What is sign of acceleration?

Negative acceleration does not always mean "slowing down". A particle is slowing down when v and a have opposite signs and speeding up when they have the same sign.

NESA 2018 HSC Q12-style practice: A particle has velocity v(t) = 4 t - t^2 m/s for t ≥ 0. Given that the particle starts at the origin, find its displacement at t = 3 seconds.

Integrate the velocity to recover displacement. x(t) = int (4 t - t^2) \, dt = 2 t^2 - (t^3)/(3) + C. Initial condition x(0) = 0 gives C = 0. x(3) = 2(9) - (27)/(3) = 18 - 9 = 9 m. Markers expect the explicit antiderivative, use of the initial condition to fix C, and a final answer with units.

§-Syllabus dot point

NSWMaths AdvancedSyllabus dot point

How do we use calculus to analyse the motion of a particle moving in a straight line?

Apply calculus to motion in a straight line, with displacement, velocity and acceleration as derivatives and integrals with respect to time

A focused answer to the HSC Maths Advanced dot point on rectilinear motion. Velocity as the derivative of displacement, acceleration as the derivative of velocity, and recovering displacement from velocity by integration.

Generated by Claude Opus 4.814 min answerUpdated 2026-06-20

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

NESA wants you to model a particle moving along a straight line and link the three quantities (displacement, velocity, acceleration) by calculus. Differentiate with respect to time to step from displacement to velocity to acceleration, and integrate to step back.

The answer

The kinematic chain

Let $x(t)$ be the displacement of the particle at time $t$ from a fixed origin.

v(t) = \frac{dx}{dt}, \qquad a(t) = \frac{dv}{dt} = \frac{d^2 x}{dt^2}.

Going the other way:

v(t) = \int a(t) \, dt + C_1, \qquad x(t) = \int v(t) \, dt + C_2.

The constants of integration are fixed by initial conditions (typically $x(0)$ and $v(0)$ ). The constant matters: integrating gives the family of all possible motions with the right velocity, and the initial condition picks out the one the particle actually follows. Skip it and your displacement is wrong by an unknown constant.

Sign conventions

Choose a positive direction on the line. Then:

$x > 0$ means the particle is on the positive side of the origin.
$v > 0$ means the particle is moving in the positive direction.
$a > 0$ means the velocity is increasing (becoming more positive).

The particle is at rest when $v(t) = 0$ . The particle is at the origin when $x(t) = 0$ . These are different events and questions often hinge on the distinction.

Speed versus velocity

Speed is the magnitude of velocity: $\text{speed} = |v(t)|$ . Velocity carries a sign; speed does not.

Distance versus displacement

If the particle changes direction during an interval, the total distance travelled is not $|x(b) - x(a)|$ . Instead, split the interval at the times when $v(t) = 0$ and sum the magnitudes of the displacement changes on each subinterval, or compute $\int_a^b |v(t)| \, dt$ .

The reason the naive $|x(b) - x(a)|$ fails is that it only measures net displacement: if the particle goes out and comes back, the trip out and the trip back partly cancel, just as positive and negative areas cancel in a definite integral. Total distance refuses that cancellation by taking the magnitude on each leg, which is why you must break the journey at every reversal first. This is the same split-at-the-zeros idea used to turn a signed integral into a geometric area.

Total distance from a velocity sign diagram, stage by stage

Finding total distance (as opposed to net displacement) is the most procedural skill in this topic, and a sign diagram is the cleanest way to do it. Below the method is built up for the worked-example velocity $v(t) = t^2 - 4t + 3 = (t-1)(t-3)$ with starting position $x(0) = 2$ .

Stage 1, find when the particle is at rest. Solve $v(t) = 0$ . The factored form $(t-1)(t-3) = 0$ gives $t = 1$ and $t = 3$ : the only times the velocity is zero, so the only times the particle can change direction.

Stage 2, read the sign of $v$ on each interval. Between the rest times the velocity keeps one sign. Test a point in each interval (or read it off the parabola): $v > 0$ on $(0, 1)$ , $v < 0$ on $(1, 3)$ , and $v > 0$ on $(3, \infty)$ .

Stage 3, turn signs into directions. A positive velocity means the particle moves in the positive direction, a negative velocity means it moves in the negative direction. So the particle goes forward, reverses at $t = 1$ , comes back, then reverses again at $t = 3$ and goes forward. The two reversals are exactly the rest times.

Stage 4, add the leg lengths. Compute the position at the start, at each rest time and at the end: $x(0) = 2$ , $x(1) = \tfrac{10}{3}$ , $x(3) = 2$ , $x(5) = \tfrac{26}{3}$ . The total distance is the sum of the magnitudes of the changes on each leg, $\left|\tfrac{10}{3} - 2\right| + \left|2 - \tfrac{10}{3}\right| + \left|\tfrac{26}{3} - 2\right| = \tfrac{4}{3} + \tfrac{4}{3} + \tfrac{20}{3} = \tfrac{28}{3}$ m, which is much more than the net displacement $x(5) - x(0) = \tfrac{20}{3}$ m.

Reading a displacement-time graph

Many HSC items present a graph rather than a formula. On a displacement-time graph, the gradient at any point is the velocity, so the particle is at rest where the curve is momentarily horizontal (a turning point) and is at the origin where the curve crosses the time axis. A steeper section means a faster particle, and the steepest sections correspond to the greatest speeds. On a velocity-time graph, the gradient is the acceleration and the signed area between the curve and the time axis is the displacement. Being fluent at translating between these two graphs is examined directly.

Maximum displacement and turning points

To find how far a particle gets before turning back, you find when $v(t) = 0$ (a stationary point of displacement) and evaluate $x$ there. This connects motion to the curve-sketching skills from applications of differentiation: the times of rest are the stationary points of the displacement function, and the second derivative (acceleration) tells you whether the displacement is a maximum or a minimum at that instant.

Speeding up or slowing down

A favourite "explain" question asks whether the particle is speeding up or slowing down at a given instant, and the sign of $a$ alone does not answer it. Speed is $|v|$ , so the particle speeds up when speed is increasing and slows down when speed is decreasing. The clean test is the sign of the product $v \cdot a$ : if $v$ and $a$ have the same sign ( $va > 0$ ) the particle is speeding up, and if they have opposite signs ( $va < 0$ ) it is slowing down. A negative acceleration on its own can mean either, depending on which way the particle is already moving.

How exam questions ask about motion

The phrasing maps directly onto "differentiate or integrate, and which function":

"When is the particle at rest / stationary / momentarily at rest?" Solve $v(t) = 0$ . Do not solve $x(t) = 0$ .
"When is the particle at the origin?" Solve $x(t) = 0$ . This is a different equation from "at rest".
"Find the acceleration at $t = \ldots$ ". Differentiate twice (or differentiate $v$ ) and substitute; give units m/s $^2$ .
"Given the velocity, find the displacement / position." Integrate $v$ and use the initial position to fix the constant $C$ .
"Find the total distance travelled from $t = a$ to $t = b$ ." Split at the rest times, sum the magnitudes of the leg displacements (or integrate $|v|$ ). The sign-diagram stages above are exactly this method.
"Find the displacement / how far from the start." This is net displacement, $x(b) - x(a)$ , with its sign; it is not the total distance.
"Is the particle speeding up or slowing down?" Compare the signs of $v$ and $a$ : same sign speeds up, opposite slows down.
"From the displacement-time (or velocity-time) graph..." Read gradients and areas: gradient of $x$ - $t$ is $v$ , gradient of $v$ - $t$ is $a$ , and signed area under $v$ - $t$ is displacement.

The recurring decision is always "distance or displacement?" Distance refuses cancellation and needs the split at reversals; displacement keeps the sign and is just the change in $x$ .

Worked example

This is the velocity used in the stage-by-stage sign diagram above, worked through in full.

Analyse the motion of a particle with velocity $v(t) = t^2 - 4t + 3$

Read the question and identify what is asked: A particle moves so that its velocity is $v(t) = t^2 - 4 t + 3$ m/s for $t \geq 0$ , with $x(0) = 2$ . Find when the particle is at rest, the displacement at $t = 5$ , and the total distance travelled from $t = 0$ to $t = 5$ . Note straight away that "at rest" is a $v = 0$ condition, while displacement and distance are different questions about $x$ .
Find when it is at rest: At rest means $v(t) = 0$ . Factor: $v(t) = (t - 1)(t - 3) = 0$ , so $t = 1$ s and $t = 3$ s. These are the two reversal times (stage 1 of the diagram).
Integrate for the displacement: Integrate the velocity and fix the constant with the initial position:

x(t) = \int v(t) \, dt = \frac{t^3}{3} - 2 t^2 + 3 t + C.

With $x(0) = 2$ , we get $C = 2$ , so

x(5) = \frac{125}{3} - 50 + 15 + 2 = \frac{125 - 99}{3} = \frac{26}{3} \approx 8.67 \text{ m}.

Total distance via the direction changes: The particle reverses at $t = 1$ and $t = 3$ , so split the journey into three legs and sum their magnitudes (stages 3 and 4 of the diagram). Compute $x(0) = 2$ , $x(1) = \frac{10}{3}$ , $x(3) = 2$ , $x(5) = \frac{26}{3}$ . Then

\text{distance} = |x(1) - x(0)| + |x(3) - x(1)| + |x(5) - x(3)| = \frac{4}{3} + \frac{4}{3} + \frac{20}{3} = \frac{28}{3} \approx 9.33 \text{ m}.

Compare this with the net displacement $x(5) - x(0) = \frac{26}{3} - 2 = \frac{20}{3} \approx 6.67$ m: the total distance is larger precisely because the middle leg doubled back.

Speeding up or slowing down at an instant: To decide whether the particle is speeding up at, say, $t = 2$ , compare the signs of $v$ and $a$ . There $v(2) = 4 - 8 + 3 = -1 < 0$ and $a(t) = 2t - 4$ gives $a(2) = 0$ , the turning point of velocity; just after $t = 2$ , $a > 0$ while $v < 0$ (opposite signs), so the particle is slowing down as it approaches its return. This is the $va < 0$ test in action.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 HSC Q134 marksA particle moves in a straight line with displacement

x = t^3 - 6 t^2 + 9 t

metres at time

t

seconds. Find the times when the particle is at rest and its acceleration at

t = 4

seconds.

Show worked answer →

Velocity is the derivative of displacement.

$v(t) = \frac{dx}{dt} = 3 t^2 - 12 t + 9 = 3(t^2 - 4 t + 3) = 3(t - 1)(t - 3)$ .

The particle is at rest when $v(t) = 0$ : $t = 1$ s and $t = 3$ s.

Acceleration is the derivative of velocity.

$a(t) = \frac{dv}{dt} = 6 t - 12$ .

$a(4) = 24 - 12 = 12$ m/s $^2$ .

Markers reward correct factorisation, identification that "at rest" means $v = 0$ (not $x = 0$ ), and an acceleration value with units.

2018 HSC Q123 marksA particle has velocity

v(t) = 4 t - t^2

m/s for

t \geq 0

. Given that the particle starts at the origin, find its displacement at

t = 3

seconds.

Show worked answer →

Integrate the velocity to recover displacement.

$x(t) = \int (4 t - t^2) \, dt = 2 t^2 - \frac{t^3}{3} + C$ .

Initial condition $x(0) = 0$ gives $C = 0$ .

$x(3) = 2(9) - \frac{27}{3} = 18 - 9 = 9$ m.

Markers expect the explicit antiderivative, use of the initial condition to fix $C$ , and a final answer with units.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksA particle has displacement

x = 2t^3 - 9t^2 + 12t

metres at time

t

seconds. Find its velocity and acceleration at

t = 2

seconds.

Show worked solution →

Velocity is the derivative of displacement.

v(t) = \frac{dx}{dt} = 6t^2 - 18t + 12, \qquad v(2) = 6(4) - 18(2) + 12 = 24 - 36 + 12 = 0 \text{ m/s}.

Acceleration is the derivative of velocity.

a(t) = \frac{dv}{dt} = 12t - 18, \qquad a(2) = 12(2) - 18 = 6 \text{ m/s}^2.

Marker's note: one mark for $v(2) = 0$ m/s from a correct $v(t)$ , one for $a(2) = 6$ m/s $^2$ from a correct $a(t)$ , each with units. Differentiating only once and reporting $v$ as the acceleration is the usual slip.

foundation3 marksA particle moves so that its velocity is

v(t) = 3t^2 - 12t + 9

m/s for

t \geq 0

, and it starts at

x(0) = 1

m. Find an expression for its displacement

x(t)

and the times at which the particle is momentarily at rest.

Show worked solution →

Integrate the velocity and fix the constant with the initial position.

x(t) = \int (3t^2 - 12t + 9)\, dt = t^3 - 6t^2 + 9t + C.

Using $x(0) = 1$ gives $C = 1$ , so

x(t) = t^3 - 6t^2 + 9t + 1.

At rest means $v(t) = 0$ . Factor the velocity:

v(t) = 3t^2 - 12t + 9 = 3(t - 1)(t - 3) = 0 \quad\Rightarrow\quad t = 1 \text{ s and } t = 3 \text{ s}.

Marker's note: one mark for the antiderivative, one for using $x(0) = 1$ to get $C = 1$ , one for solving $v = 0$ to $t = 1, 3$ . Solving $x(t) = 0$ instead of $v(t) = 0$ for the rest times is the built-in trap.

core4 marksA particle moving in a straight line has velocity

v(t) = t^2 - 5t + 6

m/s for

t \geq 0

, and starts at the origin. (a) Find the net displacement over the first

4

seconds. (b) Find the total distance travelled over the first

4

seconds.

Show worked solution →

Integrate for the position, with $x(0) = 0$ .

x(t) = \int (t^2 - 5t + 6)\, dt = \frac{t^3}{3} - \frac{5t^2}{2} + 6t + C, \qquad x(0) = 0 \Rightarrow C = 0.

Part (a): net displacement is $x(4) - x(0)$ .

x(4) = \frac{64}{3} - \frac{5(16)}{2} + 24 = \frac{64}{3} - 40 + 24 = \frac{64}{3} - 16 = \frac{64 - 48}{3} = \frac{16}{3} \text{ m}.

Since $x(0) = 0$ , the net displacement is $\dfrac{16}{3} \approx 5.33$ m.

Part (b): split at the rest times for total distance. Solve $v = 0$ : $t^2 - 5t + 6 = (t - 2)(t - 3) = 0$ , so the particle reverses at $t = 2$ and $t = 3$ . Evaluate the position at the start, each reversal and the end:

x(0) = 0, \quad x(2) = \frac{8}{3} - 10 + 12 = \frac{14}{3}, \quad x(3) = 9 - \frac{45}{2} + 18 = \frac{9}{2}, \quad x(4) = \frac{16}{3}.

Sum the leg magnitudes:

\text{distance} = \left|\tfrac{14}{3} - 0\right| + \left|\tfrac{9}{2} - \tfrac{14}{3}\right| + \left|\tfrac{16}{3} - \tfrac{9}{2}\right| = \frac{14}{3} + \frac{1}{6} + \frac{5}{6} = \frac{28}{6} + \frac{1}{6} + \frac{5}{6} = \frac{34}{6} = \frac{17}{3} \approx 5.67 \text{ m}.

Marker's note: one mark for $x(t)$ with $C = 0$ , one for the net displacement $\tfrac{16}{3}$ m in (a), one for identifying the rest times $t = 2, 3$ and one for summing the leg magnitudes to $\tfrac{17}{3}$ m in (b). Reporting $\tfrac{16}{3}$ m for the total distance (treating it as net) forfeits the last two marks.

core4 marksThe velocity-time graph of a particle over

0 \leq t \leq 8

seconds is a straight line: it starts at

v = -6

m/s when

t = 0

, rises steadily, crosses the time axis at

t = 3

, and reaches

v = 10

m/s when

t = 8

. The particle starts at the origin. (a) Describe the direction of motion before and after

t = 3

. (b) State the acceleration. (c) Find the net displacement over the

8

seconds using the signed area under the graph.

Show worked solution →

Part (a): read the direction from the sign of $v$ . For $0 \leq t < 3$ the line is below the axis, so $v < 0$ and the particle moves in the negative direction; for $t > 3$ it is above the axis, so $v > 0$ and the particle moves in the positive direction. The reversal is at $t = 3$ , where $v = 0$ .

Part (b): acceleration is the gradient of the velocity-time line. The line rises from $v = -6$ at $t = 0$ to $v = 10$ at $t = 8$ , so

a = \frac{\Delta v}{\Delta t} = \frac{10 - (-6)}{8 - 0} = \frac{16}{8} = 2 \text{ m/s}^2.

Part (c): net displacement is the signed area between the line and the time axis. Below the axis is a triangle on $0 \leq t \leq 3$ with height $6$ ; above the axis is a triangle on $3 \leq t \leq 8$ with height $10$ :

\text{area below} = \frac{1}{2}(3)(6) = 9, \qquad \text{area above} = \frac{1}{2}(5)(10) = 25.

\text{net displacement} = -9 + 25 = 16 \text{ m}.

Marker's note: one mark for the direction change at $t = 3$ in (a), one for the acceleration $2$ m/s $^2$ as the gradient in (b), and two in (c) for treating the area below the axis as negative and combining to $16$ m. Adding the two triangle areas as $9 + 25 = 34$ (ignoring the sign of the first) is the classic error and would give the total distance, not the net displacement.

exam5 marksA particle moves in a straight line so that its displacement from the origin is

x(t) = t + 2\cos t

metres for

0 \leq t \leq 2\pi

, where

t

is in seconds and the argument of the cosine is in radians. (a) Show that the particle is momentarily at rest when

\sin t = \tfrac{1}{2}

, and hence find the two times of rest in the interval. (b) Find the exact displacement at each of these times. (c) Determine whether the particle is speeding up or slowing down at

t = \pi

, justifying your answer.

Show worked solution →

Part (a): at rest means $v(t) = 0$ . Differentiate the displacement:

v(t) = \frac{dx}{dt} = 1 - 2\sin t.

Setting $v(t) = 0$ :

1 - 2\sin t = 0 \quad\Rightarrow\quad \sin t = \frac{1}{2},

as required. In $0 \leq t \leq 2\pi$ the solutions of $\sin t = \tfrac{1}{2}$ are

t = \frac{\pi}{6} \text{ s} \quad\text{and}\quad t = \frac{5\pi}{6} \text{ s}.

Part (b): substitute each time into $x(t)$ . Using $\cos\tfrac{\pi}{6} = \tfrac{\sqrt{3}}{2}$ and $\cos\tfrac{5\pi}{6} = -\tfrac{\sqrt{3}}{2}$ :

x\!\left(\tfrac{\pi}{6}\right) = \frac{\pi}{6} + 2\cdot\frac{\sqrt{3}}{2} = \frac{\pi}{6} + \sqrt{3} \text{ m},

x\!\left(\tfrac{5\pi}{6}\right) = \frac{5\pi}{6} + 2\cdot\left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6} - \sqrt{3} \text{ m}.

Part (c): compare the signs of $v$ and $a$ at $t = \pi$ . The acceleration is

a(t) = \frac{dv}{dt} = -2\cos t.

At $t = \pi$ , with $\sin\pi = 0$ and $\cos\pi = -1$ :

v(\pi) = 1 - 2(0) = 1 > 0, \qquad a(\pi) = -2(-1) = 2 > 0.

Since $v$ and $a$ have the same sign, $v \cdot a > 0$ , so the particle is speeding up at $t = \pi$ .

Marker's note: one mark for differentiating to $v = 1 - 2\sin t$ and reaching $\sin t = \tfrac{1}{2}$ , one for both rest times $t = \tfrac{\pi}{6}, \tfrac{5\pi}{6}$ , one for each exact displacement in (b), and one in (c) for the $v \cdot a > 0$ test concluding the particle speeds up. Quoting the sign of $a$ alone (without $v$ ) does not justify the conclusion and loses the final mark.

exam5 marksA particle starts at the origin with velocity

5

m/s and moves in a straight line with acceleration

a(t) = 2t - 6

m/s

^2

for

t \geq 0

. (a) Show that its velocity is

v(t) = t^2 - 6t + 5

. (b) Find the times at which the particle is at rest. (c) Show that the particle reaches its greatest positive displacement at the first rest time, and find that displacement. (d) Find the total distance travelled in the first

5

seconds.

Show worked solution →

Part (a): integrate the acceleration and fix the constant. Since $v = \int a\, dt$ ,

v(t) = \int (2t - 6)\, dt = t^2 - 6t + C_1.

The particle starts with velocity $5$ m/s, so $v(0) = 5$ gives $C_1 = 5$ and

v(t) = t^2 - 6t + 5,

as required.

Part (b): at rest means $v(t) = 0$ .

t^2 - 6t + 5 = (t - 1)(t - 5) = 0 \quad\Rightarrow\quad t = 1 \text{ s and } t = 5 \text{ s}.

Part (c): integrate for the position, then justify the maximum. With $x = \int v\, dt$ and $x(0) = 0$ :

x(t) = \frac{t^3}{3} - 3t^2 + 5t.

The stationary points of $x$ are exactly the rest times $t = 1$ and $t = 5$ . Test the nature with the acceleration, $x''(t) = a(t) = 2t - 6$ :

a(1) = 2(1) - 6 = -4 < 0,

so $t = 1$ is a local maximum of displacement. Evaluating,

x(1) = \frac{1}{3} - 3 + 5 = \frac{1}{3} + 2 = \frac{7}{3} \approx 2.33 \text{ m},

which is the greatest positive displacement, since after $t = 1$ the velocity turns negative and the particle moves back through the origin.

Part (d): total distance splits at the rest times. On $0 \leq t \leq 5$ the only reversal in range is at $t = 1$ (the next is at the endpoint $t = 5$ ). Evaluate the position at $t = 0, 1, 5$ :

x(0) = 0, \qquad x(1) = \frac{7}{3}, \qquad x(5) = \frac{125}{3} - 75 + 25 = \frac{125}{3} - 50 = \frac{125 - 150}{3} = -\frac{25}{3}.

Sum the leg magnitudes:

\text{distance} = \left|\tfrac{7}{3} - 0\right| + \left|-\tfrac{25}{3} - \tfrac{7}{3}\right| = \frac{7}{3} + \frac{32}{3} = \frac{39}{3} = 13 \text{ m}.

Marker's note: one mark for showing $v = t^2 - 6t + 5$ via the initial velocity in (a), one for the rest times $t = 1, 5$ in (b), one for justifying the maximum with $a(1) < 0$ and giving $x(1) = \tfrac{7}{3}$ m in (c), and two in (d) for splitting at $t = 1$ and summing the magnitudes to $13$ m. Taking $|x(5) - x(0)| = \tfrac{25}{3}$ m as the distance ignores the reversal and loses the (d) marks.