Notes · MathematicsUK · A-Levels

Kinematics

Kinematics describes motion without reference to its causes. The A-Level covers displacement, velocity and acceleration and the information in motion graphs, the constant-acceleration (suvat) equations, motion under gravity, the use of calculus for variable acceleration, and motion in two dimensions including projectiles.

5 sections~12 min reading time3 competenciesLevel Standard 3 · Advanced 2

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Exam profile

AO1 · Use the suvat equations and calculus to solve motion problemsAO2 · Interpret displacement-time and velocity-time graphsAO3 · Model projectile and variable-acceleration motion

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basic level

AS-Level covers motion graphs, the constant-acceleration equations and vertical motion under gravity.

higher level

The full A-Level adds variable acceleration using calculus and motion in two dimensions, including projectiles.

Depth

Text

Contents · 5 sections

Kinematics

Motion graphs#

StandardDfE GCE Mathematics — kinematicsAQA 7357 (Mechanics)

A velocity-time graph

Fig. 1The gradient of each segment is the acceleration; the area beneath the graph is the displacement.

Motion graphs display how the quantities describing motion change with time, and reading them correctly is a core skill. On a displacement-time graph the gradient is the velocity, so a straight line means constant velocity and a curve means changing velocity. A horizontal section means the object is at rest, and a negative gradient means it is moving back towards its starting point.

On a velocity-time graph two features carry meaning: the gradient is the acceleration, and the area between the graph and the time axis is the displacement. A straight, sloping line therefore represents constant acceleration, and the displacement over an interval is found as the area of the trapezium or triangle beneath the line. This 'gradient gives acceleration, area gives displacement' pairing is used constantly.

Because area gives displacement, motion in which the velocity becomes negative must be handled with care: area below the time axis counts as negative displacement (motion in the reverse direction). To find the total distance travelled, as opposed to the net displacement, you add the magnitudes of the areas above and below the axis separately.

Sketching or interpreting a motion graph is often the quickest route through a multi-stage problem: a journey with phases of acceleration, constant speed and deceleration becomes a simple sequence of straight-line segments whose gradients and areas answer the question. Being fluent in translating between a described journey and its graph is heavily rewarded.

v = \frac{ds}{dt} \ (\text{gradient of } s\text{-}t), \qquad s = \int v\,dt \ (\text{area under } v\text{-}t)

What the graphs give

Displacement-time gradient is velocity; velocity-time area is displacement.

Worked example

Displacement from a velocity-time graph

For the graph rising from 0 to $8\ \text{m s}^{-1}$ over 4 s, holding for 6 s, then falling to 0 over 4 s, find the total displacement.

Rising phase (triangle)
Area $= \tfrac{1}{2}(4)(8) = 16$ m.
Constant phase (rectangle)
Area $= 6 \times 8 = 48$ m.
Falling phase (triangle)
Area $= \tfrac{1}{2}(4)(8) = 16$ m; total $= 16 + 48 + 16$ .
$16 + 48 + 16 = 80\ \text{m}$

Result: The total displacement is $80$ m.

Typical mistakes

Confusing displacement-time and velocity-time graphs (a horizontal line means at rest on the first, constant velocity on the second).
Adding areas above and below the axis when total distance is required (they should be added as magnitudes).

Active revision

A car accelerates uniformly from rest to $20\ \text{m s}^{-1}$ in 8 s, travels at that speed for 12 s, then brakes uniformly to rest in 5 s. Sketch the velocity-time graph and find the total distance travelled.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

The constant-acceleration equations#

StandardDfE GCE Mathematics — kinematicsAQA 7357 (Mechanics)

Constant acceleration on a velocity-time graph

Fig. 2For constant acceleration the velocity-time graph is a straight line; the shaded area is the displacement.

When acceleration is constant, five equations — the 'suvat' equations — connect displacement

s

, initial velocity

u

, final velocity

v

, acceleration

a

and time

t

. They are

v = u + at

s = ut + \tfrac{1}{2}at^2

s = \tfrac{1}{2}(u + v)t

v^2 = u^2 + 2as

and

s = vt - \tfrac{1}{2}at^2

. Each equation omits one of the five quantities, so you choose the one that avoids the variable you neither know nor want.

The equations are not independent axioms but follow from the definitions of velocity and acceleration when

a

is constant:

v = u + at

is 'final = initial + gain', and

s = \tfrac{1}{2}(u + v)t

is average velocity times time. The others come from combining these two, so understanding their origin makes them easy to recall and to select.

The reliable method is to list the five quantities, marking which are known and which is required, and then pick the equation containing exactly those and not the unknown you are avoiding. Consistency of sign is essential: choose a positive direction at the outset and give every displacement, velocity and acceleration its correct sign relative to it.

The equations apply only while the acceleration is constant, so a journey with different phases must be split, applying the equations separately to each phase with the end of one phase supplying the start of the next. Recognising when the constant-acceleration model applies, and when it does not, is part of using them correctly.

v = u + at, \quad s = ut + \tfrac{1}{2}at^2, \quad v^2 = u^2 + 2as

Three of the suvat equations

The others are $s = \tfrac{1}{2}(u+v)t$ and $s = vt - \tfrac{1}{2}at^2$ ; each omits one variable.

Worked example

Selecting a suvat equation

A particle starts at $5\ \text{m s}^{-1}$ and accelerates at $2\ \text{m s}^{-2}$ . Find its velocity after travelling 24 m.

List the quantities
$u = 5$ , $a = 2$ , $s = 24$ , $v = ?$ (t not involved), so use $v^2 = u^2 + 2as$ .
Substitute
$v^2 = 5^2 + 2(2)(24) = 25 + 96 = 121$ .
$v^2 = 25 + 96 = 121$
Solve
$v = \sqrt{121} = 11\ \text{m s}^{-1}$ (taking the positive root for forward motion).

Result: The velocity is $11\ \text{m s}^{-1}$ .

Typical mistakes

Using a suvat equation across a phase where the acceleration changes.
Sign errors, especially treating a deceleration or a downward quantity as positive.

Active revision

A car travelling at $30\ \text{m s}^{-1}$ brakes with constant deceleration and stops in 60 m. Find the deceleration and the time taken to stop.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Motion under gravity#

StandardDfE GCE Mathematics — kinematicsAQA 7357 (Mechanics)

Height of a projectile thrown vertically

Fig. 3

h = 20t - 4.9t^2

: the object rises to a maximum where

v = 0

, then falls back.

An object moving freely under gravity (with air resistance neglected) has constant acceleration

g = 9.8\ \text{m s}^{-2}

directed downwards, so the suvat equations apply with

a = \pm g

according to the chosen positive direction. This vertical motion covers objects dropped, thrown up, or thrown down, and is one of the most common mechanics contexts.

Sign convention is decisive. If you take upwards as positive, then the acceleration is

-9.8\ \text{m s}^{-2}

throughout the motion, including on the way up, at the top, and on the way down — gravity does not change direction. An object thrown upwards decelerates, stops instantaneously at its highest point (where

v = 0

), and then accelerates downwards.

The condition

v = 0

at the highest point is the key to many questions: it gives the time to the top and, via

v^2 = u^2 + 2as

, the maximum height. By symmetry (in the absence of air resistance) the time up equals the time down to the same level, and the speed on returning to the launch height equals the launch speed.

Care is needed with the meaning of 'ground level' and the sign of a downward displacement. An object thrown up from a cliff and landing at the base has a negative final displacement relative to the launch point; choosing and stating the positive direction, and interpreting the sign of the answer, avoids the frequent errors in these problems.

a = -g = -9.8\ \text{m s}^{-2} \ (\text{taking up as positive}), \qquad v = 0 \text{ at the highest point}

Vertical motion under gravity

Gravity gives constant downward acceleration; the velocity is momentarily zero at the top.

Worked example

Maximum height

A stone is thrown vertically upwards at $21\ \text{m s}^{-1}$ . Taking $g = 9.8\ \text{m s}^{-2}$ , find the maximum height reached.

Condition at the top
At the highest point $v = 0$ ; take upwards positive so $a = -9.8$ , $u = 21$ .
Use v² = u² + 2as
$0 = 21^2 + 2(-9.8)s = 441 - 19.6s$ .
$0 = 441 - 19.6s$
Solve for s
$s = \dfrac{441}{19.6} = 22.5$ m.

Result: The maximum height is $22.5$ m above the launch point.

Typical mistakes

Changing the sign of the acceleration on the way down (gravity acts downwards throughout).
Mishandling the sign of a downward final displacement (e.g. landing below the launch point).

Active revision

A ball is thrown vertically upwards at $14\ \text{m s}^{-1}$ from a point 2 m above the ground. Taking $g = 9.8\ \text{m s}^{-2}$ , find the maximum height above the ground and the time before it hits the ground.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Variable acceleration and calculus#

AdvancedDfE GCE Mathematics — kinematicsAQA 7357 (Mechanics)

Velocity of a particle with variable acceleration

Fig. 4

v = t^2 - 6t + 8 = (t-2)(t-4)

: the particle is instantaneously at rest at

t = 2

and

t = 4

When the acceleration is not constant, the suvat equations no longer apply and calculus takes over. Because velocity is the rate of change of displacement and acceleration the rate of change of velocity, differentiation moves down the chain:

v = \dfrac{ds}{dt}

and

a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}

. Given displacement as a function of time, you differentiate to find velocity and again for acceleration.

Integration moves back up the chain:

v = \displaystyle\int a\,dt

and

s = \displaystyle\int v\,dt

, each introducing a constant of integration that is fixed by an initial condition (the velocity or position at a known time). Given the acceleration as a function of time and the starting conditions, you integrate to recover the velocity and the position.

The turning points of motion are found from these relationships. The particle is instantaneously at rest when

v = 0

, and its velocity is greatest or least when

a = \dfrac{dv}{dt} = 0

. Solving these conditions locates key moments, and combining them with the graph interpretations from earlier gives a full description of the motion.

Distance travelled over an interval, as distinct from displacement, again requires care: you must find any times where the velocity changes sign and integrate the speed piecewise, because integrating a velocity that goes negative would subtract the reverse motion. This ties the kinematics of variable acceleration directly to the integration topic.

v = \frac{ds}{dt}, \quad a = \frac{dv}{dt}, \qquad v = \int a\,dt, \quad s = \int v\,dt

Calculus of motion

Differentiate to go from displacement to velocity to acceleration; integrate to go back.

Worked example

From acceleration to velocity and displacement

A particle starts at the origin with velocity $2\ \text{m s}^{-1}$ and has acceleration $a = 6t\ \text{m s}^{-2}$ . Find its velocity and displacement at time $t$ .

Integrate for velocity
$v = \displaystyle\int 6t\,dt = 3t^2 + c$ ; $v = 2$ at $t = 0$ gives $c = 2$ , so $v = 3t^2 + 2$ .
Integrate for displacement
$s = \displaystyle\int (3t^2 + 2)\,dt = t^3 + 2t + k$ ; $s = 0$ at $t = 0$ gives $k = 0$ .
$s = t^3 + 2t$
State
$v = 3t^2 + 2$ and $s = t^3 + 2t$ .

Result: $v = 3t^2 + 2\ \text{m s}^{-1}$ and $s = t^3 + 2t\ \text{m}$ .

Typical mistakes

Forgetting the constant of integration, or not using the initial condition to find it.
Integrating a velocity that changes sign to find distance, instead of splitting at $v = 0$ .

Active revision

A particle moves with velocity $v = 3t^2 - 12t + 9\ \text{m s}^{-1}$ . Find the times when it is at rest and its acceleration at $t = 1$ .

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Projectiles#

AdvancedDfE GCE Mathematics — kinematicsAQA 7357 (Mechanics)

Parabolic trajectory of a projectile

Fig. 5Launched at

20\ \text{m s}^{-1}

45^\circ

: the path is a parabola with maximum height

\approx 10.2

m and range

\approx 40.8

A projectile moves in two dimensions under gravity alone, and the central idea is that the horizontal and vertical motions are independent. Horizontally there is no acceleration (air resistance neglected), so the horizontal velocity is constant; vertically the acceleration is

-g

as in one-dimensional motion under gravity. You treat the two directions with separate suvat equations, linked only by the common time

t

For a launch speed

u

at angle

\theta

to the horizontal, the initial components are

u\cos\theta

horizontally and

u\sin\theta

vertically. The horizontal displacement is then

x = (u\cos\theta)\,t

, and the vertical displacement is

y = (u\sin\theta)\,t - \tfrac{1}{2}gt^2

. Resolving the initial velocity into components is the essential first step of every projectile problem.

The standard results follow from these. The time of flight (to return to launch height) comes from setting

y = 0

; the maximum height comes from the vertical velocity being zero at the top; and the range is the horizontal distance at the time of flight. The trajectory is a parabola, as eliminating

t

between the two displacement equations shows.

Common questions ask for the time to reach a given height, the velocity (as a vector, with magnitude and direction) at a moment, or where the projectile lands on sloping ground. The reliable approach is always the same: resolve into components, apply the suvat equations to each direction with the shared time, and recombine at the end. Stating the modelling assumptions (particle, no air resistance) is expected.

x = (u\cos\theta)\,t, \qquad y = (u\sin\theta)\,t - \tfrac{1}{2}gt^2

Projectile displacements

Constant horizontal velocity; vertical motion under gravity, sharing the time $t$ .

Worked example

Time of flight and range

A projectile is launched at $20\ \text{m s}^{-1}$ at $30^\circ$ from ground level ( $g = 9.8\ \text{m s}^{-2}$ ). Find the time of flight and the horizontal range.

Resolve
Horizontal: $20\cos 30^\circ = 17.32\ \text{m s}^{-1}$ ; vertical: $20\sin 30^\circ = 10\ \text{m s}^{-1}$ .
Time of flight
Set $y = 0$ : $10t - 4.9t^2 = 0 \Rightarrow t(10 - 4.9t) = 0$ , so $t = \dfrac{10}{4.9} = 2.04$ s.
$t = \frac{10}{4.9} = 2.04\ \text{s}$
Range
$x = 17.32 \times 2.04 = 35.3$ m.

Result: Time of flight $\approx 2.04$ s; range $\approx 35.3$ m.

Typical mistakes

Applying gravity to the horizontal motion (there is no horizontal acceleration).
Forgetting to resolve the launch velocity, using $u$ instead of $u\cos\theta$ or $u\sin\theta$ .

Active revision

A ball is kicked at $15\ \text{m s}^{-1}$ at $30^\circ$ to the horizontal from ground level. Taking $g = 9.8\ \text{m s}^{-2}$ , find the time of flight, the range, and the maximum height.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

References & sources

Sources

Department for Education

Mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Mathematics 7357 specification

Kinematics

5 sections~12 min reading time3 competenciesLevel Standard 3 · Advanced 2

Kinematics

Key points

Displacement from a velocity-time graph

Key points

Selecting a suvat equation

Key points

Maximum height

Key points

From acceleration to velocity and displacement

Key points

Time of flight and range

Sources

Kinematics

Key points

Displacement from a velocity-time graph

Key points

Selecting a suvat equation

Key points

Maximum height

Key points

From acceleration to velocity and displacement

Key points

Time of flight and range

Sources