Motion: Speed, Velocity & Acceleration

Distance and Displacement

When something moves, we often want to know how far it has travelled. There are actually two ways of answering that question, and in physics they mean different things.

Distance is the total length of the path travelled. It does not care about direction, so it is a scalar quantity. Displacement is the straight-line distance from start to finish in a stated direction, so it is a vector quantity. Both are measured in metres (m).

Imagine you walk 3 m east, then 4 m west. The distance you travelled is $3 + 4 = 7\text{ m}$. But your displacement is only $1\text{ m}$ west, because that is where you ended up relative to where you started.

Key terms Scalar — a quantity with size (magnitude) only, e.g. distance, speed, mass, time.

Vector — a quantity with both size and direction, e.g. displacement, velocity, acceleration, force.

Speed and Velocity

Speed tells you how fast an object is moving, without saying which way. It is a scalar. Velocity is speed in a given direction, so it is a vector. Both are measured in metres per second (m/s).

A car going round a roundabout at a steady $30\text{ m/s}$ has a constant speed, but its velocity is constantly changing because its direction keeps changing.

The everyday equation you will use most is:

$$\text{average speed} = \frac{\text{distance}}{\text{time}} \qquad v = \frac{d}{t}$$

Here $v$ is speed in m/s, $d$ is distance in m, and $t$ is time in s. Rearranged, $d = v\,t$ and $t = \dfrac{d}{v}$.

Worked example A runner completes a $400\text{ m}$ track in $50\text{ s}$. Calculate her average speed.

$v = \dfrac{d}{t} = \dfrac{400}{50} = 8\text{ m/s}$

Notice that because she finishes back where she started, her displacement is zero, so her average velocity for the whole lap is $0\text{ m/s}$, even though her average speed is $8\text{ m/s}$.

It helps to have a feel for some typical everyday speeds.

Exam tip Always include units and state direction when a question asks for a vector. "5 m/s" is a speed; "5 m/s north" is a velocity. Markers look for this.

Acceleration

Acceleration is how quickly velocity changes. It is a vector, measured in metres per second squared (m/s²).

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

where $u$ is the initial velocity, $v$ is the final velocity, $t$ is the time taken, and $a$ is the acceleration.

If an object speeds up, $a$ is positive. If it slows down, $a$ is negative — this is often called deceleration. An object that changes direction is also accelerating, even at constant speed, because velocity is a vector.

Worked example A car speeds up from $8\text{ m/s}$ to $20\text{ m/s}$ in $6\text{ s}$. Find its acceleration.

$a = \dfrac{v - u}{t} = \dfrac{20 - 8}{6} = \dfrac{12}{6} = 2\text{ m/s}^2$

Distance–Time Graphs

A distance–time graph plots distance on the vertical axis against time on the horizontal axis. The key idea is that the gradient (steepness) equals the speed.

A flat (horizontal) line means the object is stationary — distance is not changing.
A straight sloped line means constant speed.
A steeper line means a faster speed.
A curved line means the speed is changing (accelerating if getting steeper, decelerating if levelling off).

To find the speed from a straight section, calculate the gradient:

$$\text{speed} = \text{gradient} = \frac{\text{change in distance}}{\text{change in time}}$$

Worked example On a distance–time graph a line rises from $0\text{ m}$ to $60\text{ m}$ between $0\text{ s}$ and $12\text{ s}$. Find the speed.

$\text{speed} = \dfrac{60 - 0}{12 - 0} = \dfrac{60}{12} = 5\text{ m/s}$

Velocity–Time Graphs

A velocity–time graph plots velocity (vertical) against time (horizontal). Two pieces of information can be read off it:

The gradient equals the acceleration.
The area under the line equals the distance travelled.

A horizontal line means constant velocity (zero acceleration).
A straight sloped line means constant (uniform) acceleration.
A line sloping downwards means deceleration.

To find the distance, work out the area under the graph. Split awkward shapes into rectangles and triangles.

Worked example A train accelerates uniformly from rest to $30\text{ m/s}$ in $20\text{ s}$, then travels at $30\text{ m/s}$ for a further $40\text{ s}$.

Acceleration $= \text{gradient} = \dfrac{30 - 0}{20} = 1.5\text{ m/s}^2$

Distance in first stage (triangle) $= \frac{1}{2} \times 20 \times 30 = 300\text{ m}$

Distance in second stage (rectangle) $= 30 \times 40 = 1200\text{ m}$

Total distance $= 300 + 1200 = 1500\text{ m}$

Watch out Do not confuse the two graphs. On a distance–time graph a horizontal line means stopped. On a velocity–time graph a horizontal line means moving at constant velocity. Read the axis labels before you decide.

The Equations of Motion

For motion with constant acceleration, the Edexcel specification gives two useful equations:

$$v = u + at$$ $$v^2 = u^2 + 2as$$

where $u$ = initial velocity (m/s), $v$ = final velocity (m/s), $a$ = acceleration (m/s²), $t$ = time (s) and $s$ = distance/displacement (m).

Use $v = u + at$ when you know the time. Use $v^2 = u^2 + 2as$ when time is not given but distance is.

Worked example A motorbike accelerates from rest at $4\text{ m/s}^2$ over a distance of $50\text{ m}$. Find its final velocity.

Use $v^2 = u^2 + 2as$ with $u = 0$, $a = 4$, $s = 50$.

$v^2 = 0 + (2 \times 4 \times 50) = 400$

$v = \sqrt{400} = 20\text{ m/s}$

Worked example A ball is thrown upwards at $u = 15\text{ m/s}$ and slows under gravity at $a = -10\text{ m/s}^2$. How long until it momentarily stops ($v = 0$) at the top?

Use $v = u + at$: $\quad 0 = 15 + (-10)t \quad\Rightarrow\quad t = \dfrac{15}{10} = 1.5\text{ s}$

Exam tip Before substituting, write down each known value with its symbol ($u$, $v$, $a$, $t$, $s$) and tick off which one you are looking for. This makes choosing the right equation much easier and earns method marks even if the arithmetic slips.

Investigating Speed and Acceleration

You can measure speed and acceleration in the lab in two common ways.

Light gates. A light gate has a beam of light and a sensor. When a trolley with a card of known length passes through, the beam is broken. A connected timer records how long the beam is broken for. Then:

$$\text{speed} = \frac{\text{length of card}}{\text{time beam is blocked}}$$

Using two light gates a known distance apart lets you find the velocity at each gate, and the acceleration between them with $a = \dfrac{v - u}{t}$. Light gates are accurate because they remove human reaction-time error.

Ticker tape. A ticker-timer prints dots on a paper tape at a fixed rate (often 50 dots per second, so $0.02\text{ s}$ between dots). The tape is attached to a moving trolley.

Dots close together mean slow movement.
Dots spreading apart mean the trolley is accelerating.
Even spacing means constant speed.

By measuring the gap between dots and knowing the time per dot, you can calculate the speed over each interval, and how it changes to find acceleration.

Real world Driving instructors talk about "stopping distance", which is thinking distance plus braking distance. The braking part is exactly $v^2 = u^2 + 2as$ in action: doubling your speed $u$ roughly quadruples the braking distance $s$, because $s$ depends on $v^2$. That is why small increases in speed make a big difference to road safety.