Kinematics — Edexcel A-Level Mathematics
Test yourself on Kinematics with PEARSON EDEXCEL A-Level practice questions.
7 days Premium · Then free forever · No card, no charge
Kinematics explained
Kinematics describes motion without considering its causes.
Read the full explanation
Position locates a particle relative to a fixed origin, often given as a coordinate or vector. Displacement is the change in position, a vector with both magnitude and direction; distance travelled is the total scalar path length and is never negative. Velocity is the rate of change of displacement, a vector, so a negative velocity indicates motion in the negative direction. Speed is the magnitude of velocity, a scalar, so it cannot be negative. Acceleration is the rate of change of velocity, a vector, with unit m s⁻²; a particle can accelerate while moving at constant speed if its direction changes, though in straight-line motion acceleration changes the speed. For example, a particle moving 5 m east then 3 m west has displacement 2 m east but distance travelled 8 m.
7.2 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.
For straight-line motion, a displacement–time graph plots displacement on the vertical axis against time. Its gradient gives velocity: a positive gradient means motion in the positive direction, a negative gradient means motion in the negative direction, and a horizontal line means the particle is at rest. A velocity–time graph plots velocity vertically against time. Its gradient gives acceleration, since acceleration is the rate of change of velocity. The area between the graph and the time axis gives displacement, with areas above the axis counted positive and areas below counted negative; the total distance travelled is the sum of the magnitudes of these areas. For example, a velocity rising steadily from 0 m s⁻¹ to 10 m s⁻¹ over 5 s has acceleration 2 m s⁻² and displacement equal to the triangular area ½ × 5 × 10 = 25 m.
7.3 Understand, use and derive the formulae for constant acceleration for motion in a straight line. Extend to 2 dimensions using vectors.
For constant acceleration a along a straight line, velocity changes linearly with time: v = u + at. Displacement accumulates as s = ut + ½at², and eliminating t gives v² = u² + 2as; the average-velocity form s = ½(u + v)t follows from the trapezium area under a velocity–time graph. Derivation uses a = dv/dt and v = ds/dt with a constant, integrating from t = 0 with v = u and s = 0. In two dimensions each formula applies independently to the x- and y-components, so r = r₀ + u t + ½a t² and v = u + a t become vector equations; a projectile has a = −g ĵ while the horizontal component of velocity stays constant. Choose the equation omitting the unwanted quantity, fix a positive direction, and keep signs consistent.
7.4 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.
Position r, velocity v and acceleration a are linked by differentiation and integration with respect to time. Velocity is the time derivative of displacement, v = dr/dt, and acceleration is the derivative of velocity, a = dv/dt = d²r/dt². Reversing the process, displacement follows from integrating velocity, r = ∫ v dt, and velocity from integrating acceleration, v = ∫ a dt; each indefinite integral needs a constant of integration fixed by initial conditions, while a definite integral between two times gives the change in the quantity. In two dimensions the same operations apply to each component, so r = x i + y j, v = ẋ i + ẏ j and a = ẍ i + ÿ j. This calculus approach handles variable acceleration, where constant-acceleration formulae do not apply.
7.5 Model motion under gravity in a vertical plane using vectors; projectiles.
A projectile moves in a vertical plane under gravity alone, so its acceleration is constant and directed downwards with magnitude g, approximately 9.8 m s⁻². Taking horizontal and vertical axes, the horizontal acceleration is zero and the vertical acceleration is −g, giving r = r₀ + u t + ½a t² with a = −g ĵ. The horizontal velocity component u cos θ stays constant while the vertical component u sin θ − gt changes linearly. This yields time of flight from the vertical displacement equation, maximum height when the vertical velocity is zero, and horizontal range from the horizontal velocity multiplied by the time of flight. The model neglects air resistance and treats g as constant, so it applies only while the projectile is in free flight.
Your focus
- Define position, displacement, distance travelled, velocity, speed and acceleration accurately.
- Classify each quantity as vector or scalar and give its S.I. unit.
- Calculate displacement and distance travelled for a described journey.
Show all 15 objectives
- Find velocity from the gradient of a displacement–time graph.
- Find acceleration from the gradient and displacement from the area of a velocity–time graph.
- Distinguish displacement from distance travelled using signed and unsigned areas.
- Derive the constant-acceleration formulae from a = dv/dt and v = ds/dt with stated initial conditions.
- Select and apply the appropriate formula to a straight-line motion problem with consistent signs and units.
- Extend the formulae to two dimensions by resolving motion into perpendicular components and using vector notation.
- Differentiate position with respect to time to find velocity and acceleration in one and two dimensions.
- Integrate acceleration and velocity with respect to time, applying initial conditions to determine constants.
- Solve variable-acceleration problems by calculus and interpret the results in context.
- Resolve initial velocity into components and set up vector equations for projectile motion.
- Calculate time of flight, maximum height and range for a projectile launched in a vertical plane.
- State and justify the assumptions of the projectile model and recognise when they fail.
Kinematics exam tips
Marking Points
- Define position as location relative to a fixed origin and displacement as change in position, a vector.
- Distinguish distance travelled as the total scalar path length from displacement as a vector change in position.
- Define velocity as the rate of change of displacement and speed as its magnitude, noting velocity is a vector and speed a scalar.
- Define acceleration as the rate of change of velocity, a vector with unit m s⁻².
- Apply the definitions to a simple journey, computing displacement and distance travelled separately.
- Interpret the gradient of a displacement–time graph as velocity, including its sign as direction.
- Interpret the gradient of a velocity–time graph as acceleration.
- Interpret the area under a velocity–time graph as displacement, treating areas below the axis as negative.
- Distinguish displacement from distance travelled by summing signed areas for displacement and magnitudes for distance.
- Use graph features such as intercepts, turning points and horizontal sections to describe the motion in words.
- States the five constant-acceleration formulae and identifies which variable each omits, so the correct equation can be selected for a given problem.
- Derives v = u + at by integrating a = dv/dt with constant a and initial condition v = u at t = 0.
- Derives s = ut + ½at² by integrating v = u + at with s = 0 at t = 0, and obtains v² = u² + 2as by eliminating t.
- Applies the formulae componentwise in two dimensions, writing r = r₀ + u t + ½a t² and v = u + a t as vector equations.
- Interprets gradient and area of velocity–time and displacement–time graphs as acceleration and displacement respectively.
- Maintains consistent sign conventions and units, for example taking upwards as positive so that a = −9.8 m s⁻² under gravity.
- Differentiates a given position vector or displacement function to obtain velocity, and differentiates again to obtain acceleration.
- Integrates an acceleration function to obtain velocity and integrates velocity to obtain displacement, including the constant of integration.
- Uses initial conditions to determine constants of integration or evaluates definite integrals to find changes over a time interval.
- Applies differentiation and integration componentwise to vector functions, keeping i and j components separate.
- Recognises when acceleration varies with time and therefore uses calculus rather than constant-acceleration formulae.
- Interprets stationary points and turning points of displacement or velocity functions in the context of the motion.
- Resolves the initial velocity into horizontal and vertical components using u cos θ and u sin θ.
- Sets up the vector equation r = r₀ + u t + ½a t² with a = −g ĵ and applies it componentwise.
- Finds time of flight by solving the vertical displacement equation for the time when the projectile returns to the required height.
- Finds maximum height by setting the vertical velocity component to zero and substituting the resulting time.
- Finds horizontal range from the constant horizontal velocity multiplied by the total time of flight.
- States the modelling assumptions, including negligible air resistance and constant g, and comments on their effect.
Examiner Tips
- 💡Sketch a simple diagram with a marked origin and direction arrows before calculating.
- 💡State clearly whether an answer is a vector or scalar and include direction where required.
- 💡Check the sign of displacement and velocity against your chosen positive direction.
- 💡Label axes with quantity and unit before sketching or reading a graph.
- 💡Split the area under a velocity–time graph into simple shapes such as triangles, rectangles and trapezia.
- 💡State the sign convention and use it consistently when interpreting gradients and areas.
- 💡List the known and unknown quantities with signs before choosing an equation, so the omitted variable identifies the formula to use.
- 💡Show the derivation from a = dv/dt and v = ds/dt when a question asks you to derive rather than quote a formula.
- 💡For vector motion, write the x- and y-equations on separate lines and keep the unit vectors throughout to avoid losing direction marks.
- 💡Write the definitions v = dr/dt and a = dv/dt before starting, so the required operation is clear.
- 💡Substitute initial conditions immediately after integrating to find the constant, rather than leaving it undetermined.
- 💡Check dimensions and signs of each component after differentiating or integrating a vector expression.
- 💡Draw a diagram with axes and mark the positive direction before writing any equations.
- 💡Keep horizontal and vertical equations separate and label each with its component to avoid mixing them.
- 💡Check whether the projectile lands at the same height as it was launched, since this changes the time-of-flight calculation.
Common Mistakes
- Using distance travelled and displacement interchangeably; the correction is that displacement is the vector change in position while distance travelled is the scalar total path length.
- Giving a negative value for speed; the correction is that speed is the magnitude of velocity and is non-negative.
- Assuming acceleration must change speed; the correction is that acceleration is the rate of change of velocity, so a change in direction also produces acceleration.
- Reading the height of a velocity–time graph as displacement; the correction is that displacement is the area under the graph, while the height gives velocity.
- Treating area below the time axis as positive when finding displacement; the correction is that such area contributes negatively to displacement but positively to distance travelled.
- Confusing the gradient of a displacement–time graph with acceleration; the correction is that this gradient gives velocity, while acceleration comes from the gradient of a velocity–time graph.
- Treating acceleration as variable and applying constant-acceleration formulae outside their range; the correction is to check that a is constant before using them, otherwise use calculus.
- Mixing up u and v in v² = u² + 2as; the correction is to label initial and final velocities explicitly before substituting.
- Applying a single scalar equation to a two-dimensional problem; the correction is to resolve into perpendicular components and apply the formulae to each component separately.
- Omitting the constant of integration when finding velocity or displacement; the correction is to add the constant and evaluate it from the given initial conditions.
- Differentiating or integrating only one component of a vector function; the correction is to treat each component separately and recombine with unit vectors.
- Using constant-acceleration formulae when a is a function of time; the correction is to integrate the given acceleration function instead.
- Using g as positive in the vertical equation when upwards has been taken as positive; the correction is to keep a = −g consistent with the chosen positive direction.
- Assuming the horizontal velocity changes during flight; the correction is to treat the horizontal component as constant because horizontal acceleration is zero.
- Confusing time to maximum height with total time of flight; the correction is to note that for level ground the total time is twice the time to maximum height.