Displacement, Velocity and Acceleration

AS · M1 · 10 min

Kinematics is the description of motion: where something is, how fast it is going and how quickly that is changing, without yet asking what causes it. Paper 4 deals only with motion in a straight line, so every quantity is a single number with a sign. This note sets up the language: the difference between distance and displacement, speed and velocity, and what acceleration and deceleration mean. Almost every kinematics mark lost in the exam traces back to confusing one of these pairs.

Scalars and vectors in one dimension

A scalar has size only. A vector has size and direction. In a straight line there are only two directions, so a vector is just a number whose sign gives the direction: choose one direction as positive and the other is negative.

Definition

Distance is the total length of path travelled. It is a scalar and never decreases.
Displacement is the position relative to a fixed origin, measured along the line, with a sign for direction. It is a vector.
Speed is the rate at which distance is covered. It is a scalar and is never negative.
Velocity is the rate of change of displacement. It is a vector: its size is the speed and its sign gives the direction of motion.
Acceleration is the rate of change of velocity. It is a vector.

The syllabus states this precisely: distance and speed are scalar quantities; displacement, velocity and acceleration are vector quantities.

Quantity	Type	Symbol	SI unit
distance	scalar	$d$	$\text{m}$
displacement	vector	$s$ (or $x$ )	$\text{m}$
speed	scalar		$\text{m s}^{-1}$
velocity	vector	$v$ (initial $u$ )	$\text{m s}^{-1}$
acceleration	vector	$a$	$\text{m s}^{-2}$
time	scalar	$t$	$\text{s}$

Distance versus displacement

If a particle always moves in the same direction, distance travelled and the size of the displacement are the same. They differ as soon as the particle turns back.

The distance walked is 400 + 150 = 550 m, but the final displacement from O is only 250 m east.

A walker goes $400\ \text{m}$ east from $O$ to $A$ , then $150\ \text{m}$ west to $B$ . Taking east as positive:

distance travelled $= 400 + 150 = 550\ \text{m}$ ;
displacement from $O$ $= +400 - 150 = +250\ \text{m}$ , that is $250\ \text{m}$ east of $O$ .

If she then walks all the way back to $O$ , the distance becomes $800\ \text{m}$ but the displacement returns to $0$ .

Speed versus velocity

A car going round a roundabout at a steady $10\ \text{m s}^{-1}$ has constant speed but changing velocity. In a straight line the difference is just the sign: a particle moving at $5\ \text{m s}^{-1}$ in the negative direction has velocity $-5\ \text{m s}^{-1}$ and speed $5\ \text{m s}^{-1}$ .

Averages are where the two really differ:

Averages

\text{average speed} = \frac{\text{total distance travelled}}{\text{total time taken}}, \qquad \text{average velocity} = \frac{\text{change in displacement}}{\text{total time taken}}

Average speed is not the mean of the speeds unless equal times are spent at each. Example 4 shows the trap.

For motion at constant velocity, displacement grows at a steady rate:

s = vt

Acceleration and deceleration

Acceleration measures how quickly velocity changes:

a = \frac{\text{change in velocity}}{\text{time taken}} = \frac{v - u}{t} \quad\text{(when the acceleration is constant)}

An acceleration of $3\ \text{m s}^{-2}$ means the velocity increases by $3\ \text{m s}^{-1}$ every second.

The sign of the acceleration compares with the sign of the velocity:

If $v$ and $a$ have the same sign, the particle is speeding up.
If $v$ and $a$ have opposite signs, the particle is slowing down.

So a negative acceleration does not always mean slowing down. A ball falling downwards (taking up as positive) has negative velocity and negative acceleration, and it is speeding up.

Definition

Deceleration (or retardation) means acceleration that reduces speed. The syllabus notes that "deceleration" may be used for decreasing speed. A deceleration of $2\ \text{m s}^{-2}$ in the direction of motion is an acceleration of $-2\ \text{m s}^{-2}$ .

Watch out

If a question says "decelerates at $2\ \text{m s}^{-2}$ ", substitute $a = -2$ (with the direction of motion positive). Do not write $a = 2$ and also subtract; the minus sign goes in exactly once.

Units and conversions

Paper 4 uses SI units: metres, seconds, $\text{m s}^{-1}$ , $\text{m s}^{-2}$ . Questions sometimes give speeds in $\text{km h}^{-1}$ to test conversion.

Key result

1\ \text{km h}^{-1} = \frac{1000\ \text{m}}{3600\ \text{s}} = \frac{1}{3.6}\ \text{m s}^{-1}

Divide by $3.6$ to go from $\text{km h}^{-1}$ to $\text{m s}^{-1}$ ; multiply by $3.6$ to go back. So $72\ \text{km h}^{-1} = 20\ \text{m s}^{-1}$ .

Choosing a positive direction

Before any calculation, decide which direction is positive and write it down ("taking the direction of motion as positive", "taking upwards as positive"). Then:

every displacement, velocity and acceleration gets a sign according to that choice;
a negative answer for a velocity means motion in the negative direction;
a negative answer for a displacement means the particle is on the negative side of the origin.

Changing the choice changes every sign but not the physics. What you must not do is change it halfway through a question.

Distance, displacement and the two averages

A walker goes $400\ \text{m}$ due east in $320\ \text{s}$ , then $150\ \text{m}$ due west in $130\ \text{s}$ . Find her average speed and her average velocity for the whole walk.

Solution

Total distance $= 550\ \text{m}$ ; total time $= 450\ \text{s}$ ; final displacement $= 250\ \text{m}$ east.

\text{average speed} = \frac{550}{450} = 1.22\ \text{m s}^{-1}

\text{average velocity} = \frac{250}{450} = 0.556\ \text{m s}^{-1} \text{ due east}

A velocity needs a direction: " $0.556\ \text{m s}^{-1}$ " on its own is incomplete.

Acceleration with unit conversion

A car increases its speed from $36\ \text{km h}^{-1}$ to $90\ \text{km h}^{-1}$ in $8\ \text{s}$ , with constant acceleration. Find the acceleration in $\text{m s}^{-2}$ .

Solution

Convert first: $36 \div 3.6 = 10\ \text{m s}^{-1}$ and $90 \div 3.6 = 25\ \text{m s}^{-1}$ .

a = \frac{25 - 10}{8} = 1.875\ \text{m s}^{-2} = 1.88\ \text{m s}^{-2} \text{ (3 s.f.)}

Mixing units (subtracting km/h and dividing by seconds) is a common way to lose all the marks.

Signs of velocity and acceleration

A ball is thrown vertically upwards. Taking upwards as positive, state the sign of the velocity and of the acceleration (a) on the way up, (b) at the highest point, (c) on the way down. In each case say whether the ball is speeding up or slowing down.

Solution

Gravity gives an acceleration of $10\ \text{m s}^{-2}$ downwards throughout, so $a = -10$ at every stage (including the top).

(a) Velocity positive, acceleration negative: opposite signs, so the ball slows down.

(b) Velocity zero (momentarily), acceleration still $-10\ \text{m s}^{-2}$ . The ball is not "stopped"; its velocity is changing from positive to negative.

The ball's acceleration is not zero at the top. This is a favourite misconception, and it matters: if the acceleration were zero there, the ball would stay at the top for ever.

Average speed is not the average of the speeds

A cyclist rides $6\ \text{km}$ at $12\ \text{km h}^{-1}$ and then another $6\ \text{km}$ at $24\ \text{km h}^{-1}$ along the same straight road. Find the average speed for the whole journey.

Solution

Times: $\dfrac{6}{12} = 0.5\ \text{h}$ and $\dfrac{6}{24} = 0.25\ \text{h}$ .

\text{average speed} = \frac{12}{0.75} = 16\ \text{km h}^{-1}

Not $18\ \text{km h}^{-1}$ : the cyclist spends twice as long at the slower speed, so the average is pulled towards $12$ .

Two runners meeting

Points $A$ and $B$ are $300\ \text{m}$ apart on a straight track. Runner $P$ leaves $A$ and runs towards $B$ at a constant $6\ \text{m s}^{-1}$ . Ten seconds later runner $Q$ leaves $B$ and runs towards $A$ at a constant $4\ \text{m s}^{-1}$ . Find when and where they meet.

Solution

Let $t$ be the time in seconds after $P$ starts. $P$ has run $6t$ metres; $Q$ has run $4(t - 10)$ metres (for $t \ge 10$ ). They meet when together they have covered the $300\ \text{m}$ gap:

6t + 4(t - 10) = 300 \quad\Rightarrow\quad 10t = 340 \quad\Rightarrow\quad t = 34\ \text{s}

They meet $34\ \text{s}$ after $P$ starts, at $6 \times 34 = 204\ \text{m}$ from $A$ . Check: $Q$ has run $4 \times 24 = 96\ \text{m}$ and $204 + 96 = 300$ .

The key modelling step is giving each runner an expression for distance in terms of the same time variable, then writing the condition for meeting.

Common mistakes

Kinematics vocabulary errors

Giving a distance when asked for a displacement, or the reverse. Read which one the question wants.
Velocity without a direction. In words (" $4\ \text{m s}^{-1}$ towards $A$ ") or by sign, but always give it.
Average speed as the mean of speeds. Always total distance over total time.
Zero acceleration at the top of a throw. Velocity is zero there; acceleration is still $g$ downwards.
Unit mixing between $\text{km h}^{-1}$ and $\text{m s}^{-1}$ , or minutes and seconds.

Exam technique

Exam tip

State your positive direction at the start of any kinematics answer.
Answers to "find the speed" must be positive; answers to "find the velocity" need a sign or direction.
When the question says "decelerates" or "retards", decide the sign once and stick to it.
Final answers to 3 significant figures unless exact. Times in seconds unless the question works in minutes or hours.

Summary

Distance and speed are scalars; displacement, velocity and acceleration are vectors. In one dimension, a vector is a signed number.
Distance equals the size of the displacement only if the particle never turns back.
Average speed $=$ total distance $\div$ total time; average velocity $=$ displacement $\div$ time.
Acceleration is the rate of change of velocity. Same sign as velocity: speeding up; opposite sign: slowing down.
Deceleration of $d$ means $a = -d$ in the direction of motion.
$1\ \text{km h}^{-1} = \tfrac{1}{3.6}\ \text{m s}^{-1}$ .
Fix a positive direction before you start and keep it.

Practice

Question

A particle moves $12\ \text{m}$ in the positive direction and then $20\ \text{m}$ in the negative direction. Find the distance travelled and the final displacement.
A train slows from $108\ \text{km h}^{-1}$ to $54\ \text{km h}^{-1}$ in $12\ \text{s}$ with constant deceleration. Find the deceleration in $\text{m s}^{-2}$ .
A boat travels $1200\ \text{m}$ upstream in $200\ \text{s}$ , waits for $40\ \text{s}$ , then travels $800\ \text{m}$ back downstream in $160\ \text{s}$ . Find its average speed and its average velocity for the whole trip.
Taking upwards as positive, a lift has velocity $-3\ \text{m s}^{-1}$ and acceleration $+0.5\ \text{m s}^{-2}$ . Describe its motion.
Two trains are $8\ \text{km}$ apart on parallel straight tracks and travel towards each other at constant speeds of $25\ \text{m s}^{-1}$ and $15\ \text{m s}^{-1}$ . Find how long it takes for them to meet and how far the faster train has travelled.
A car travelling at a constant $30\ \text{m s}^{-1}$ passes a stationary police motorbike. Two seconds later the motorbike sets off in pursuit at a constant $35\ \text{m s}^{-1}$ (ignore its acceleration phase). Find how long after the car passed the motorbike sets off it catches the car, and how far from its starting point.
A runner goes up a straight hill path at $12\ \text{km h}^{-1}$ and comes straight back down the same path at $20\ \text{km h}^{-1}$ . Find the average speed for the round trip, and the average velocity.
A particle starts at the origin $O$ and moves along a straight line. For $5\ \text{s}$ it moves at $4\ \text{m s}^{-1}$ in the positive direction; for the next $3\ \text{s}$ it is at rest; for the next $6\ \text{s}$ it moves at $5\ \text{m s}^{-1}$ in the negative direction. Find its final displacement, the total distance travelled, its average speed and its average velocity.

Answers

Distance $= 12 + 20 = 32\ \text{m}$ . Displacement $= 12 - 20 = -8\ \text{m}$ ( $8\ \text{m}$ in the negative direction).
$108 \div 3.6 = 30$ and $54 \div 3.6 = 15\ \text{m s}^{-1}$ . $a = \dfrac{15 - 30}{12} = -1.25$ , so the deceleration is $1.25\ \text{m s}^{-2}$ .
Total distance $2000\ \text{m}$ , total time $400\ \text{s}$ (including the wait): average speed $5\ \text{m s}^{-1}$ . Displacement $1200 - 800 = 400\ \text{m}$ upstream: average velocity $1\ \text{m s}^{-1}$ upstream.
The lift is moving downwards at $3\ \text{m s}^{-1}$ and its acceleration is upwards, opposite to the velocity, so it is slowing down: it is decelerating as it descends.
The gap closes at $25 + 15 = 40\ \text{m s}^{-1}$ , so they meet after $8000 \div 40 = 200\ \text{s}$ . The faster train has travelled $25 \times 200 = 5000\ \text{m}$ .
Let $t$ be the time since the car passed. Car: $30t$ ; motorbike: $35(t - 2)$ . Equal when $35t - 70 = 30t$ , so $t = 14\ \text{s}$ ; that is $12\ \text{s}$ after the motorbike sets off. Distance $= 30 \times 14 = 420\ \text{m}$ .
Let the path have length $d\ \text{km}$ . Time $= \dfrac{d}{12} + \dfrac{d}{20} = \dfrac{8d}{60} = \dfrac{2d}{15}$ . Average speed $= \dfrac{2d}{2d/15} = 15\ \text{km h}^{-1}$ . The runner ends where she started, so the average velocity is $0$ .
Displacement $= 4 \times 5 + 0 - 5 \times 6 = 20 - 30 = -10\ \text{m}$ . Distance $= 20 + 30 = 50\ \text{m}$ . Total time $14\ \text{s}$ . Average speed $= \dfrac{50}{14} = 3.57\ \text{m s}^{-1}$ ; average velocity $= \dfrac{-10}{14} = -0.714\ \text{m s}^{-1}$ (that is $0.714\ \text{m s}^{-1}$ in the negative direction).

Scalars and vectors in one dimension

Distance versus displacement

Speed versus velocity

Acceleration and deceleration

Units and conversions

Choosing a positive direction

Common mistakes

Exam technique

Summary

Practice

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