A particle moving with constant acceleration passes three points $A$, $B$ and $C$, in that order, on a straight line. The distance $AB$ is $20 \text{ m}$ and the distance $AC$ is $84 \text{ m}$. The speed of the particle at $B$ is $12 \text{ m s}^{-1}$ and its speed at $C$ is $20 \text{ m s}^{-1}$. Find the acceleration of the particle and its speed at $A$.

$a = 2 \text{ m s}^{-2}$; speed at $A = 8 \text{ m s}^{-1}$

Car A travels along a straight road at a constant speed of $25 \text{ m s}^{-1}$. Car B starts from rest at the same point and accelerates uniformly at $3 \text{ m s}^{-2}$ until it reaches a speed of $30 \text{ m s}^{-1}$, after which it maintains this speed. Find the distance Car B travels before it catches up to Car A.

M1 for finding the time taken for Car B to reach $30 \text{ m s}^{-1}$: $t = \frac{30}{3} = 10 \text{ s}$. M1 for finding the distances covered in the first $10 \text{ s}$: Car B covers $\frac{1}{2} \times 10 \times 30 = 150 \text{ m}$. Car A covers $25 \times 10 = 250 \text{ m}$. M1 for setting up an equation for the remaining distance where Car B catches A. Let $t_2$ be the additional time: $150 + 30t_2 = 250 + 25t_2$. M1 for solving to find $t_2$: $5t_2 = 100 \implies t_2 = 20 \text{ s}$. Tot

Edexcel IAL M1 Kinematics | A Level Math Revision

M1 Kinematics notes

The suvat equations

For constant acceleration: $v = u + a t$ , $s = u t + \frac{1}{2} a t^{2}$ , $s = v t - \frac{1}{2} a t^{2}$ , $s = \frac{1}{2} (u + v) t$ and $v^{2} = u^{2} + 2 a s$ . List the three quantities you know and the one you want, then pick the equation without the fifth. They do not apply when the acceleration changes.

Vertical motion under gravity

Take upwards as positive, so $a = - 9.8$ m s $^{- 2}$ . A ball thrown up at $19.6$ m s $^{- 1}$ reaches its highest point (where $v = 0$ ) after $2$ s, at a height of $19.6$ m, and is back at the start after $4$ s.

Velocity-time and displacement-time graphs

On a velocity-time graph the gradient is the acceleration and the area under it is the displacement. A car that accelerates from rest to $12$ m s $^{- 1}$ in 4 s, holds that speed for 10 s and then slows uniformly to rest in 6 s travels $180$ m. On a displacement-time graph the gradient is the velocity.

Two particles and catching up

Write each particle's displacement in terms of the same time $t$ , measured from the same point, then set them equal. Check whether one starts later: then its time is $t - T$ .

Key formulas

suvat Learn it: $v = u + a t, s = u t + \frac{1}{2} a t^{2}, s = v t - \frac{1}{2} a t^{2}, s = \frac{1}{2} (u + v) t, v^{2} = u^{2} + 2 a s$
Velocity–time graph: gradient = acceleration, area = displacement Learn it

In the formula booklet = printed in Pearson's IAL formulae booklet, so know where to find it. Learn it = not printed there, so learn it by heart. All M1 formulas on one page (free PDF) →

M1 Kinematics worked examples

Each mark scheme is checked line by line before it is published here. M marks are for method, A marks for accurate answers that follow from them, B marks for independent results.

Worked example 1 · 5 marks · hard

A particle moving with constant acceleration passes three points $A$ , $B$ and $C$ , in that order, on a straight line. The distance $A B$ is $20 m$ and the distance $A C$ is $84 m$ . The speed of the particle at $B$ is $12 {m s}^{- 1}$ and its speed at $C$ is $20 {m s}^{- 1}$ . Find the acceleration of the particle and its speed at $A$ .

Mark scheme

M1 for $B C = 84 - 20 = 64 m$ .
M1 for $v^{2} = u^{2} + 2 a s$ from $B$ to $C$ : $20^{2} = 12^{2} + 2 a (64)$ .
A1 for $128 a = 256 \Rightarrow a = 2 {m s}^{- 2}$ .
M1 for $v^{2} = u^{2} + 2 a s$ from $A$ to $B$ : $12^{2} = u_{A}^{2} + 2 (2) (20)$ .
A1 for $u_{A}^{2} = 64 \Rightarrow u_{A} = 8 {m s}^{- 1}$ .

5 steps

Worked example 2 · 5 marks · hard

Car A travels along a straight road at a constant speed of $25 {m s}^{- 1}$ . Car B starts from rest at the same point and accelerates uniformly at $3 {m s}^{- 2}$ until it reaches a speed of $30 {m s}^{- 1}$ , after which it maintains this speed. Find the distance Car B travels before it catches up to Car A.

Mark scheme

M1 for finding the time taken for Car B to reach $30 {m s}^{- 1}$ : $t = \frac{30}{3} = 10 s$ .
M1 for finding the distances covered in the first $10 s$ : Car B covers $\frac{1}{2} \times 10 \times 30 = 150 m$ . Car A covers $25 \times 10 = 250 m$ .
M1 for setting up an equation for the remaining distance where Car B catches A. Let $t_{2}$ be the additional time: $150 + 30 t_{2} = 250 + 25 t_{2}$ .
M1 for solving to find $t_{2}$ : $5 t_{2} = 100 ⟹ t_{2} = 20 s$ . Total time = $30 s$ .
A1 for finding the total distance: $25 \times 30 = 750 m$ .

5 steps · 1 to fill in

Worked example 3 · 4 marks · hard

A motorcycle starts from rest, accelerates uniformly to $30 {m s}^{- 1}$ in $20$ seconds, maintains this speed for $t$ seconds, and then decelerates uniformly to rest in $15$ seconds. The total distance covered is $1500 m$ . Find the total time for the journey.

Mark scheme

M1 for expressing the total distance as the area of a trapezium or sum of shapes: $\frac{1}{2} \times 20 \times 30 + 30 t + \frac{1}{2} \times 15 \times 30 = 1500$ .
M1 for simplifying the equation to $300 + 30 t + 225 = 1500$ .
M1 for solving for the constant velocity time $t$ : $30 t = 975 ⟹ t = 32.5$ .
A1 for calculating the total time: $20 + 32.5 + 15 = 67.5 s$ .

4 steps · 1 to fill in

Common M1 kinematics mistakes

Mixed signs in vertical motion. If up is positive, $g$ is $- 9.8$ and a point below the start has negative displacement.
suvat with changing acceleration. Split the journey into stages with constant acceleration and apply suvat to each stage separately.
Distance versus displacement. Area below the time axis on a velocity-time graph is a negative displacement but still adds to the distance travelled.
Units. Convert km h $^{- 1}$ to m s $^{- 1}$ first: multiply by $\frac{1000}{3600}$ .

Every M1 mistake from the examiner reports, with the paper it came from →

M1 Kinematics FAQ

Which suvat equation should I use?

Write down s, u, v, a and t, fill in the three you know and mark the one you want. The equation that leaves out the remaining fifth quantity is the one to use.

Is variable acceleration in IAL M1?

No. M1 kinematics uses constant acceleration only; motion with variable acceleration (using calculus) and projectiles are in M2.

Is a calculator allowed in IAL M1?

Yes. A calculator is allowed in every Edexcel IAL maths unit, M1 included. Method marks are for the working you write down, so show it.

How long is the IAL M1 exam?

Each IAL maths unit, M1 included, is one written paper of 1 hour 30 minutes worth 75 marks.

How often does kinematics come up in M1?

In our mapping of the 4 Pearson M1 papers from June 2024 to June 2025, kinematics was tested in 4 of them (5 questions in all). Any topic can be examined in any series.

Next steps

More M1: Forces and Newton's Laws · Moments · Vectors (Mechanics context) · Dynamics of a Particle · Statics of a Particle · Momentum and Impulse

Edexcel IAL M1 Kinematics: exam questions and answers, notes and worked examples

Try it first

M1 Kinematics notes

The suvat equations

Vertical motion under gravity

Velocity-time and displacement-time graphs

Two particles and catching up

Key formulas

M1 Kinematics worked examples

Worked example 1 · 5 marks · hard

Mark scheme

Worked example 2 · 5 marks · hard

Mark scheme

Worked example 3 · 4 marks · hard

Mark scheme

Common M1 kinematics mistakes

M1 Kinematics FAQ

Which suvat equation should I use?

Is variable acceleration in IAL M1?

Is a calculator allowed in IAL M1?

How long is the IAL M1 exam?

How often does kinematics come up in M1?

Next steps

Your M1 revision kit