Pearson Edexcel · A-Level · Mathematics
Course version: 9MA0 · Version from Mar 2026
Use calculus in kinematics for motion in a straight line: v = dr/dt, a =…
Full specification wording
Kinematics: 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.
Checked against the Pearson Edexcel 9MA0 specification (opens in a new tab) on .
- Point 4 of 5
- 4 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
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.
What earns the marks
- 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.
Common mistakes and how to put them right
Omitting the constant of integration when finding velocity or displacement.
Put it right: Add the constant and evaluate it from the given initial conditions.
Differentiating or integrating only one component of a vector function.
Put it right: Treat each component separately and recombine with unit vectors.
Using constant-acceleration formulae when a is a function of time.
Put it right: Integrate the given acceleration function instead.
Exam tips
- 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.
Try the lesson
The start of the Kinematics: 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. lesson, made from checked teaching notes for this specification point.
Have a go first
A particle moves along a straight line so that its displacement from a fixed point is r = 3t² + 2t metres at time t seconds. What is its velocity at t = 4 s?
Listen
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. A strong answer earns marks for these points. 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. A common slip: Omitting the constant of integration when finding velocity or displacement. Instead, add the constant and evaluate it from the given initial conditions. Exam tip: Write the definitions v = dr/dt and a = dv/dt before starting, so the required operation is clear.
Key words
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Differentiation and integration as inverse operations
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Initial conditions and constants of integration
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Componentwise treatment of vector motion
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Calculus as the general language of kinematics
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See every definition
- Differentiation and integration as inverse operations
- Velocity is the time derivative of displacement and acceleration the derivative of velocity, while displacement follows from integrating velocity and velocity from integrating acceleration.
- Initial conditions and constants of integration
- 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.
- Componentwise treatment of vector motion
- 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.
- Calculus as the general language of kinematics
- This calculus approach handles variable acceleration, where constant-acceleration formulae do not apply.
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Test your understanding of Kinematics: 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.
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