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Kinematics and rates of change

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The kinematic chain

Displacement, velocity and acceleration are linked by calculus in both directions:

$s \;\xrightarrow{\;\text{differentiate}\;}\; v \;\xrightarrow{\;\text{differentiate}\;}\; a$

$a \;\xrightarrow{\;\text{integrate}\;}\; v \;\xrightarrow{\;\text{integrate}\;}\; s$

Each part:
- $s$ — displacement, position relative to a starting point (metres)
- $v = \dfrac{ds}{dt}$ — velocity, the rate of change of displacement (m s⁻¹)
- $a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}$ — acceleration (m s⁻²)
Each integration needs a constant, and each constant needs its own condition.

Displacement versus distance

Displacement is a signed quantity — it can be negative, meaning the object is on the other side of the origin.
Distance travelled is always positive and counts every metre covered, including any doubling back.
They differ whenever the object changes direction, which happens when $v = 0$ and the velocity changes sign.
To find total distance travelled:
- Find the times when $v = 0$ inside the interval.
- Split the integral at each of those times.
- Take magnitudes and add.
This is exactly the same procedure as finding area when a curve crosses the $x$ -axis — and for the same reason.

Velocity versus speed

Speed is the magnitude of velocity: $|v|$ .
A velocity of $-8$ m s⁻¹ is a speed of 8 m s⁻¹ in the negative direction.
The object is momentarily at rest when $v = 0$ , which is also where it may turn around.

Distance as an area

$\displaystyle\int_{t_1}^{t_2} v\,dt$ is the area under the velocity–time graph, and it gives the displacement over that interval.
Total distance is the area counted positively on both sides of the time axis.
Similarly $\displaystyle\int_{t_1}^{t_2} a\,dt$ gives the change in velocity.

Reading the initial condition correctly

"Initially at rest" means $v = 0$ when $t = 0$ .
"Starts from the origin" means $s = 0$ when $t = 0$ .
"Initially 5 m from the origin" means $s = 5$ when $t = 0$ .
The warning from the 2025 report, in full: "In Question Two (d), a kinematics question, a significant number of candidates assumed that as $s = 0$ when $t = 0$ , then the constant of integration, $c$ , would also be $c = 0$ . The displacement equation was an exponential function, and consequently this was not the case."
Substitute and solve for the constant. Never assume it.

Rates of change generally

The same structure applies to any rate, not only motion:
- Given a rate, integrate to find the total. If water flows in at $\frac{dV}{dt} = 3t + 2$ litres per minute, the volume added between $t = 0$ and $t = 4$ is $\displaystyle\int_0^4 (3t+2)\,dt$ .
- Given a total, differentiate to find the rate.
Watch the units. Integrating a rate in "litres per minute" over a time in minutes gives litres — the time unit cancels.

Setting out a kinematics question

State which quantity you are given and which you want.
Decide whether to differentiate or integrate, and how many times.
Integrate, adding a constant each time.
Apply each condition to the appropriate function.
Answer with units, and interpret the sign.

Tips and Tricks

The constant is not zero just because the quantity is. This is the specific error the 2025 report calls out on this exact topic. Substitute the initial condition into the whole expression — including any exponential term, which is $1$ at $t = 0$ , not $0$ — and solve.

"Distance travelled" is not "displacement". If the question asks for distance and the object changes direction, you must find where $v = 0$ , split, and add the magnitudes. The report lists "understood the relationship between acceleration, velocity and displacement" and "interpreted given information regarding velocity and displacement correctly" as Merit behaviours — the interpretation is the graded part.

Match each condition to the right function. An initial velocity is a condition on $v$ ; an initial position is a condition on $s$ . Applying one to the other produces a wrong answer that looks reasonable.

Give units on every answer, and state the direction where the sign carries meaning: "the particle is 12 m to the left of the origin", not just " $-12$ ".

When integrating a rate to find a total, check the units multiply out sensibly. Litres per minute times minutes gives litres; if your units do not cancel, the setup is wrong.

Worked ExampleDisplacement and total distance

A particle moves in a straight line with velocity $v = 3t^2 - 12t + 9$ m s⁻¹, where $t$ is in seconds. It starts at the origin. Find (a) its displacement after 4 seconds and (b) the total distance it travels in that time.

Step 1 — Find the displacement function

Integrate the velocity:

$s = \int \left(3t^2 - 12t + 9\right)dt = t^3 - 6t^2 + 9t + c$

Apply the initial condition. "Starts at the origin" means $s = 0$ when $t = 0$ :

$0 = 0 - 0 + 0 + c \quad \Rightarrow \quad c = 0$

$s = t^3 - 6t^2 + 9t$

Step 2 — Part (a): displacement after 4 seconds

Substitute $t = 4$ :

$s = 64 - 96 + 36 = 4$

Step 3 — Part (b): find when the particle changes direction

Distance and displacement differ only if the particle turns around, which happens where $v = 0$ :

$3t^2 - 12t + 9 = 0$

Divide by 3 and factorise:

$t^2 - 4t + 3 = 0$

$(t - 1)(t - 3) = 0 \quad \Rightarrow \quad t = 1 \text{ and } t = 3$

Both are inside the interval $0 \leq t \leq 4$ , so the motion has three phases.

Step 4 — Determine the direction in each phase

Test the sign of $v$ in each:

$t = 0.5$ : $v = 0.75 - 6 + 9 = 3.75 > 0$ — moving forwards
$t = 2$ : $v = 12 - 24 + 9 = -3 < 0$ — moving backwards
$t = 3.5$ : $v = 36.75 - 42 + 9 = 3.75 > 0$ — moving forwards again

The particle goes out, comes back, then goes out again.

Step 5 — Find the position at each turning point

Using $s = t^3 - 6t^2 + 9t$ :

$s(0) = 0$
$s(1) = 1 - 6 + 9 = 4$
$s(3) = 27 - 54 + 27 = 0$
$s(4) = 64 - 96 + 36 = 4$

Step 6 — Compute the distance for each phase

Take the magnitude of each change in position:

Phase 1 ( $0 \to 1$ ): from 0 to 4, distance $= |4 - 0| = 4$ m
Phase 2 ( $1 \to 3$ ): from 4 to 0, distance $= |0 - 4| = 4$ m
Phase 3 ( $3 \to 4$ ): from 0 to 4, distance $= |4 - 0| = 4$ m

Step 7 — Add the magnitudes

$\text{Total distance} = 4 + 4 + 4 = 12 \text{ m}$

Step 8 — Compare the two answers

Displacement: 4 m — the particle ends 4 m from the origin.
Distance travelled: 12 m — it actually covered three times that, going out, back, and out again.

The particle's journey is $0 \to 4 \to 0 \to 4$ . Integrating the velocity straight from 0 to 4 gives only the net result and would have answered the wrong question.