# Motion Kinematics

## Displacement

Suppose an object $P$ moves along a straight line so that its position $s$ from an origin $O$ is given as some function of time $t$. We write $x=x(t)$ where $t \ge 0$.
$x(t)$ is a displacement function and for any value of $t$ it gives the displacement from the origin.
On the horizontal axis through $O$:

• if $x(t) \gt 0$, $P$ is located to the right of $O$
• if $x(t) = 0$, $P$ is located at $O$
• if $x(t) \lt 0$, $P$ is located to the left of $O$

## Motion Graphs

Consider $x(0)=2$, $x(1)=8$ and $x(2)=16$.
To appreciate the motion of $P$ we draw a motion graph.

## Velocity

The average velocity of an object moving in a straight line in the time interval from $t=t_1$ to $t=t_2$ is the ratio of the change in displacement to the time taken.

If $x(t)$ is the displacement function then:
$$\text{average velocity}=\dfrac{x(t_2)-x(t_1)}{t_2 – t_1}$$
On a graph of $x(t)$ against time $t$ for the time interval from $t=t_1$ to $t=t_2$, the average velocity is the gradient of a chord through the points $(t_1,x(t_1))$ and $(t_2,x(t_2))$.

If $x(t)$ is the displacement function of an object moving in a straight line, then:
$$v(t)=x'(t)=\lim_{h \rightarrow 0} \dfrac{x(t+h)-x(t)}{h}$$
is the instantaneous velocity or velocity function of the object at time $t$.
On a graph of $x(t)$ against time $t$, the instantaneous velocity at a particular time is the gradient of the tangent to the graph at that point.

## Acceleration

If an object moves in a straight line with velocity function $v(t)$ then its average acceleration for the time interval from $t=t_1$ to $t=t_2$ is the ratio of the change in velocity to the time taken.
$$\text{average acceleration}=\dfrac{v(t_2)-v(t_1)}{t_2 – t_1}$$
If a particle moves in a straight line with velocity function $v(t)$, then the instantaneous acceleration at time $t$ is:
$$a(t)=v'(t)=\lim_{h \rightarrow 0} \dfrac{v(t+h)-v(t)}{h}$$

### Example 1

A particle moves in a straight line with displacement from $O$ given by $x(t)=t^2-2t$ metres at time $t$ seconds. Find the average velocity for the time interval from $t=3$ to $t=5$ seconds.

\begin{align} \displaystyle \text{average velocity} &= \dfrac{x(5) – x(3)}{5-3} \\ &= \dfrac{(5^2 – 2 \times 5)-(3^2 – 2 \times 3)}{2} \\ &= \dfrac{15-3}{2} \\ &= 6 \text{ m s}^{-1} \end{align}

### Example 2

A particle moves in a straight line with displacement from $O$ given by $x(t)=t^2-2t$ metres at time $t$ seconds. Find the average velocity for the time interval from $t=3$ to $t=3+h$ seconds.

\begin{align} \displaystyle \text{average velocity} &= \dfrac{x(3+h) – x(3)}{3+h-3} \\ &= \dfrac{((3+h)^2 – 2 \times (3+h))-(3^2 – 2 \times 3)}{h} \\ &= \dfrac{9+6h+h^2 – 6 -2h – 3}{h} \\ &= \dfrac{4h+h^2}{h} \\ &= 4+h \text{ m s}^{-1}\\ \end{align}

### Example 3

A particle moves in a straight line with velocity $v(t)=3 \sqrt{t}+1$ metres at time $t$ m s$-1$. Find the average acceleration for the time interval from $t=1$ to $4$ seconds.

\begin{align} \displaystyle \text{average acceleration} &= \dfrac{v(4) – v(1)}{4-1} \\ &= \dfrac{(3\sqrt{4}+1)-(3\sqrt{1}+1)}{3} \\ &= \dfrac{(3\times 2+1)-(3\times 1+1)}{3} \\ &= \dfrac{7-2}{3} \\ &= \dfrac{5}{3} \text{ m s}^{-2}\\ \end{align}