Movement and Position

1.4 · Average speed

Speed and velocity — same formula, different idea

Specification 1.4 asks you to know and use

Speed is a scalar: how fast, no direction. “$15\,\text{m/s}$” is a speed. Velocity is a vector: speed in a stated direction. “$15\,\text{m/s}$ due north” is a velocity. I teach both now because later graphs use the word velocity, and because Lesson B needs vectors.

Distance is the length of the path. If you walk $40\,\text{m}$ east and $40\,\text{m}$ west back to the start, the distance is $80\,\text{m}$ but you are back where you began. Average speed uses the $80\,\text{m}$. (Displacement — the straight-line change of position — is Physics-only point 1.2. You may meet the word; you are not examined on it in Double Award.)

Average speed is not “the speed in the middle”

Average speed is always

If a cyclist rides $12\,\text{km}$ in $30\,\text{min}$ then $12\,\text{km}$ in $50\,\text{min}$, you do not average the two speeds. You add: $24\,\text{km}$ in $80\,\text{min} = 80/60 = 1.333\,\text{h}$, so $v = 24/1.333 = 18\,\text{km/h}$.

Typical speeds I expect you to have a feel for

• Walking $\approx 1.5\,\text{m/s}$ · running $\approx 3\,\text{m/s}$ · cycling $\approx 6\,\text{m/s}$
• Car in a town $\approx 13\,\text{m/s}$ ($47\,\text{km/h}$) · motorway $\approx 30\,\text{m/s}$
• Sound in air $\approx 340\,\text{m/s}$ · if your answer for a person is hundreds of m/s, stop and check units

1.6 and 1.10

Acceleration — rate of change of velocity

If velocity is “how fast, and which way,” acceleration is “how quickly that velocity is changing.” Speeding up, slowing down, or changing direction are all accelerations. For this paper we almost always work in a straight line, so you can treat it as speeding up or slowing down.

$u$ = initial velocity, $v$ = final velocity, $t$ = time for that change. Unit: m/s$^2$. I say it out loud as “metres per second, per second” — each second, the speed changes by that many m/s.

Positive, negative, and the word deceleration

If we take the direction of travel as positive:

• $v > u$ → $a$ is positive → speeding up.
• $v < u$ → $a$ is negative → slowing down. You may call this a deceleration of the positive value. A deceleration of $3\,\text{m/s}^2$ is an acceleration of $-3\,\text{m/s}^2$.

A car going from $20\,\text{m/s}$ to $8\,\text{m/s}$ in $6.0\,\text{s}$:

In words: deceleration $2.0\,\text{m/s}^2$. Either form is acceptable if you are consistent.

The second equation — when time is missing

Specification 1.10 gives you a version that does not need time:

I teach students a 10-second check before they write anything:

1. Write $u$, $v$, $a$, $s$, $t$. Tick what you have. Question-mark what you want.
2. If time is given or wanted, start with $a = (v-u)/t$.
3. If time is missing and you have (or want) distance, use $v^2 = u^2 + 2as$.
4. Released from rest means $u = 0$. Comes to rest means $v = 0$.

A ball dropped from rest through $5.0\,\text{m}$ with $a = 10\,\text{m/s}^2$:

You take the positive root because we asked for speed downwards. Do not write $\pm 10$ unless the question is about direction both ways.

1.3 · Distance–time graphs

A distance–time graph is a story of a journey

The vertical axis is distance from the start (m). The horizontal axis is time (s). The one rule that must be automatic:

I make students say “rise over run” and then immediately “change in distance over change in time,” so they do not grab the wrong pair of numbers from the axes.

The five shapes you must recognise instantly

A line that comes back down towards the time axis means the object is returning towards the starting point (distance from start is falling). On a true distance graph some papers keep distance always rising because distance travelled cannot decrease. Read the axis label. If it says “distance from start” or “displacement,” a downward slope is possible. If it says “distance travelled,” the line should not go down.

Calculating speed from a straight section — do it like this

On the graph above, section A goes from $(0\,\text{s},\, 0\,\text{m})$ to $(10\,\text{s},\, 80\,\text{m})$ if the top of A is $80\,\text{m}$. Draw a large triangle on that straight line. Do not use a tiny triangle — small triangles magnify reading errors.

Section B is horizontal: speed $= 0$. Section C is steeper, so faster than A. That is all “explain this graph” questions want: name the motion in each time interval, and calculate a gradient where they ask for a value.

Curved graphs and a tangent

If the line is curved, the speed is changing. The speed at one instant is the gradient of a tangent — a straight line that just touches the curve at that time. Draw the tangent with a ruler, pick two far-apart points on the tangent, then $\Delta d / \Delta t$. You will not be asked this on every paper, but Cognito shows it and higher-demand questions do use it.

1.7, 1.8, 1.9

Velocity–time graphs — two skills, every year

I tell my classes: if you only remember two sentences from this entire topic, make them these.

1. Gradient of a $v$–$t$ graph = acceleration. $a = \Delta v / \Delta t$.
2. Area between the line and the time axis = distance travelled.

Reading the shape

• Straight line sloping up: constant acceleration.
• Horizontal line: constant velocity, so $a = 0$. The object is still moving if the line is above the axis.
• Straight line sloping down: constant deceleration.
• Steeper slope: larger magnitude of acceleration.
• Curve: acceleration is changing. You estimate a tangent if they ask for $a$ at a point.
• Line on the time axis: at rest ($v = 0$).

A horizontal line at $12\,\text{m/s}$ is not stationary. Stationary is a horizontal line on a distance–time graph, or $v = 0$ on a velocity–time graph. Mix those two up and you lose an easy mark.

Worked area — I will not let you skip the split

Suppose the graph goes $0 \to 16\,\text{m/s}$ in $4.0\,\text{s}$, stays at $16\,\text{m/s}$ until $8.0\,\text{s}$, then falls to $0$ at $12\,\text{s}$.

Acceleration $0$–$4\,\text{s}$:

Distance — split into three shapes:

• Triangle: $\tfrac{1}{2} \times 4.0 \times 16 = 32\,\text{m}$
• Rectangle: $4.0 \times 16 = 64\,\text{m}$
• Triangle: $\tfrac{1}{2} \times 4.0 \times 16 = 32\,\text{m}$

Total distance $= 128\,\text{m}$. A trapezium for the whole shape is fine if you prefer: average velocity $\times$ time $= 8.0 \times 12 = 96\,\text{m}$ — wait. That only works if you use the correct average. The mean of $0$ and $16$ is $8$, but the object spent extra time at $16$, so the trapezium of the whole journey is not a single triangle. That is why I insist you split. For $0$–$12\,\text{s}$ the honest method is three areas.

1.5 · Required practical

Investigate the motion of everyday objects

The specification names toy cars or tennis balls. The examiner is not testing whether you own light gates. They are testing whether you can write a method that would actually produce a speed or an acceleration, with variables, repeats, and a sensible error.

I teach three versions. Learn one in full and be able to mention the other two as improvements.

Version 1 — Ramp, toy car, metre rule, stopwatch

Aim.

To find how average speed (or acceleration) of a toy car depends on the distance along a ramp, or on the release height.

Apparatus.

Wooden ramp, toy car, metre rule or tape, stopwatch, marker pen or chalk, books to change the height, a soft barrier at the end.

Variables.

• Independent: distance $s$ from a start line to a finish line, or height of the top of the ramp.
• Dependent: time $t$ to travel $s$, then calculated average speed $v = d/t$ (or acceleration if you also measure a final speed).
• Control: same car, same surface, same release method (let go from rest — do not push), same start position relative to the top if you are changing distance.

Method I accept for 3 or 4 marks.

1. Set the ramp at a fixed, gentle angle. Mark a start line and a finish line a measured distance $s$ apart (for example $0.80\,\text{m}$).
2. Hold the car so its front is on the start line. Release it from rest — do not give it a push.
3. Start the stopwatch as you release (or as the front crosses the start) and stop it as the front crosses the finish.
4. Record $t$. Repeat two more times. Discard an obvious anomaly if one time is wildly different, then calculate a mean $t$.
5. Calculate average speed $v = s / t_{\text{mean}}$.
6. Repeat for several values of $s$ or several ramp heights. Plot a graph of $v$ against $s$ or against height.

If they ask for acceleration from rest.

Use $v^2 = u^2 + 2as$ with $u = 0$ and $v$ as the average speed only if the question allows that estimate — it is crude because average speed is not the final speed. A better classroom method: measure time over two short intervals, or use a ticker timer / light gates so you have $u$ and $v$.

Version 2 — Ticker timer (the one I want you to understand)

A ticker timer makes a dot on paper tape at a fixed frequency. In the UK and Singapore the mains is $50\,\text{Hz}$, so there are $50$ dots each second.

Attach the tape to the toy car. As the car moves, the dots spread out:

• Equal spacing → constant speed.
• Dots getting further apart → accelerating.
• Dots getting closer → decelerating.

To find a speed from a section of tape, count the gaps (not the dots). Five gaps is $5 \times 0.02 = 0.10\,\text{s}$. Measure that length $s$ in metres. Then $v = s / 0.10$. Do this for an early section ($u$) and a later section ($v$), find the time between the midpoints of those sections, then $a = (v-u)/t$.

Version 3 — Light gates (the improvement you should name)

A light gate is a beam and a detector. When a card on the car breaks the beam, a data logger records the time the beam is blocked. If the card has length $l$ and the blocked time is $t$,

Two light gates a known distance apart give $u$ and $v$ and the time between them, so you can calculate $a$ without a human stopwatch. That is the answer to “how could the student improve the accuracy of the time measurement?”

Errors, evaluation, and the sentences that score

Safety: do not stand in the path of the car; catch it with a padded stop; keep the ticker-timer voltage as specified (typically a low-voltage AC supply).

Check you can — then look

Cover the answers. These are the standard of a mid-topic homework.

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