M7 Kinematics — notes on one page

Every Edexcel A-level Maths 9MA0 specification point in M7, read straight through: 5 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

Checked against Edexcel 9MA0 section M7. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

M7.1 · Understand and use the language of kinematics: position; displacement; distance travelled; velocity; speed; acceleration.

Position locates a particle relative to an origin, displacement is the signed change in position, and distance travelled is the total path length and is always non-negative.
Velocity is the rate of change of displacement, speed is the magnitude of velocity, and acceleration is the rate of change of velocity; choose a positive direction before assigning signs.
For motion from $x_1$ to $x_2$ , displacement is $x_2-x_1$ ; split reversals into separate path lengths for distance, and use change in velocity divided by time for average acceleration.
A common error is to treat distance and displacement, or speed and velocity, as interchangeable; distance and speed cannot be negative.

Worked example

Taking east as positive, a cyclist's velocity changes uniformly from $+8\,\text{m s}^{-1}$ to $-4\,\text{m s}^{-1}$ in $6\,\text{s}$ . Find the acceleration, the time when the cyclist is instantaneously at rest, and the speed at the end.

1.The acceleration is the change in velocity divided by time: $a=(-4-8)/6=-2\,\text{m s}^{-2}$ .
2.Using $v=u+at$ , rest occurs when $0=8-2t$ , so $t=4\,\text{s}$ .
3.The final velocity is $-4\,\text{m s}^{-1}$ , whose magnitude gives speed $4\,\text{m s}^{-1}$ .

Answer: Acceleration $=-2\,\text{m s}^{-2}$ ; The cyclist is at rest after $4\,\text{s}$ ; Final speed $=4\,\text{m s}^{-1}$

Common mistakes

Don't calculate displacement by adding journey lengths without assigning directions.
Don't report a negative velocity as a negative speed and ignore the chosen positive direction.

Exam tip. Keep signs for displacement, velocity and acceleration, but report distance and speed as non-negative magnitudes.

M7.2 · Understand, use and interpret graphs in kinematics for motion in a straight line: displacement against time and interpretation of gradient; velocity against time and interpretation of gradient and area under the graph.

The gradient of a displacement-time graph is velocity; the gradient of a velocity-time graph is acceleration, with signs determined by the chosen positive direction.
Find displacement from a velocity-time graph using signed area, splitting the graph into rectangles, triangles or trapezia as needed.
For a straight segment from velocity $u$ to $v$ over time $t$ , the area is $\dfrac12(u+v)t$ , which gives the displacement during that interval.
A common error is to add all areas as positive: area below the time axis is negative for displacement, though its magnitude contributes positively to distance travelled.

On a velocity-time graph, gradient gives acceleration and signed area gives displacement.

Worked example

A particle's velocity increases uniformly from $2\,\text{m s}^{-1}$ to $8\,\text{m s}^{-1}$ during the first $3\,\text{s}$ , then remains at $8\,\text{m s}^{-1}$ for $4\,\text{s}$ . Find its acceleration during the first stage and its displacement over all $7\,\text{s}$ .

1.The first gradient is $(8-2)/3=2\,\text{m s}^{-2}$ .
2.The first-stage area is $\tfrac12(2+8)(3)=15\,\text{m}$ and the constant-velocity area is $8(4)=32\,\text{m}$ , giving displacement $15+32=47\,\text{m}$ .

Answer: Acceleration $=2\,\text{m s}^{-2}$ ; Displacement $=47\,\text{m}$

Common mistakes

Don't treat the area under a displacement-time graph as displacement instead of using its gradient for velocity.
Don't read displacement from the final velocity instead of the signed area under the velocity-time graph.

Exam tip. On kinematics graphs, state explicitly whether a gradient or signed area gives the requested quantity.

M7.3 · Understand, use and derive the formulae for constant acceleration for motion in a straight line; extend to 2 dimensions using vectors.

Definition. Specification M7.3: understand, use and derive the formulae for constant acceleration for motion in a straight line, $v=u+at$ , $s=ut+\frac12at^2$ , $s=\frac12(u+v)t$ , $v^2=u^2+2as$ and $s=vt-\frac12at^2$ ; extend to two dimensions using vectors.

List $s$ , $u$ , $v$ , $a$ , $t$ with a stated positive direction, choose the formula containing the three known quantities and the one required, and substitute with signs. The formulae derive from a straight-line velocity–time graph: the gradient is $a$ , giving $v=u+at$ , and the area is the displacement, giving $s=\frac12(u+v)t$ ; eliminating $t$ or $v$ gives the others.
On a speed–time graph the area under each section is the distance, and a trapezium or triangle is usually quicker than the formulae.
In two dimensions the same formulae hold for vectors, $\mathbf v=\mathbf u+\mathbf at$ and $\mathbf r=\mathbf r_0+\mathbf ut+\frac12\mathbf at^2$ , worked component by component; speed is the magnitude $\sqrt{v_x^2+v_y^2}$ and a direction needs a bearing or an angle.
The formulae apply only while the acceleration is constant; a velocity given as a function of $t$ needs calculus.
Give units with every answer.

Worked example

A particle

P

moves with constant acceleration

(\mathbf i+2\mathbf j)

m s

^{-2}

. At time

t=0

the velocity of

P

(2\mathbf i-5\mathbf j)

m s

^{-1}

. Find the speed of

P

when

t=4

(4)

(Total for Question 1 is 4 marks)

Working

1.Use $\mathbf v=\mathbf u+\mathbf at$ : $\mathbf v=(2\mathbf i-5\mathbf j)+4(\mathbf i+2\mathbf j)$ .
2.Collect components: $\mathbf v=(6\mathbf i+3\mathbf j)$ m s $^{-1}$ .
3.Speed is the magnitude: $\sqrt{6^2+3^2}=\sqrt{45}=3\sqrt5=6.71$ m s $^{-1}$ .

Answer: $3\sqrt5$ m s $^{-1}$ ( $6.7$ m s $^{-1}$ )

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	Uses $\mathbf v=\mathbf u+\mathbf at$ with the given vectors and $t=4$	M1
	$\mathbf v=6\mathbf i+3\mathbf j$	A1
	Applies Pythagoras, including the square root, to their velocity	M1
	$\sqrt{45}=3\sqrt5$ or $6.7$ or better	A1
Notes
A1: Units required

(4 marks)

Mark-scheme wording to use

“v_B = (−16i − 3j) + 5(2.4i + j) M1 v_B = (−4i + 2j) A1 √((−4)² + 2²) M1 √20 = 2√5 , 4.5 or better ( m s⁻¹) A1 … M1 Use of Pythagoras on their v_B to give a magnitude (need the root)” — Pearson 9MA0, June 2023 Paper 32, Q4(a)
“M1 Any complete suvat method to find a … A1 Any equivalent number with correct units. … 2.5, 5/2 , 10/4 m s⁻² units needed.” — Pearson 9MA0, June 2024 Paper 32, Q2(a)
“s = (1/2) × 3.2 × 5² OR s = ((0 + 16)/2) × 5 OR s = (16 × 5) − (1/2) × 3.2 × 5² OR 16² = 2 × 3.2 × s OR from a v-t graph, s = (1/2) × 5 × 16 M1 s = 40 (m) A1 … M1 Complete method to find an equation in s only, possibly using their ‘16’ … A1 cao. Must be positive.” — Pearson 9MA0, June 2023 Paper 32, Q1(b)

What the examiner wants. Units, and the right tool. The June 2024 Paper 32 report says some lost the second mark on the acceleration for incorrect or missing units, that the area approach to the speed–time graph was more successful than the formulae, and that a few used a constant-acceleration approach where the acceleration was not constant and scored nothing.

Common mistakes

Don't mix a displacement sign convention with an acceleration value treated as an unsigned magnitude.
Don't select a constant-acceleration formula containing an extra unknown and introduce avoidable algebra.

Exam tip. List the known kinematic quantities with signs, then choose the equation containing only the required unknown.

M7.4 · Use calculus in kinematics for motion in a straight line: v = dr/dt, a = dv/dt = d²r/dt², r = ∫v dt, v = ∫a dt; extend to 2 dimensions using vectors.

Definition. Specification M7.4: use calculus in kinematics for motion in a straight line: $v=\dfrac{dr}{dt}$ , $a=\dfrac{dv}{dt}=\dfrac{d^2r}{dt^2}$ , $r=\int v\,dt$ , $v=\int a\,dt$ ; extend to two dimensions using vectors.

When velocity or acceleration is given as a function of $t$ , the constant-acceleration formulae do not apply:
differentiate displacement to get velocity and velocity to get acceleration, or integrate the other way, adding a constant of integration fixed by the initial conditions. For vectors do this component by component, keeping the $\mathbf i$ and $\mathbf j$ separate and writing the answer in $\mathbf i$ , $\mathbf j$ form.
Speed is the magnitude of the velocity, $\sqrt{v_x^2+v_y^2}$ , so a speed question is not finished at the velocity; a direction of motion is the angle of the velocity vector.
The particle is momentarily at rest when $v=0$ , moving parallel to a given vector when the components are in that ratio, and has maximum speed where $a=0$ .
Distance travelled is not displacement when the direction changes, so find the times where $v=0$ and add the separate sections. Reject negative times.

Worked example

A particle

P

moves along a straight line. At time

t

seconds,

t\geq0

, the velocity of

P

v

m s

^{-1}

, where

v=3t^2-12t+9

. When

t=0

P

is at the point

A

, where

OA=2

m. Find the total distance travelled by

P

in the interval

0\leq t\leq4

(5)

(Total for Question 1 is 5 marks)

Working

1.Integrate to find displacement from $O$ : $r=t^3-6t^2+9t+c$ , and $r=2$ at $t=0$ gives $c=2$ .
2. $P$ changes direction where $v=0$ : $3(t-1)(t-3)=0$ , so at $t=1$ and $t=3$ .
3.Positions: $r(0)=2$ , $r(1)=6$ , $r(3)=2$ , $r(4)=6$ .
4.Distance $=|6-2|+|2-6|+|6-2|=4+4+4=12$ m.

Answer: $12$ m

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	Integrates $v$ with respect to $t$ , $t^n\to t^{n+1}$ in at least two terms	M1
	$r=t^3-6t^2+9t+2$	A1
	Sets $v=0$ to find when the direction changes, $t=1$ and $t=3$	M1
	Evaluates $r$ at $t=0,1,3,4$ and adds the moduli of the changes	dM1
	$12$ m	A1
Notes
A1: Uses $r=2$ at $t=0$ for the constant
A1: Displacement $4$ m scores A0

(5 marks)

Mark-scheme wording to use

“M1 Both powers of t decreasing by 1 … A1 Correct unsimplified derivative or two correct components … M1 Put t = 4 in their v (must be using an attempted derivative of r) and then use Pythagoras with the root, allow a missing − sign … A1 Accept 3.1 or better” — Pearson 9MA0, June 2024 Paper 32, Q4(b)
“Differentiate v wrt t to give a vector. M1 (2t − 3)i + 4tj A1 … M1 At least one power decreasing by 1 in each component in their v … A1 cao (must be a vector)” — Pearson 9MA0, June 2023 Paper 32, Q3(c)
“t² − 3t + 7 = 2t² − 3 OR (t² − 3t + 7)/(2t² − 3) = 1/1 = 1 M1 t = 2 only A1 … M1 Equating i and j components of v or a ratio of 1:1 to obtain a quadratic in t only.” — Pearson 9MA0, June 2023 Paper 32, Q3(b)

What the examiner wants. Finishing the question and using calculus where the acceleration varies. The June 2024 Paper 32 report says most differentiated correctly and used Pythagoras for the speed, but a minority found the velocity and stopped, losing two marks, and a few tried a constant-acceleration approach without realising the acceleration was not constant.

Common mistakes

Don't differentiate position once and label the result acceleration.
Don't integrate velocity to find displacement but equate net displacement with total distance despite direction changes.

Exam tip. For total distance, find every zero of velocity in the interval and add the magnitudes of the separate displacements.

M7.5 · Model motion under gravity in a vertical plane using vectors; projectiles.

Definition. Specification M7.5: model motion under gravity in a vertical plane using vectors; projectiles. A projectile is modelled as a particle moving freely under gravity, so its horizontal acceleration is zero and its vertical acceleration is $g$ downwards.

Resolve the initial velocity into $u\cos\alpha$ horizontally and $u\sin\alpha$ vertically, then treat the two directions separately with the constant-acceleration formulae: horizontally $x=u\cos\alpha\,t$ ; vertically $y=u\sin\alpha\,t-\tfrac12gt^2$ with upwards positive. Eliminating $t$ gives the equation of the path, $y=x\tan\alpha-\dfrac{gx^2}{2u^2\cos^2\alpha}$ , which is derived every time it is used, never quoted.
Time of flight to the same level comes from $y=0$ ; range from substituting that time into $x$ ; greatest height from the vertical velocity being zero, $0=(u\sin\alpha)^2-2gh$ , while the horizontal velocity is unchanged there.
Speed at any instant combines both components by Pythagoras.
State the value of $g$ used, $9.8$ unless told otherwise, and give answers to two or three significant figures.
Limitations of the model that score are air resistance, spin, wind, size and shape of the object, and treating $g$ as constant.

A projectile is resolved into horizontal

u\cos\theta

and vertical

u\sin\theta

components.

Worked example

A ball is projected from a point

O

on horizontal ground with speed

20

m s

^{-1}

at an angle of

30^\circ

above the horizontal. The ball is modelled as a particle moving freely under gravity. Find the horizontal distance from

O

to the point where the ball first hits the ground.

(4)

(Total for Question 1 is 4 marks)

Working

1.Resolve the initial velocity: horizontally $20\cos30^\circ=10\sqrt3$ m s $^{-1}$ , vertically $20\sin30^\circ=10$ m s $^{-1}$ .
2.Vertically, taking upwards as positive, $s=ut+\tfrac12at^2$ with $s=0$ : $0=10t-4.9t^2$ , so $t=\dfrac{10}{4.9}=2.04$ s.
3.Horizontally there is no acceleration: $x=10\sqrt3\times\dfrac{10}{4.9}=35.3$ m.

Answer: $35$ m (to 2 s.f.); $35.3$ m to 3 s.f.

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	Vertical motion with $s=0$ : $0=20\sin30^\circ\,t-\tfrac12(9.8)t^2$	M1
	$t=\dfrac{10}{4.9}$ or $2.04$ s	A1
	Horizontal motion $x=20\cos30^\circ\times$ their $t$	M1
	$35$ or $35.3$ m	A1
Notes
M1: Condone sign errors; $g=9.8$ used
A1: awrt 35

(4 marks)

Mark-scheme wording to use

“Using horizontal motion, s = ut, with 35 resolved … M1 … x = 35cos α × t … A1 … Using vertical motion, s = ut + (1/2)at², with 35 resolved … M1 … y = 35sin α × t − (1/2)gt² … A1 … Eliminate t : y = 35sin α × x/(35cos α) − (1/2)g(x/(35cos α))² … DM1 … A1* Given answer correctly obtained, with at least one further line of working with trig ratios and g = 9.8 explicitly seen or 4.9 oe used for 0.5g” — Pearson 9MA0, June 2024 Paper 32, Q5(a)
“M1 ALT 1: Use of y = 0 in equation of path and solve for x. … (OA = )120 (m) … A1 cao” — Pearson 9MA0, June 2024 Paper 32, Q5(b)
“B1 Any single correct answer. Acceptable answer: the inaccuracy of using 9.8 m s⁻² for g or wind or weather effects or the spin of the stone or the size (or shape or surface area) of the stone oe or the stone is still modelled as a particle … Unacceptable answers: The model (it) does not take account of: the mass or weight of the stone … air resistance.” — Pearson 9MA0, June 2024 Paper 32, Q5(e)

What the examiner wants. The path derived from components, with $g=9.8$ visible. The June 2024 Paper 32 report says the final mark of the show-that required $9.8$ to be seen used explicitly, that quoting the general trajectory equation and substituting the angle achieved no credit, and that some found the greatest height but gave $60$ , the $x$ -value, instead of $22.5$ .

Common mistakes

Don't use a same-level horizontal-range formula for a projectile that lands at a different vertical level.
Don't use the full launch speed in both directions instead of resolving it into horizontal and vertical components.

Exam tip. Resolve the initial velocity first, use zero vertical displacement for same-level flight, and keep horizontal acceleration zero.