If a particle $P$ moves in a straight line and its position is given by the displacement function $x(t)$, then:

- the velocity of $P$ at time $t$ is given by $v(t) = x'(t)$
- the acceleration of $P$ at time $t$ is given by $a(t)=v'(t)=x^{\prime \prime}(t)$
- $x(0)$, $v(0)$ and $a(0)$ give the position, velocity and acceleration of the particle at time $t=0$, and these are called the initial conditions.

$$x(t) \xrightarrow{\text{differentiate}}v(t)=\dfrac{dx}{dt}\xrightarrow{\text{differentiate}} a(t)=\dfrac{dv}{dt}=\dfrac{d^2x}{dt^2}$$

## Sign Interpretation

Suppose a particle $P$ moves in a straight line with displacement function $s(t)$ relative to an origin $O$. Its velocity function is $v(t)$ and its acceleration function is $a(t)$.

The sign diagram is being used to interpret:

- where the particle is located relative to the origin
- the direction of motion and where a change of direction occurs
- when the particle’s velocity is increasing or decreasing

## Sign of Displacement $x(t)$

\( \begin{array}{|c|c|} \hline

x(t)=0 & \text{the particle is at the origin} \\ \hline

x(t) \gt 0 & \text{the particle is located at the right of the origin} \\ \hline

x(t) \lt 0 & \text{the particle is located at the left of the origin} \\ \hline

\end{array} \)

## Sign of Velocity $v(t)$

\( \begin{array}{|c|c|} \hline

v(t)=0 & \text{the particle is at rest} \\ \hline

v(t) \gt 0 & \text{the particle is moving to the right} \\ \hline

v(t) \lt 0 & \text{the particle is moving to the left} \\ \hline

\end{array} \)

## Sign of Acceleration $a(t)$

\( \begin{array}{|c|c|} \hline

a(t)=0 & \text{velocity is increasing} \\ \hline

a(t) \gt 0 & \text{velocity is decreasing} \\ \hline

a(t) \lt 0 & \text{velocity is constant} \\ \hline

\end{array} \)

## Speed

Velocities have magnitude and direction. In contrast, speed simply measures how fast something is travelling, regardless of the direction of travel. Speed is a scalar quantity that has size but no sign. Speed is always positive, but cannot be negative.

\( \begin{array}{|c|c|c|c|} \hline

v(t) & a(t) & \text{movement} & \text{speed} \\ \hline

+ & + & v=1, 2, 3, \cdots & s=1, 2, 3, \cdots \text{ increasing} \\ \hline

– & – & v=-1, -2, -3, \cdots & s=1, 2, 3, \cdots \text{ increasing} \\ \hline

+ & – & v=4, 3, 2, \cdots & s=4, 3, 2, \cdots \text{ decreasing} \\ \hline

– & + & v=-4, -3, -2, \cdots & s=4, 3, 2, \cdots \text{ decreasing} \\ \hline

\end{array} \)

### Example 1

A particle moves in a straight line with position relative to the origin given by $x(t)=2t^3 + 3t^2 -6$ metres, where $t$ is the time in seconds.

(a) Find the expression for the particle’s velocity.

\( \begin{align} \displaystyle

v(t) &= x'(t) \\

&= 6t^2 + 6t

\end{align} \)

(b) Find the expression for the particle’s acceleration.

\( \begin{align} \displaystyle

a(t) &= 12t + 6

\end{align} \)

(c) Describe the motion of its initial conditions.

\( \begin{align} \displaystyle

x(0) &= 2 \times 0^3 + 3 \times 0^2 -6 \\

&= -6 \\

v(2) &= 6 \times 0^2 + 6 \times 0 \\

&= 0 \\

a(0) &= 12 \times 0 + 6 \\

&= 6 \\

\end{align} \)

The particle is 6 metres to the left of the origin, at rest initially then moving to the right.

### Example 2

Represent the acceleration expression in terms of displacement and velocity.

\( \begin{align} \displaystyle a &= \frac{dx}{dt} \times \frac{dv}{dx} \\ &= v \times \frac{dv}{dx} &\require{ASMsymbols} \color{green}{\text{as } v = \frac{dx}{dt}} \\ &= \frac{d}{dv} \frac{v^2}{2} \times \frac{dv}{dx} \\ &= \frac{d}{dx} \frac{v^2}{2} \end{align} \)

### Example 3

Represent the acceleration expression in terms of displacement and time.

\( \begin{align} \displaystyle a &= \frac{dx}{dt} \times \frac{dv}{dx} \\ &= \frac{dx}{dt} \times \frac{d}{dx}v \\ &= \frac{dx}{dt} \times \frac{d}{dx}\frac{dx}{dt} & \require{ASMsymbols} \color{green}{\text{as } v = \frac{dx}{dt}} \\ &= \frac{d}{dt}\frac{dx}{dt} \\ &= \frac{d^2 x}{(dt)^2} \\ &= \frac{d^2 x}{dt^2} \end{align} \)

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