Kinematics (AI HL)
Kinematics is the study of motion using calculus - displacement, velocity and acceleration are all connected by differentiation and integration. Differentiate displacement to get velocity, and velocity to get acceleration; integrate the other way to recover displacement from velocity, or velocity from acceleration. The trickiest part is distinguishing displacement (a signed position) from the total distance actually travelled, especially when a particle changes direction.
What the syllabus says
This topic maps onto one AHL syllabus point, which draws together the differentiation and integration content covered elsewhere in Calculus.
| Code | Syllabus content |
|---|---|
| AHL5.13 | Kinematic problems involving displacement \(s\), velocity \(v\) and acceleration \(a\): \(v=\dfrac{ds}{dt}\); \(a=\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2}=v\dfrac{dv}{ds}\). Displacement \(=\displaystyle\int_{t_1}^{t_2} v(t)\,dt\). Total distance travelled \(=\displaystyle\int_{t_1}^{t_2} |v(t)|\,dt\). Speed is the magnitude of velocity. Use of \(\dot x=\dfrac{dx}{dt}\) and \(\ddot x=\dfrac{d^2x}{dt^2}\) notation. |
In SL examinations, kinematics questions are not set - it's examinable only at AHL, building on the SL differentiation and integration content.
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is displacement?
Displacement \(s\) is a particle's position relative to a fixed starting point, and it can be negative - a negative value just means the particle is on the other side of where it started. It's a function of time, \(s(t)\).
e.g. For \(s(t)=t^2-4t\), \(s(3)=9-12=-3\) m: 3 m on the negative side of the start.
What is velocity?
Velocity \(v\) is the rate of change of displacement, \(v=\dfrac{ds}{dt}\) - it tells you both speed and direction. A positive velocity means displacement is increasing; a negative velocity means the particle is moving back the way it came.
e.g. With \(s(t)=t^2-4t\), \(v(t)=2t-4\), so \(v(1)=2(1)-4=-2\) m/s: moving backward.
What is acceleration?
Acceleration \(a\) is the rate of change of velocity, \(a=\dfrac{dv}{dt}\) - equivalently the second derivative of displacement. A constant positive acceleration means velocity is steadily increasing, regardless of the sign of velocity itself.
e.g. With \(v(t)=2t-4\), \(a(t)=\dfrac{dv}{dt}=2\) m/sĀ² (constant).
What's the difference between displacement and distance travelled?
Displacement is a signed net change in position; distance travelled is always positive and adds up every bit of ground covered, including any doubling back. They're only equal when velocity never changes sign over the interval.
e.g. For \(v(t)=2t-4\) on \([0,3]\): displacement \(=\int_0^3 v\,dt=-3\) m, but distance \(=|{-4}|+|1|=5\) m (split at \(t=2\), where \(v=0\)).
What is speed?
Speed is simply the magnitude of velocity - it's never negative, and it tells you how fast the particle is moving without saying which direction. "Speeding up" means \(|v|\) is increasing, which is not quite the same as acceleration being positive.
e.g. With \(v(t)=2t-4\), at \(t=1\), \(v=-2\) m/s, so the speed is \(|{-2}|=2\) m/s.
Key formulas
The whole topic rests on a small set of calculus relationships between \(s\), \(v\) and \(a\). The tables below summarise them at a glance - the explanations underneath go into more depth on each one.
Formula reference
The core kinematics relationships, and the displacement/distance integrals, are all on the formula booklet.
| Formula | Used for | Booklet? |
|---|---|---|
| \(v=\dfrac{ds}{dt}\) | Velocity from displacement | ā Yes |
| \(a=\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2}=v\dfrac{dv}{ds}\) | Acceleration from velocity or displacement | ā Yes |
| Displacement \(=\displaystyle\int_{t_1}^{t_2} v(t)\,dt\) | Recovering displacement from velocity | ā Yes |
| Total distance \(=\displaystyle\int_{t_1}^{t_2} |v(t)|\,dt\) | Total ground covered | ā Yes |
| \(\dot x=\dfrac{dx}{dt},\ \ddot x=\dfrac{d^2x}{dt^2}\) | Dot notation for time-derivatives | ā Yes |
Displacement vs distance travelled
These two quantities answer different questions about the same motion, and mixing them up is the single biggest source of lost marks on this topic.
| Feature | Displacement | Total distance travelled |
|---|---|---|
| Can it be negative? | Yes | No - always \(\geq0\) |
| Formula | \(\int_{t_1}^{t_2} v\,dt\) | \(\int_{t_1}^{t_2} |v|\,dt\) |
| If \(v\) changes sign | Forward and backward motion can cancel | Split at each zero of \(v\), add the magnitudes |
| Answers the question | "Where did it end up?" | "How far did it actually travel?" |
The core relations
Everything in this topic is either differentiating down this chain, or integrating back up it.
Differentiate to go forward
\[v=\dfrac{ds}{dt}, \qquad a=\dfrac{dv}{dt}\]
Displacement ā velocity ā acceleration, each step is one differentiation.
ā In the formula bookletIntegrate to go back
\[v=\int a\,dt, \qquad s=\int v\,dt\]
Use a given boundary condition (an initial velocity or position) to fix the constant of integration each time.
ā In the formula bookletAlternative acceleration form
\[a=v\dfrac{dv}{ds}\]
Useful when acceleration is more naturally expressed as a function of displacement rather than time.
ā In the formula bookletFinding displacement and distance
The method is the same integral either way - the only question is whether you need \(v\) or \(|v|\).
Displacement over an interval
Integrate \(v(t)\) directly between the two times.
Regions where \(v<0\) subtract from the total - that's what makes it "net" displacement.
ā In the formula bookletTotal distance travelled
Find where \(v(t)=0\) within the interval, split the integral there, and add the magnitude of each piece.
Equivalent to integrating \(|v(t)|\), but splitting by hand avoids sign errors.
ā In the formula bookletMaximum displacement
Occurs when \(v(t)=0\) - the same stationary-point logic as any other function.
Check the sign of \(v\) either side (or use \(a\)) to confirm it's a maximum, not a minimum.
Not in the formula booklet - applies the SL5.6 stationary-point methodWorked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
A particle has displacement \(s(t)=2t^3-9t^2+12t\) (m).
(a) Find \(v(t)\).
(b) Find \(a(t)\).
(c) Find the velocity at \(t=1\).
Worked solution
(a) \(v(t) = 6t^2 - 18t + 12.\) M1
\(v(t) = 6t^2 - 18t + 12.\) A1
(b) \(a(t) = 12t - 18.\) A1
(c) \(v(1) = 0\) m/s. M1
\(v(1) = 0\) m/s. A1
A GDC is permitted on this paper, so you may evaluate or verify this result directly on the calculator.
A particle has acceleration \(a(t) = 2t - 6\) m/sĀ², with \(v(0) = 5\) m/s and \(s(0) = 0.\)
(a) Find \(v(t).\)
(b)(i) Find the smaller time at which the particle is at rest.
(b)(ii) Find the larger time.
(c) Find \(s(t)\) and the position at \(t = 5.\)
Worked solution
(a) \(v = t^2 - 6t + C;\ v(0) = 5 \Rightarrow v(t)\) M1
\(= t^2 - 6t + 5.\) A1
(b)(i) \((t-1)(t-5) = 0 \Rightarrow t = 1\) s. M1
(b)(ii) \(t=5\) s. A1
(c) \(s = \tfrac{t^3}{3} - 3t^2 + 5t;\ s(0)\) M1
\(=0.\) A1
\(s(5) = \tfrac{125}{3} - 50\) M1
\(\approx -8.33\) m. A1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Confusing displacement with distance travelled. If velocity changes sign anywhere in the interval, \(\int v\,dt\) alone gives you displacement, not distance - forward and backward motion partly cancel out.
- Forgetting to split the integral at \(v=0\) for total distance. You need to find every time the particle is momentarily at rest within the interval, integrate each section separately, and add the magnitudes.
- Misreading a negative velocity as "stopped" rather than "moving backward". Zero velocity means momentarily at rest; a negative value means motion in the opposite direction, which still counts toward distance travelled.
- Mixing up maximum displacement with maximum speed. Displacement is greatest when \(v=0\); maximum speed happens at a different time in general, and needs its own stationary-point analysis on \(|v(t)|\) or on \(a(t)=0\).
Using your GDC
Kinematics questions usually come down to two GDC tasks you already know from elsewhere in Calculus: solving \(v(t)=0\) to find when a particle is momentarily at rest, using your calculator's equation solver, and evaluating a definite integral numerically to get displacement or total distance once you've set up \(\int v\,dt\) or \(\int|v|\,dt\) correctly. Some calculators also have a graphing "distance travelled" or absolute-value integral option that handles the direction-change splitting for you - check your model's menus if the integral involves \(|v(t)|\). As always, the setup - deciding what to integrate, and between which limits - is the part you need to do by hand.
See the full GDC guide for model-specific button sequences across every topic.
Ready to practise properly?
Kinematics questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
What's the difference between displacement and distance travelled?
Displacement is the net change in position - it can be negative, and it's found by integrating velocity directly. Distance travelled is always positive and measures the total ground covered, which means splitting the integral wherever velocity changes sign and adding the magnitudes of each piece.
How do I move between displacement, velocity and acceleration?
Differentiate to go from displacement to velocity to acceleration: \(v = \dfrac{ds}{dt}\) and \(a = \dfrac{dv}{dt}\). Integrate to go the other way: \(v\) is the antiderivative of \(a\), and \(s\) is the antiderivative of \(v\) - using any given boundary condition to find the constant of integration each time.
When is displacement at its maximum?
Displacement is at a maximum (or minimum) exactly when velocity is zero - the same logic as finding a stationary point of any function. Don't confuse this with maximum speed, which happens at a completely different time in general.
Can I use my GDC for kinematics problems on the exam?
Yes. Solving \(v(t)=0\) to find when a particle is at rest, and evaluating a definite integral to get displacement or distance, are both routine GDC tasks - use the equation solver and the numerical integration feature once you've set the problem up correctly by hand.
Sub-topics
Kinematics broken down into its individual skills, each with its own focused page.
Related topics
More Calculus topics from the same AI HL syllabus unit, in case you want to keep going.