Kinematics of a particle moving in a straight line or plane

Edexcel International A-Level Mathematics· M2 §1.1 to §1.4, 2018 Specification Issue 3· 25 min read

1. Motion in a Vertical Plane with Constant Acceleration★★☆☆☆⏱ 5 min

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📘 Definition

Constant acceleration in vertical plane

Motion of a particle under constant vector acceleration (most often gravity, acting vertically downwards), where horizontal and vertical components of motion can be analysed independently.

When solving problems for motion in a vertical plane with constant acceleration, split all vector quantities (displacement, velocity, acceleration) into horizontal and vertical components. Use M1 suvat equations for each component separately, as acceleration is constant for both axes: horizontal acceleration is usually 0 (no air resistance), while vertical acceleration is $g = 9.8 m s^{- 2}$ downwards.

📐 Worked Example

A ball is thrown from ground level with a velocity of $20 m s^{- 1}$ at an angle of 30° above the horizontal, moving freely under gravity. Find the maximum height reached by the ball.

1
Split initial velocity into vertical component, take upwards as positive:
$u_y = 20 \sin 30^\circ = 10 \text{ m s}^{-1}$
2
At maximum height, vertical velocity $v_{y} = 0$ , vertical acceleration $a = - g = - 9.8 m s^{- 2}$ . Use suvat equation $v^{2} = u^{2} + 2 a s$ for vertical motion:
$0 = 10^2 + 2(-9.8)s$
3
Rearrange to solve for s (maximum height):
$s = \frac{100}{19.6} \approx 5.10 \text{ m (3 s.f.)}$

Exam tip:

Always explicitly state which direction you are taking as positive for vertical motion at the start of your working to avoid sign errors.

2. Projectile Motion★★★☆☆⏱ 7 min

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📘 Definition

Projectile

A particle launched with an initial velocity, then moving freely under the sole influence of gravity, with no air resistance or other forces acting.

All projectile problems in M2 are solved by splitting motion into independent horizontal and vertical components: horizontal velocity remains constant (no acceleration) while vertical velocity changes due to gravity. You must derive all results from first principles; memorised range or time of flight formulas will not earn full method marks.

📐 Worked Example

Using the same launch parameters as the previous example (20 m s⁻¹ at 30° above horizontal, ground level launch), find the total horizontal distance travelled by the ball before it hits the ground again.

1
Find total time of flight using vertical motion: vertical displacement $s = 0$ when the ball lands. Use suvat equation $s = u t + \frac{1}{2} a t^{2}$ :
$0 = 10t - 4.9t^2$
2
Factorise to solve for t (ignore t=0 as this is launch time):
$t(10 - 4.9t) = 0 \implies t = \frac{10}{4.9} \approx 2.0408 \text{ s}$
3
Calculate constant horizontal velocity:
$u_x = 20 \cos 30^\circ \approx 17.32 \text{ m s}^{-1}$
4
Horizontal distance = horizontal velocity × time of flight:
$x = 17.32 \times 2.0408 \approx 35.3 \text{ m (3 s.f.)}$

Exam tip:

Keep intermediate values stored in your calculator to avoid rounding errors; only round your final answer to 3 significant figures.

3. Kinematics with Displacement as a Function of Time★★★☆☆⏱ 6 min

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📘 Definition

Time-based kinematic calculus

Use of differentiation and integration with respect to time to relate displacement, velocity and acceleration for particles moving with non-constant acceleration.

For straight line motion with non-constant acceleration, use the following relationships: velocity is the first derivative of displacement with respect to time ( $v = \frac{d x}{d t}$ ), and acceleration is the first derivative of velocity with respect to time ( $a = \frac{d v}{d t}$ ). To reverse the process, integrate acceleration to find velocity, and integrate velocity to find displacement.

📐 Worked Example

A particle moves in a straight line such that its displacement from a fixed origin at time t seconds is given by $x (t) = 2 t^{3} - 4 t^{2} + 3 t$ metres, t ≥ 0. Find the acceleration of the particle at t = 2 s.

1
Differentiate displacement to find velocity:
$v(t) = \frac{dx}{dt} = 6t^2 - 8t + 3$
2
Differentiate velocity to find acceleration:
$a(t) = \frac{dv}{dt} = 12t - 8$
3
Substitute t = 2:
$a(2) = 12(2) - 8 = 16 \text{ m s}^{-2}$

Exam tip:

Always remember to add a constant of integration when integrating velocity or acceleration, and use initial conditions to find its value.

4. Differentiation and Integration of Kinematic Vectors★★★★☆⏱ 7 min

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📘 Definition

Vector kinematics

Analysis of 2D motion using vectors for position, velocity and acceleration, where calculus operations are performed component-wise on the vectors.

When working with kinematic vectors, differentiate or integrate each component (i and j) separately with respect to time. The derivative of position vector $r$ with respect to time is velocity $v = \dot{r}$ , and the second derivative is acceleration $a = \ddot{r}$ . When integrating, add a constant vector, determined using initial conditions.

📐 Worked Example

A particle has position vector at time t seconds given by $r (t) = t^{2} i + \frac{3}{2} t^{2} j$ metres, t ≥ 0. Find the speed of the particle at t = 1 s, and the magnitude of its acceleration.

1
Differentiate position vector component-wise to find velocity:
$\mathbf{v}(t) = \dot{\mathbf{r}}(t) = 2t \mathbf{i} + 3t \mathbf{j}$
2
Substitute t = 1, then calculate speed as the magnitude of velocity:
$\mathbf{v}(1) = 2\mathbf{i} + 3\mathbf{j} \implies \text{speed} = \sqrt{2^2 + 3^2} = \sqrt{13} \approx 3.61 \text{ m s}^{-1} (3 s.f.)$
3
Differentiate velocity to find acceleration vector, then calculate its magnitude:
$\mathbf{a}(t) = \dot{\mathbf{v}}(t) = 2\mathbf{i} + 3\mathbf{j} \implies |\mathbf{a}| = \sqrt{2^2 + 3^2} = \sqrt{13} \approx 3.61 \text{ m s}^{-2} (3 s.f.)$

Exam tip:

Read questions carefully to check if you are asked for a vector (velocity/acceleration) or scalar (speed/magnitude of acceleration) answer to avoid losing easy marks.

5. Common Pitfalls

Wrong move:

Using memorised projectile range/time of flight formulas without derivation

Why:

Edexcel mark schemes require first-principle derivation of projectile results, so you will lose all method marks for using un-derived formulas

Correct move:

Split motion into horizontal/vertical components, use suvat equations to derive all required results step-by-step

Wrong move:

Forgetting to define your positive vertical direction, leading to sign errors for g

Why:

Gravity acts downwards, so if you take upwards as positive, acceleration is negative, which is a common source of incorrect answers

Correct move:

Explicitly state your positive vertical direction at the start of every vertical motion or projectile problem

Wrong move:

Using $a = v \frac{d v}{d x}$ to solve acceleration problems in M2

Why:

This method is part of M3, and is not permitted for M2 kinematics questions which require time-based calculus only

Correct move:

Differentiate displacement with respect to time to get velocity, then differentiate velocity with respect to time to get acceleration

Wrong move:

Rounding intermediate calculation values

Why:

Rounding early introduces cumulative error, leading to final answers outside the acceptable 3 significant figure tolerance

Correct move:

Store all intermediate values in your calculator, only round your final answer to 3 significant figures

Wrong move:

Giving a vector answer when a scalar is asked for, or vice versa

Why:

Mark schemes award separate marks for vector and scalar answers, so you will lose marks for providing the wrong type

Correct move:

Check the question carefully: if it asks for speed, calculate the magnitude of velocity; if it asks for velocity, give full i/j component form

6. Quick Reference Cheatsheet

Concept	Formula/Rule	Key Exam Note
Vertical plane constant acceleration	Split vectors into x/y components, apply suvat to each axis	Horizontal a=0, vertical a = g = 9.8 m s⁻² downwards
Projectile motion	Derive all results from component suvat equations	No memorised range/time of flight formulas allowed
Straight line time-based calculus	$v = \frac{d x}{d t}$ , $a = \frac{d v}{d t}$ ; integrate to reverse	No $v \frac{d v}{d x}$ permitted in M2
Vector kinematics	Differentiate/integrate i and j components separately; $v = \dot{r}$ , $a = \ddot{r}$	Speed = $∣ v ∣$ , magnitude of acceleration = $∣ a ∣$

7. Frequently Asked

Do I need to memorise projectile range formulas for M2?

No, Edexcel requires all projectile results to be derived from first principles using component suvat equations. Memorised range/time of flight formulas will not earn full method marks if not derived.

Can I use $v \frac{d v}{d x}$ to solve acceleration problems in M2?

No, this technique is part of M3. All calculus in M2 kinematics must be performed with respect to time only, using $\frac{d x}{d t}$ and $\frac{d v}{d t}$ .

Going deeper

What's Next

Now that you have mastered M2 kinematics, you are ready to move on to the next core M2 topic: centres of mass, which builds on your understanding of vectors and forces from M1. You should also practice past paper kinematics questions to familiarise yourself with Edexcel's exam style and mark scheme requirements, as this topic appears on almost every M2 paper. Make sure you are confident with P4 calculus and vector operations, as these are foundational for all remaining M2 topics, including work, energy, power and impulse, momentum. Regular practice of mixed kinematics questions will help you quickly identify which method to use for different problem types.