frac{\color{red}{\cancel{dv}}}{dx} \\ \require{AMSsymbols} \therefore a &= \frac{d}{dx} \left( \frac{1}{2} v^2 \right) \end{align}

## Part 2

Show that $v^2 = 16x^2(1+x)^2$.

\displaystyle \begin{align} \frac{d}{dx} \left( \frac{1}{2}v^2 \right) &= a \\ &= 32x^3 + 48 x^2 + 16x \\ \frac{1}{2}v^2 &= \int (32x^3 + 48 x^2 + 16x) dx \\ &= 8x^4 + 16x^3 + 8x^2 + C \\ \text{When } x &= 1, v = -8 \\ \frac{1}{2} \times (-8)^2 &= 8 \times 1^4 + 16 \times 1^3 + 8 \times 1^2 + C \\ C &= 0 \\ \frac{1}{2} v^2 &= 8x^4 + 16x^3 + 8x^2 \\ v^2 &= 16x^4 + 32 x^3 + 16 x^2 \\ &= 16x^2 (x^2 + 2x + 1) \\ \require{AMSsymbols} \therefore v^2 &= 16x^2 (x+1)^2 \end{align}

## Part 3

Show that $v = -4x(x+1)$.

\displaystyle \begin{align} v^2 &= 16x^2 (x+1)^2 &\color{green}{\cdots \text{Part 2}} \\ v &= 4x(x+1) \text{ or } v = -4x(x+1) \\ \text{For } v &= 4x(x+1) \\ -8 &\ne 4 \times 1 \times (1+1) &\color{green}{\text{Substitute } x=1 \text{ and } v=-8} \\ \text{So } v &\ne 4x(x+1) \\ \text{For } v &= -4x(x+1) \\ -(-8) &= -4 \times 1 \times (1+1) &\color{green}{\text{Substitute } x=1 \text{ and } v=-8} \\ \require{AMSsymbols} \therefore v &= -4x(x+1) \end{align}

## Part 4

Hence, show that $\displaystyle \int \frac{1}{x(x+1)} dx = -4t$.

\displaystyle \begin{align} v &= -4x(x+1) &\color{green}{\cdots \text{Part 3}} \\ \frac{dx}{dt} &= -4x(x+1) \\ \frac{dt}{dx} &= \frac{1}{-4x(x+1)} \\ t &= \int \frac{1}{-4x(x+1)} dx \\ \therefore -4t &= \int \frac{1}{x(x+1)} dx \end{align}

## Part 5

Find $x$ as a functions of $t$, using $\displaystyle \log_e \left( 1 + \frac{1}{x} \right) = 4t + d$, for some constant $d$.

\displaystyle \begin{align} \log_e \left(1+\frac{1}{x} \right) &= 4t + d \\ \log_e \left(1+\frac{1}{1} \right) &= 4 \times 0 + d &\color{green}{\text{Substitute } x=1 \text{ and } t=0} \\ d &= \log_e 2 \\ \log_e \left(1+\frac{1}{x} \right) &= 4t + \log_e 2 \\ 1 + \frac{1}{x} &= e^{4t+ \ln 2} \\ &= e^{4t} \times e^{\ln 2} \\ &= e^{4t} \times 2 \\ \frac{1}{x} &= 2e^{4t} – 1 \\ \require{AMSsymbols} \therefore x &= \frac{1}{2e^{4t}-1} \end{align}

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