Position, displacement, velocity and acceleration

Kinematics is the language used to describe motion. Before choosing an equation, you must know whether a number represents a location, a change of location, a path length, a rate, or a directed rate.

This lesson concentrates on motion along a straight line. In one dimension, direction is represented by a sign.

Prerequisites

You should be able to:

• use negative numbers and coordinates on a number line;
• calculate a change as final value minus initial value;
• calculate an average rate as change divided by elapsed time;
• convert units such as kilometres, hours, metres and seconds using quantities and units in mechanics.

Establishing position

To describe position on a line, choose:

1. a fixed origin, labelled $O$;
2. a positive direction;
3. a coordinate, often $x$ or $s$, measured from the origin.

For example, suppose east is positive and a particle is at position $x=-4\ \mathrm m$. The particle is $4\ \mathrm m$ west of the origin. The minus sign describes which side of the origin it occupies. It does not mean that the particle has travelled a negative distance.

Position alone does not reveal motion. A particle at $x=-4\ \mathrm m$ could be moving east, moving west, or instantaneously at rest.

Particle

In mechanics, a particle is an object whose size is ignored by the model. It can still have mass and move. This modelling convention lets one coordinate describe its position.

Displacement and distance

If a particle moves from position $x_1$ to position $x_2$, its displacement is

$$\boxed{\text{displacement}=\text{final position}-\text{initial position}}$$

or

$$\Delta x=x_2-x_1.$$

Displacement is directed, so it may be positive, negative or zero. Distance is the total length of the path followed, so it is never negative.

Quantity	What it measures	Can be negative?	Depends on the whole path?
position	location relative to an origin	yes	no
displacement	change in position	yes	no, only the endpoints
distance	total path length	no	yes

For a single movement without reversal,

$$\text{distance}=|\text{displacement}|.$$

After a reversal, the distance is usually greater than the magnitude of the displacement.

Worked example 1: crossing the origin

East is positive. A particle moves from $x=-7\ \mathrm m$ to $x=5\ \mathrm m$ without changing direction.

Its displacement is

$$\Delta x=5-(-7)=12\ \mathrm m.$$

Thus it has displacement $12\ \mathrm m$ east. Since it did not reverse, the distance travelled is also

$$12\ \mathrm m.$$

The initial position was negative, but the displacement is positive. Position and displacement are different quantities.

Worked example 2: a journey with a reversal

On the same axis, a particle starts at $x=-3\ \mathrm m$, moves to $x=8\ \mathrm m$, then finishes at $x=2\ \mathrm m$.

The overall displacement uses only the starting and finishing positions:

$$\Delta x=2-(-3)=5\ \mathrm m.$$

For distance, add the lengths of both stages:

$$|-3\mathbin{\to}8|+|8\mathbin{\to}2| =11+6 =17\ \mathrm m.$$

The particle’s displacement is $5\ \mathrm m$ in the positive direction, while its distance travelled is $17\ \mathrm m$.

Zero displacement does not mean no journey

If a runner completes one lap and returns to the starting point, the final and initial positions are equal. The displacement is zero, but the distance is one lap length.

Self-check 1

North is positive. A lift starts $6\ \mathrm m$ above a reference floor, descends to $10\ \mathrm m$ below it, then rises to $2\ \mathrm m$ below it. Find its final position, displacement and distance travelled.

Answer

The coordinates are $6$, $-10$ and $-2$ metres.

Final position:

$$x=-2\ \mathrm m.$$

Displacement:

$$\Delta x=-2-6=-8\ \mathrm m,$$

so the lift’s displacement is $8\ \mathrm m$ south.

Distance:

$$|-10-6|+|-2-(-10)|=16+8=24\ \mathrm m.$$

Speed and velocity

Average speed is total distance divided by elapsed time:

$$\boxed{\text{average speed}=\frac{\text{total distance}}{\text{total time}}}.$$

Average velocity is displacement divided by elapsed time:

$$\boxed{\text{average velocity}=\frac{\text{displacement}}{\text{total time}}} =\frac{\Delta x}{\Delta t}.$$

Speed has magnitude but no direction. Velocity includes direction, so in one dimension it may be signed. The SI unit of both is $\mathrm{m\,s^{-1}}$.

At a particular instant, the instantaneous speed is the magnitude of the instantaneous velocity:

$$\boxed{\text{speed}=|v|}.$$

Thus $v=-9\ \mathrm{m\,s^{-1}}$ means motion at a speed of $9\ \mathrm{m\,s^{-1}}$ in the negative direction.

Worked example 3: average speed and average velocity

A cyclist travels $300\ \mathrm m$ east in $50\ \mathrm s$, then $180\ \mathrm m$ west in $30\ \mathrm s$. Take east as positive.

The total distance is

$$300+180=480\ \mathrm m,$$

and the total time is

$$50+30=80\ \mathrm s.$$

Therefore

$$\text{average speed}=\frac{480}{80}=6\ \mathrm{m\,s^{-1}}.$$

The displacement is

$$300-180=120\ \mathrm m,$$

so

$$\text{average velocity}=\frac{120}{80}=1.5\ \mathrm{m\,s^{-1}}.$$

The average velocity is $1.5\ \mathrm{m\,s^{-1}}$ east. The average speed is larger because it counts motion in both directions.

Do not average speeds without considering time

A vehicle travelling at $20\ \mathrm{m\,s^{-1}}$ for one second and $10\ \mathrm{m\,s^{-1}}$ for nine seconds does not have average speed $15\ \mathrm{m\,s^{-1}}$. Always use total distance divided by total time.

Worked example 4: average speed over unequal times

A car travels at $20\ \mathrm{m\,s^{-1}}$ for $1\ \mathrm s$, then at $10\ \mathrm{m\,s^{-1}}$ for $9\ \mathrm s$, without reversing.

The distances are

$$d_1=20(1)=20\ \mathrm m, \qquad d_2=10(9)=90\ \mathrm m.$$

Hence

$$\text{average speed} =\frac{20+90}{1+9} =11\ \mathrm{m\,s^{-1}}.$$

The longer interval at the lower speed has more influence on the average.

Self-check 2

A runner completes a $400\ \mathrm m$ lap in $80\ \mathrm s$ and finishes where they started. Find the average speed and average velocity.

Answer

$$\text{average speed}=\frac{400}{80}=5\ \mathrm{m\,s^{-1}}.$$

The displacement is zero, so

$$\text{average velocity}=\frac{0}{80}=0\ \mathrm{m\,s^{-1}}.$$

Acceleration

Acceleration measures the rate of change of velocity. Average acceleration over a time interval is

$$\boxed{\text{average acceleration} =\frac{\text{final velocity}-\text{initial velocity}}{\text{elapsed time}}}$$

or, using the standard initial and final velocity symbols,

$$\boxed{a=\frac{v-u}{t}}$$

when acceleration is constant. Its SI unit is $\mathrm{m\,s^{-2}}$. An acceleration of $3\ \mathrm{m\,s^{-2}}$ means that velocity changes by $3\ \mathrm{m\,s^{-1}}$ each second.

Acceleration is directed. Its sign describes the direction in which velocity is changing:

• $a>0$ means velocity is becoming more positive;
• $a<0$ means velocity is becoming more negative.

Negative acceleration does not necessarily mean slowing down. Speed depends on the magnitude $|v|$.

Velocity $v$	Acceleration $a$	What happens to speed?
positive	positive	increases
positive	negative	decreases
negative	positive	decreases
negative	negative	increases

In short:

$$\boxed{v\text{ and }a\text{ have the same sign} \Longrightarrow \text{speeding up}}$$ $$\boxed{v\text{ and }a\text{ have opposite signs} \Longrightarrow \text{slowing down}}.$$

Worked example 5: velocity becoming more negative

East is positive. A particle’s velocity changes uniformly from

$$u=-4\ \mathrm{m\,s^{-1}}$$

to

$$v=-10\ \mathrm{m\,s^{-1}}$$

in $3\ \mathrm s$. Its acceleration is

$$a=\frac{v-u}{t} =\frac{-10-(-4)}{3} =-2\ \mathrm{m\,s^{-2}}.$$

Both velocity and acceleration are negative, so the particle is speeding up westwards. Its speed has increased from $4\ \mathrm{m\,s^{-1}}$ to $10\ \mathrm{m\,s^{-1}}$.

Worked example 6: changing direction

East is positive. A particle’s velocity changes uniformly from $6\ \mathrm{m\,s^{-1}}$ to $-2\ \mathrm{m\,s^{-1}}$ in $4\ \mathrm s$.

$$a=\frac{-2-6}{4}=-2\ \mathrm{m\,s^{-2}}.$$

At first $v>0$ and $a<0$, so the particle slows down while moving east. It reaches $v=0$, then $v<0$ while $a<0$, so it speeds up moving west.

Because the velocity changes by $2\ \mathrm{m\,s^{-1}}$ each second, the sequence is

$$6,\ 4,\ 2,\ 0,\ -2.$$

The particle is at rest after $3\ \mathrm s$ and then reverses direction. A negative final velocity is meaningful, not an error.

Self-check 3

Take upwards as positive. A ball has velocity $-5\ \mathrm{m\,s^{-1}}$ and acceleration $-9.8\ \mathrm{m\,s^{-2}}$. State its direction of motion and whether it is speeding up or slowing down.

Answer

The negative velocity means the ball is moving downwards. Velocity and acceleration have the same sign, so its speed is increasing. It is speeding up as it falls.

Instantaneous rates and calculus

Average velocity describes an interval. Instantaneous velocity describes motion at one instant. If position $x$ varies smoothly with time $t$, then

$$v=\frac{dx}{dt}.$$

Acceleration is the instantaneous rate of change of velocity:

$$a=\frac{dv}{dt}=\frac{d^2x}{dt^2}.$$

These derivatives also have graphical meanings:

• velocity is the gradient of a displacement-time graph;
• acceleration is the gradient of a velocity-time graph;
• displacement is the signed area under a velocity-time graph.

The full calculus treatment appears in calculus in kinematics, while kinematics graphs develops the graph interpretations.

Worked example 7: reading meaning from a position function

A particle’s position is

$$x=t^2-6t+5,\qquad t\geq0,$$

where $x$ is in metres and $t$ is in seconds.

Differentiate to find velocity:

$$v=\frac{dx}{dt}=2t-6.$$

Differentiate again to find acceleration:

$$a=\frac{dv}{dt}=2\ \mathrm{m\,s^{-2}}.$$

At $t=2$,

$$x(2)=4-12+5=-3\ \mathrm m,$$ $$v(2)=4-6=-2\ \mathrm{m\,s^{-1}}.$$

The particle is on the negative side of the origin and is moving in the negative direction. Since $v<0$ but $a>0$, it is slowing down at this instant.

At $t=3$, $v=0$. For $t>3$, velocity becomes positive, so the particle changes direction at $t=3$.

A reliable translation method

When reading a mechanics question:

1. Draw a line and mark the positive direction.
2. Mark the origin if positions are involved.
3. Attach signs to directed quantities: displacement, velocity and acceleration.
4. Keep distance, speed and time non-negative.
5. Distinguish a position $x$ from a displacement $\Delta x$ or $s$.
6. Check the final answer in words. State a direction when the sign carries physical meaning.

Common language translations

Words in a question	Mathematical meaning
starts from rest	$u=0$
comes to rest	$v=0$ at that instant
returns to its starting point	overall displacement $=0$
moves with constant speed	$
moves with constant velocity	magnitude and direction of $v$ are constant
uniform acceleration	acceleration is constant
decelerates while moving in the positive direction	$v>0$ and $a<0$
decelerates while moving in the negative direction	$v<0$ and $a>0$

The word deceleration can obscure signs. It is safer to choose a positive direction and assign the acceleration its correct sign.

Mixed self-check

A particle moves on a straight line with east positive. It starts at $x=4\ \mathrm m$. It travels $10\ \mathrm m$ west in $5\ \mathrm s$, then $6\ \mathrm m$ east in $3\ \mathrm s$.

1. Find its final position.
2. Find its displacement and distance travelled.
3. Find its average velocity and average speed for the complete journey.

Answer

The signed changes in position are $-10\ \mathrm m$ and $+6\ \mathrm m$. Therefore

$$x_{\text{final}}=4-10+6=0\ \mathrm m.$$

The displacement is

$$0-4=-4\ \mathrm m,$$

which is $4\ \mathrm m$ west. The distance is

$$10+6=16\ \mathrm m.$$

The total time is $8\ \mathrm s$, so

$$\text{average velocity}=\frac{-4}{8}=-0.5\ \mathrm{m\,s^{-1}},$$

and

$$\text{average speed}=\frac{16}{8}=2\ \mathrm{m\,s^{-1}}.$$

What to learn next

You should now be able to distinguish every core kinematics quantity and interpret its sign.

• Use these quantities in the constant acceleration and SUVAT equations.
• Connect gradients and areas to motion in kinematics graphs.
• Extend the definitions to variable acceleration in calculus in kinematics.
• For motion with components in two directions, continue to two-dimensional motion.