# Displacement, Velocity and Distance Travelled by Natural Logarithmic Equations

A particle is moving in a straight line, starting from the origin. At time $t$ seconds, the particle has a displacement of $x$ metres from the origin and a velocity $v \ \text{m s}^{-1}$.
The displacement is given by $x = 3t-6 \log_e (t+1)$.

## Part 1

Find the expression for the velocity, $v$.

\begin{align} \displaystyle v &= \frac{dx}{dt} \\ &= \frac{d}{dx} 3t-6 \frac{d}{dx} \log_e (t+1) \\ &= 3-6 \times \frac{1}{t+1} \\ &= 3-\frac{6}{t+1} \end{align}

## Part 2

Find the initial velocity.

\begin{align} \displaystyle v (t=0) &= 3-\frac{6}{0+1} \\ &= -3 \end{align}

## Part 3

Determine the initial direction of the particle.

Moving backward as $v = -3 \lt 0$

## Part 4

Find when the particle comes to rest.

Particle comes to rest when $v=0$
\displaystyle \begin{align} 3-\frac{6}{t+1} &= 0 \\ -\frac{6}{t+1} &= -3 \\ \frac{6}{t+1} &= 3 \\ t+1 &= 2 \\ t &= 1 \end{align}

## Part 5

Find the distance travelled for the first second.

\require{AMSsymbols} \displaystyle \begin{align} x(t=0) &= 0 \\ x(t=1) &= 3 \times 1-6 \log_e (1+1) \\ & = 3-6 \log_e 2 \lt 0 \\ \text{distance travelled} &= 6 \log_e 2-3 &\color{green}{\cdots \text{moving backward} } \end{align}

## Part 6

Find the distance travelled that the particle comes back to the origin.

$2 \times (6 \log_e 2-3) = 12 \log_e 2-6$

## Part 7

Find the displacement at $t = 4$.

\begin{align} \displaystyle x(t=4) &= 3 \times 4-6 \log_e (4+1) \\ &= 12-6 \log_e 5 \end{align}

## Part 8

Find the distance travelled by the particle in the first four seconds.

\begin{align} (12 \log_e 2-6) + (12-6 \log_e 5) &= 6 + 12 \log_e 2 + 6 \log_e 5 \\ &= 23.974 \cdots \text{metres} \end{align}

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