1.2 Motion

Learning objectives

1.2.1Speed as distance travelled per unit• Define speed as distance travelled per unit time; recall/use: v = s/t 1.2.2Velocity as speed in a given direction• Define velocity as speed in a given direction 1.2.3Recall/use: average speed = total• Recall/use: average speed = total distance travelled / total time taken 1.2.4Distance–time and speed–time graphs• Sketch, plot and interpret distance–time and speed–time graphs 1.2.5From given data or the shape of a• Determine qualitatively from given data or the shape of a distance–time graph or speed–time graph when an object is: (a) at rest (b) moving with constant speed (c) accelerating (d) decelerating 1.2.6Speed from the gradient of a• Calculate speed from the gradient of a straight-line section of a distance–time graph 1.2.7Area under a speed–time graph to• Calculate the area under a speed–time graph to determine the distance travelled for motion with constant speed or constant acceleration 1.2.8Acceleration of free fall g for an• State: the acceleration of free fall g for an object near to the surface of the Earth is approximately constant, about 9.8 m/s² 1.2.9Acceleration as change in velocity per• Define acceleration as change in velocity per unit time; recall/use: a = Δv/Δt 1.2.10From given data or the shape of a• Determine from given data or the shape of a speed–time graph when an object is moving with: (a) constant acceleration (b) changing acceleration 1.2.11Acceleration from the gradient of a• Calculate acceleration from the gradient of a speed–time graph 1.2.12A deceleration is a negative• Know: a deceleration is a negative acceleration and use this in calculations 1.2.13Motion of objects falling in a uniform• Describe the motion of objects falling in a uniform gravitational field with and without air/liquid resistance, including reference to terminal velocity

Speed is the distance travelled per unit time. It describes how quickly distance is covered and has no direction.

$v = \frac{s}{t}$

Symbol	Meaning	Common SI unit
$v$	speed	m/s
$s$	distance travelled	m
$t$	time taken	s

Use a distance and time that refer to the same part of the journey, convert them to compatible units, substitute, and give the speed with its unit.

Speed uses distance, not displacement. Rearrange before substituting when the unknown is distance or time: $s=vt$ and $t=s/v$ .

Velocity is speed in a given direction. A velocity is complete only when both its magnitude and direction are stated.

Motion statement	Speed	Velocity
5 m/s east	5 m/s	5 m/s east
constant speed around a circle	constant	changing, because direction changes
constant speed in a fixed straight-line direction	constant	constant

Two objects can have the same speed but different velocities. A change of direction is a change of velocity even when speed stays constant.

Average speed compares the total distance travelled over the whole journey with the total time taken for that journey.

$\text{average speed} = \frac{\text{total distance travelled}}{\text{total time taken}}$

Step	Action
1	add every distance travelled
2	add every time interval, including stops when they are part of the journey
3	divide total distance by total time
4	check that distance and time units are compatible

Do not usually take the arithmetic mean of two speeds. Equal distances at different speeds take different times, so calculate each time and then use the totals.

A motion graph shows how one measured quantity changes with time. Time is on the horizontal axis; distance, speed or velocity is on the vertical axis.

Task	Reliable action
plot	choose a linear scale, label quantity and unit, plot accurately, then join as instructed
read	identify the graph type and read both coordinates with their units
compare intervals	split at corners or changes of curvature and interpret each interval separately
sketch	preserve the required starting value, sequence, straight/curved shape and relative steepness

On a distance–time graph, gradient represents speed. On a speed–time graph, gradient represents acceleration and area under the graph represents distance travelled.

The same line shape has different meanings on different graph types. Always read the vertical-axis label before interpreting a horizontal or sloping section.

Translate each graph section by asking what its height and gradient mean on that particular graph.

Motion	Distance–time graph	Speed–time graph
at rest	horizontal: distance unchanged	on the time axis: speed zero
constant speed	straight line with constant non-zero gradient	horizontal above the time axis
accelerating	gradient becomes steeper	speed rises
decelerating	gradient becomes less steep	speed falls

A horizontal distance–time line means rest, but a horizontal speed–time line above zero means motion at constant speed. Curvature shows changing gradient, not automatically one named motion without checking the axes.

The gradient of a straight section of a distance–time graph is the speed during that section.

$v = \frac{\Delta s}{\Delta t}$

Step	Action
1	choose two well-separated points on the same straight section
2	read their coordinates $(t_1,s_1)$ and $(t_2,s_2)$
3	calculate rise $\Delta s=s_2-s_1$ and run $\Delta t=t_2-t_1$
4	divide and state the unit, such as m/s

Do not use height divided by time unless the chosen straight line passes through the origin. The question limits calculation here to a straight-line section, whose gradient is constant.

For motion at constant speed or constant acceleration, the distance travelled equals the area between the speed–time graph and the time axis.

$s = \text{area under the speed--time graph}$

Graph section	Area calculation
constant speed	rectangle: $s=vt$
speed rises from zero uniformly	triangle: $s= rac{1}{2}vt$
speed changes uniformly from $u$ to $v$	trapezium: $s= rac{1}{2}(u+v)t$
several sections	split into shapes and add their areas

Multiplying m/s by s gives m, confirming that the area represents distance.

The area is distance only for a speed–time graph. Gradient, not area, gives acceleration. Use the duration of the chosen interval, not necessarily the time coordinate at its end.

Near the Earth's surface, an object in free fall has an approximately constant downward acceleration called $g$ when air resistance is ignored.

$g \approx 9.8\ \text{m/s}^{2}$

Its velocity changes by about 9.8 m/s every second in the downward direction. The same $g$ applies to different masses at the same location when resistance is negligible.

A constant acceleration does not mean constant speed. The value is approximately 9.8 m/s² near Earth's surface; use 10 m/s² only when the question supplies or permits that approximation.

Acceleration is the change in velocity per unit time. It can result from a change in speed, direction, or both.

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

Symbol	Meaning
$u$	initial velocity
$v$	final velocity
$\Delta v=v-u$	change in velocity
$t$	time over which the change occurs
$a$	acceleration, commonly in m/s $^2$

Choose a positive direction, keep velocity signs consistent, subtract initial velocity from final velocity, then divide by the elapsed time.

Acceleration is not velocity divided by time unless the initial velocity is zero. A negative answer describes acceleration opposite to the chosen positive direction.

On a speed–time graph, acceleration is represented by gradient. Compare the gradient at different times to decide whether acceleration is constant or changing.

Speed–time shape	Acceleration
straight rising line	constant positive acceleration
straight falling line	constant negative acceleration
horizontal line	zero acceleration
curved line	changing acceleration because gradient changes

For a curve, imagine tangents at successive points. A tangent that becomes steeper means the magnitude of acceleration increases; one that becomes less steep means it decreases.

A rising graph means positive acceleration, but it does not by itself prove constant acceleration. Constancy requires a straight line with constant gradient.

The gradient of a speed–time graph is acceleration. On a straight section it is found from any two well-separated points on that section.

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

Step	Action
1	select two points on the required straight section
2	calculate the vertical change $v_2-v_1$
3	calculate the horizontal change $t_2-t_1$
4	divide and state m/s $^2$ ; retain the sign

Positive gradient gives positive acceleration, zero gradient gives zero acceleration, and negative gradient gives negative acceleration.

Do not calculate acceleration from the area. If a graph is curved, a tangent is needed for instantaneous acceleration; this card's calculation method applies directly to straight sections.

Deceleration is negative acceleration: the acceleration acts opposite to the chosen positive direction and, for straight-line motion without reversal, the speed decreases.

$a = \frac{v-u}{t} < 0\quad\text{when }v<u$

Keep the sign when the question asks for acceleration. If it asks for the magnitude of deceleration, report the positive size of that negative acceleration.

A negative acceleration does not always mean an object is slowing down: it slows only when acceleration is opposite to velocity. Within a simple positive-direction braking calculation, $v<u$ gives a negative result.

A falling object's motion depends on the resultant of its downward weight and upward air or liquid resistance.

Stage	Forces and motion
released with negligible resistance	weight acts downward; acceleration is about $g$ and speed increases
speeding up in a fluid	resistance increases with speed; resultant force and acceleration decrease
terminal velocity	resistance equals weight; resultant force and acceleration are zero; downward speed is constant
parachute opens or resistance suddenly increases	resistance may exceed weight; acceleration is upward while the object still moves downward and slows
new terminal velocity	forces balance again at a lower constant downward speed

Without air or liquid resistance, the object continues to accelerate downward at approximately constant $g$ near Earth's surface and does not reach terminal velocity.

Terminal velocity does not mean rest: velocity is constant and non-zero because the forces are balanced. After a parachute opens, upward acceleration can occur while motion is still downward.

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—1.3 Mass and weightRead

1.2 Motion

Learning objectives

Calculate speed from distance and time

Distinguish velocity from speed

Calculate average speed for a whole journey

Plot and read motion graphs

Recognise motion from graph shape

Find speed from a distance–time gradient

Find distance from area under a speed–time graph

Use the acceleration of free fall near Earth

Calculate acceleration from velocity change

Tell constant from changing acceleration

Find acceleration from a speed–time gradient

Treat deceleration as negative acceleration

Explain falling motion and terminal velocity

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