What is sign of acceleration?

A retarding or restoring acceleration is negative. Track the sign carefully, especially with inverse-power laws.

NSWMaths Extension 2Syllabus dot point

How do we analyse rectilinear motion when acceleration is given as a function of displacement or of velocity rather than of time?

Apply calculus to rectilinear motion where acceleration is expressed as a function of displacement or velocity, using the forms a = v dv/dx and a = d(v^2/2)/dx

A focused answer to the HSC Maths Extension 2 dot point on rectilinear motion. Acceleration as a function of displacement or velocity, the forms a = v dv/dx and a = d(v^2/2)/dx, and integrating to recover velocity and position, with verified worked examples.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

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

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What this dot point is asking
The three forms of acceleration
Acceleration as a function of displacement
Acceleration as a function of velocity
Recovering position

What this dot point is asking

NESA wants you to handle rectilinear (straight-line) motion when acceleration is not simply a function of time. Often acceleration is given as a function of displacement $x$ or of velocity $v$ . You must choose the right expression for acceleration so that the variables separate, then integrate to find velocity and position.

The three forms of acceleration

Velocity is the rate of change of displacement, $v = \dfrac{dx}{dt}$ , and acceleration is the rate of change of velocity. The single most useful idea in this topic is that acceleration has three equivalent expressions:

a = \frac{dv}{dt}, \qquad a = v\frac{dv}{dx}, \qquad a = \frac{d}{dx}\!\left(\frac{1}{2}v^2\right).

The second follows from the chain rule, $\dfrac{dv}{dt} = \dfrac{dv}{dx}\dfrac{dx}{dt} = v\dfrac{dv}{dx}$ . The third is just the second written as a single derivative, since $\dfrac{d}{dx}(\tfrac{1}{2}v^2) = v\dfrac{dv}{dx}$ . Choosing the right form is what makes a problem solvable.

Acceleration as a function of displacement

When $a = f(x)$ , the form

\frac{d}{dx}\!\left(\frac{1}{2}v^2\right) = f(x)

is ideal, because integrating both sides with respect to $x$ gives $\tfrac{1}{2}v^2$ directly:

\frac{1}{2}v^2 = \int f(x)\,dx.

This produces $v^2$ as a function of $x$ , from which $v$ follows. This is exactly how the velocity-displacement relation for simple harmonic motion is derived.

Acceleration as a function of velocity

When $a = g(v)$ , two routes are available. To find $v$ as a function of time, use $\dfrac{dv}{dt} = g(v)$ and separate:

\int \frac{dv}{g(v)} = \int dt = t + C.

To find $v$ as a function of position, use $v\dfrac{dv}{dx} = g(v)$ and separate:

\int \frac{v\,dv}{g(v)} = \int dx = x + C.

Pick the form matching what the question asks for, $v(t)$ or $v(x)$ .

Recovering position

Once $v$ is known as a function of $t$ , position comes from $x = \int v\,dt$ . Once $v$ is known as a function of $x$ , separate $\dfrac{dx}{dt} = v(x)$ to find $t$ as a function of $x$ , then invert if a formula for $x(t)$ is required. Always apply initial conditions to fix every constant of integration.

Worked example

A particle has acceleration $a = -\dfrac{4}{x^2}$ for $x > 0$ , with $v = 3$ when $x = 1$ . Find $v^2$ as a function of $x$

Acceleration is a function of $x$ , so write

\frac{d}{dx}\!\left(\frac{1}{2}v^2\right) = -\frac{4}{x^2}.

Integrate both sides with respect to $x$ :

\frac{1}{2}v^2 = \int -4x^{-2}\,dx = \frac{4}{x} + C.

Apply the initial condition $v = 3$ at $x = 1$ : $\tfrac{1}{2}(9) = \tfrac{4}{1} + C$ , so $\tfrac{9}{2} = 4 + C$ , giving $C = \tfrac{1}{2}$ .

Therefore

\frac{1}{2}v^2 = \frac{4}{x} + \frac{1}{2} \quad\Longrightarrow\quad v^2 = \frac{8}{x} + 1.

Check the initial condition: at $x = 1$ , $v^2 = 8 + 1 = 9$ , so $v = 3$ . Correct.

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.

2024 HSC3 marksA particle of unit mass moves horizontally in a straight line. It experiences a resistive force proportional to v^2, where v m s^-1 is the speed, so that the acceleration is -k v^2. Initially the particle is at the origin with velocity 40 m s^-1 to the right. Show that v = 40 e^(-k x).

Show worked answer →

Because acceleration is given in terms of velocity and we want a relation with displacement x, use the form a = v dv/dx.

v dv/dx = -k v^2.

Divide by v (the particle is moving, so v is not zero):

dv/dx = -k v, so (1/v) dv = -k dx.

Integrate both sides: ln v = -k x + C.

Apply the initial condition x = 0, v = 40: ln 40 = 0 + C, so C = ln 40.

Then ln v - ln 40 = -k x, i.e. ln(v/40) = -k x, giving v/40 = e^(-k x).

Therefore v = 40 e^(-k x), as required.

Mark notes: 1 mark for using a = v dv/dx to form the equation, 1 mark for separating and integrating to ln v = -k x + C, 1 mark for applying v(0) = 40 to reach v = 40 e^(-k x).

2022 HSC3 marksA particle with mass 1 kg moves along the x-axis. Initially it is at the origin with speed u m s^-1 to the right. It experiences a resistive force of magnitude v + 3v^2 newtons, where v m s^-1 is the speed after t seconds. The particle is never at rest. (i) Show that dv/dx = -(1 + 3v). (ii) Hence, or otherwise, find x as a function of v.

Show worked answer →

Part (i). With mass 1, the resistive force gives acceleration -(v + 3v^2) (it opposes the motion). Use a = v dv/dx:

v dv/dx = -(v + 3v^2) = -v(1 + 3v).

Dividing by v (the particle is never at rest, so v is not zero):

dv/dx = -(1 + 3v).

Part (ii). Separate variables: dx = -dv/(1 + 3v). Integrate:

x = -(1/3) ln|1 + 3v| + C.

Apply x = 0 when v = u: 0 = -(1/3) ln(1 + 3u) + C, so C = (1/3) ln(1 + 3u).

Therefore x = (1/3) ln(1 + 3u) - (1/3) ln(1 + 3v) = (1/3) ln( (1 + 3u)/(1 + 3v) ).

Mark notes: 1 mark for using a = v dv/dx and cancelling v to show dv/dx = -(1 + 3v); then 1 mark for integrating and 1 mark for applying the initial condition to give x = (1/3) ln((1 + 3u)/(1 + 3v)).

2024 HSC4 marksAn object is dropped from rest above a magnet at distance x above it. The total acceleration is a = 27g/x^3 - g. The object is released from rest at x = 6. Show that v^2 = g(51/4 - 2x - 27/x^2).

Show worked answer →

Acceleration is a function of displacement x, so use the form a = d(v^2/2)/dx (this is the convenient form when a depends on x).

d(v^2/2)/dx = 27 g x^(-3) - g.

Integrate with respect to x:

v^2/2 = 27 g . (x^(-2)/(-2)) - g x + C = -27g/(2 x^2) - g x + C.

Multiply by 2: v^2 = -27g/x^2 - 2 g x + D, where D = 2C.

Apply the initial condition: released from rest at x = 6, so v = 0 when x = 6:
0 = -27g/36 - 12 g + D = -(3/4) g - 12 g + D, so D = (3/4) g + 12 g = 51g/4.

Therefore v^2 = -27g/x^2 - 2 g x + 51g/4 = g(51/4 - 2x - 27/x^2), as required.

Mark notes: 1 mark for using a = d(v^2/2)/dx, 1 mark for integrating correctly, 1 mark for applying v = 0 at x = 6, 1 mark for reaching the printed expression for v^2.