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When the acceleration varies with time you cannot use suvat: you differentiate and integrate instead. Questions give displacement, velocity or acceleration as a function of \(t\), in one dimension or as an \(\mathbf{i}\), \(\mathbf{j}\) vector, and ask for velocities, accelerations, times at rest, greatest values and distances travelled.
Dot notation: \(\dot{s} = v\) and \(\ddot{s} = a\) (each dot means differentiate once with respect to time).
Use calculus whenever the acceleration is not constant, e.g. \(a = 6 - 2t\). Suvat is only for constant acceleration.
Constants and key moments
diagram
Every indefinite integral needs a constant. Find it from the initial conditions: 'starts from rest at \(O\)' gives \(v = 0\) and \(s = 0\) when \(t = 0\)
Instantaneously at rest: \(v = 0\)
Greatest or least velocity: \(\dfrac{dv}{dt} = 0\), i.e. \(a = 0\). Greatest or least displacement: \(v = 0\)
The displacement between \(t = t_1\) and \(t = t_2\) is \(\int_{t_1}^{t_2} v\,dt\)
If \(v\) changes sign during the interval, the particle changes direction. Find where \(v = 0\), work out the displacement for each part, and add their magnitudes to get the total distance.
v = (t − 1)(t − 4) changes sign at t = 1 and t = 4: add the sizes of the separate areas
The functions can come from anywhere in the Pure course: polynomials, \(e^{kt}\),\(\sin kt\), \(\cos kt\)
Vectors in two dimensions
If \(\mathbf{r} = x\mathbf{i} + y\mathbf{j}\), where \(x\) and \(y\) are functions of \(t\), then \(\mathbf{v} = \dot{\mathbf{r}} = \dot{x}\mathbf{i} + \dot{y}\mathbf{j}\) and \(\mathbf{a} = \dot{\mathbf{v}} = \ddot{\mathbf{r}}\)
Integrate each component separately. The constant is a vector \(p\mathbf{i} + q\mathbf{j}\), found from the initial position or velocity.
Speed \(= |\mathbf{v}|\). Moving parallel to \(\mathbf{i}\): the \(\mathbf{j}\) component of \(\mathbf{v}\) is zero. Moving parallel to \(\mathbf{j}\): the \(\mathbf{i}\) component of \(\mathbf{v}\) is zero.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Differentiate to find velocity and acceleration
4 to 6 marksGrade D
Differentiate \(s\) to get \(v\), and \(v\) to get \(a\)
'At rest' means solve \(v = 0\); reject negative times.
Substitute the given time and state the units.
Example. A particle \(P\) moves along a straight line. At time \(t\) seconds, \(t \geqslant 0\), its displacement from a fixed point \(O\) on the line is \(s\) metres, where \(s = t^3 - 6t^2 + 9t\). (a) Find the values of \(t\) when \(P\) is instantaneously at rest. (b) Find the acceleration of \(P\) when \(t = 4\)
(b) \(a = 6t - 12\) (M1); when \(t = 4\), \(a = 12\) m s−2
(A1)
Integrate from acceleration, using initial conditions
5 to 8 marksGrade C
Integrate \(a\) to get \(v\), including \(+ c\)
Use the velocity at a known time to find \(c\)
For the greatest velocity, solve \(a = 0\) and substitute into \(v\)
Integrate \(v\) for the displacement, finding the new constant in the same way.
Example. A particle moves along a straight line. At time \(t\) seconds its acceleration is \((6 - 2t)\) m s−2. When \(t = 0\) the particle is at the point \(O\) and its velocity is 4 m s−1. (a) Find the velocity of the particle at time \(t\). (b) Find the greatest velocity of the particle. (c) Find the displacement of the particle from \(O\) when \(t = 3\)
(b) \(a = 0\) when \(t = 3\), a maximum since \(\dfrac{da}{dt} = -2 < 0\) (M1); \(v = 4 + 18 - 9 = 13\) m s−1
(A1)
(c) \(s = 4t + 3t^2 - \dfrac{t^3}{3}\), with no constant as \(s = 0\) when \(t = 0\) (M1 A1); when \(t = 3\), \(s = 12 + 27 - 9 = 30\) m
(A1)
Total distance when the particle changes direction
5 to 6 marksGrade A
Solve \(v = 0\) to find the times in the interval when the particle turns round.
Integrate \(v\) to get \(s\) (or use a definite integral for each part).
Find the displacement for each part and add the magnitudes.
Example. A particle moves along a straight line so that at time \(t\) seconds its velocity is \(v\) m s−1, where \(v = t^2 - 5t + 4\),\(t \geqslant 0\). Find the total distance travelled by the particle in the first 5 seconds.
Differentiate or integrate the \(\mathbf{i}\) and \(\mathbf{j}\) components separately.
Include a constant vector when integrating and find it from the given position or velocity.
Use the condition given: 'parallel to \(\mathbf{j}\)' means the \(\mathbf{i}\) component of \(\mathbf{v}\) is zero.
Find magnitudes using Pythagoras.
Example. A particle \(P\) moves in a horizontal plane. At time \(t\) seconds, \(t \geqslant 0\), its velocity is \(\mathbf{v}\) m s−1, where \(\mathbf{v} = (3t^2 - 12)\mathbf{i} + (4t - 2)\mathbf{j}\). When \(t = 0\), \(P\) is at the point with position vector \((2\mathbf{i} + \mathbf{j})\) m. (a) Find the magnitude of the acceleration of \(P\) when \(t = 2\). (b) Find the position vector of \(P\) at time \(t\). (c) Find the position vector of \(P\) at the instant when it is moving parallel to \(\mathbf{j}\)
Show the model answerHide the model answer
(a) \(\mathbf{a} = 6t\mathbf{i} + 4\mathbf{j}\) (M1); when \(t = 2\), \(\mathbf{a} = 12\mathbf{i} + 4\mathbf{j}\) and \(|\mathbf{a}| = \sqrt{160} = 12.6\) m s−2 (3 s.f.)
(c) \(3t^2 - 12 = 0\), so \(t = 2\) (M1 A1); \(\mathbf{r} = (-14\mathbf{i} + 5\mathbf{j})\) m
(A1)
Don’t lose marks
Using suvat when the acceleration depends on \(t\)
Leaving out the constant of integration, or assuming it is zero without checking.
Using \(\int v\,dt\) across a change of direction as the distance travelled.
Differentiating when you should integrate, or the other way round.
Forgetting the initial position vector when integrating a velocity vector.
More tips
Memory tricks
Differentiate going down the chain \(s \to v \to a\); integrate going back up \(a \to v \to s\)
Check a constant of integration by substituting \(t = 0\) back in: you should get the initial value you were given.
Use your calculator's definite integral function to check an answer, but show the integration in your working.
A quick sketch of \(v\) against \(t\) shows where the particle changes direction.
Exam technique
Show the differentiated or integrated expression before substituting numbers: the method marks depend on it.
Check the domain, usually \(t \geqslant 0\), and reject negative times.
State units: m for displacement, m s−1 for velocity, m s−2 for acceleration.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade D
Differentiate displacement to find velocity and acceleration\(v = \dfrac{ds}{dt}\) and \(a = \dfrac{dv}{dt}\)
Grade C
Integrate to find velocity or displacementIntegrate, then use the initial conditions to find the constant of integration.
Grade C
Find when a particle is at restSolve \(v = 0\); for the greatest or least velocity, solve \(a = 0\)
Grade B
Differentiate and integrate vectors in terms of \(t\)Work on the \(\mathbf{i}\) and \(\mathbf{j}\) components separately; the constant of integration is a vector.
Grade A
Use calculus with exponential or trigonometric functionsApply the Pure rules, e.g. \(v = 4e^{-0.5t}\) gives \(a = -2e^{-0.5t}\)
Grade A
Find the distance when the direction changesSplit the time interval where \(v = 0\), find each displacement and add their magnitudes.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy5 marks
A particle \(P\) moves along a straight line. At time \(t\) seconds, \(t \ge 0\), the displacement of \(P\) from a fixed point \(O\) on the line is \(s\) metres, where\[s = t^{3} - 9t^{2} - 12t\]
(a) Find an expression for the velocity of \(P\) at time \(t\) seconds.[2]
(b) Find the velocity and the acceleration of \(P\) when \(t = 5\).[3]
Show the answer and mark scheme
(a)Answer: \(v = 3t^{2} - 18t - 12\)
M1 for differentiating: at least one power of \(t\) reduced by 1
(b)Answer: Velocity \(-27\) m s−1; acceleration \(12\) m s−2
B1ft for velocity \(-27\) (m s−1)
M1 for differentiating their \(v\) to obtain \(a = 6t - 18\)
A1 for acceleration \(12\) (m s−2)
Worked solution: \(v(5) = -27\text{ m s}^{-1}\) \(a = \frac{\mathrm{d}v}{\mathrm{d}t} = 6t - 18 \Rightarrow a(5) = 12\text{ m s}^{-2}\)
Question 2Medium8 marks
A particle \(P\) moves along a straight line. At time \(t\) seconds, \(t \ge 0\), the velocity of \(P\) is \(v\) m s−1, where \[v = 3t^2 - 12t + 9\]A student is asked to find the distance travelled by \(P\) in the first 4 seconds and writes: \(\int_{0}^{4} (3t^2 - 12t + 9)\,\mathrm{d}t = \left[t^3 - 6t^2 + 9t\right]_{0}^{4} = 64 - 96 + 36 = 4\) m
(a) Explain why the student’s answer is not the distance travelled.[2]
(b) Find the total distance travelled by \(P\) in the first 4 seconds.[4]
(c) Find the greatest speed of \(P\) for \(1 \le t \le 3\).[2]
Show the answer and mark scheme
(a)Answer: \(v\) changes sign at \(t = 1\) and \(t = 3\), so \(P\) changes direction; the integral gives the displacement, in which motion in the negative direction cancels motion in the positive direction.
B1 for \(v = 0\) at \(t = 1\) and \(t = 3\) (\(v \lt 0\) between them), so \(P\) changes direction
B1 for the integral gives the (net) displacement, not the distance
Worked solution: \(v = 3(t - 1)(t - 3)\), which is negative for \(1 \lt t \lt 3\): \(P\) moves backwards during that time. The integral gives the displacement, in which the backward motion is subtracted from the forward motion.
(b)Answer: 12 m
M1 for solving \(v = 0\): \(t = 1\) and \(t = 3\)
M1 for evaluating the displacement function \(s = t^3 - 6t^2 + 9t\) at \(t = 1\), 3 and 4 (4, 0, 4)
M1 for adding the distances for each interval: \(4 + |0 - 4| + 4\)
M1 for \(\frac{\mathrm{d}v}{\mathrm{d}t} = 6t - 12 = 0 \Rightarrow t = 2\)
A1 for speed 3 (m s−1) (from \(v = -3\))
Worked solution: \(\frac{\mathrm{d}v}{\mathrm{d}t} = 6t - 12 = 0\) at \(t = 2\), where \(v = 12 - 24 + 9 = -3\). At \(t = 1\) and \(t = 3\), \(v = 0\). So the greatest speed is 3 m s−1.
Question 3Hard11 marks
A particle \(P\) moves along a straight line. At time \(t\) seconds, \(t \ge 0\), the displacement of \(P\) from a fixed point \(O\) on the line is \(s\) metres, where\[s = 2t^{3} - 12t^{2} + 18t\] In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
(a) Find the values of \(t\) for which \(P\) is instantaneously at rest.[3]
(b) Find the total distance travelled by \(P\) in the interval \(0 \le t \le 5\).[5]
(c) Find the minimum value of \(v\), justifying your answer.[3]
Show the answer and mark scheme
(a)Answer: \(t = 1\) and \(t = 3\)
M1 for differentiating: \(v = 6t^{2} - 24t + 18\)
M1 for setting their \(v = 0\) and solving a 3-term quadratic
A1 for \(t = 1\) and \(t = 3\)
Worked solution: \(v = 6t^{2} - 24t + 18 = 6(t - 1)(t - 3) = 0 \Rightarrow t = 1 \text{ or } t = 3\)
(b)Answer: \(56\) m
M1 for recognising that \(P\) changes direction at \(t = 1\) and \(t = 3\) (splitting the interval)
M1 for finding \(s\) at \(t = 1\), \(t = 3\) and \(t = 5\)
A1 for \(s(1) = 8\), \(s(3) = 0\) and \(s(5) = 40\)
M1 for adding the magnitudes of the three displacements
B1 for a justification that this is a minimum, e.g. \(\frac{\mathrm{d}^2v}{\mathrm{d}t^2} = 12 \gt 0\) or \(v\) is a positive quadratic in \(t\)
Worked solution: \(\frac{\mathrm{d}v}{\mathrm{d}t} = 12t - 24 = 0 \Rightarrow t = 2\), and \(v(2) = -6\). \(\frac{\mathrm{d}^2v}{\mathrm{d}t^2} = 12 \gt 0\), so this is a minimum: the least value of \(v\) is \(-6\) m s−1.