Velocity and acceleration as functions in HSC Mathematics Extension 2: the calculus of rectilinear motion when acceleration depends on displacement or velocity

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:

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.

From acceleration to velocity, stage by stage

When acceleration is a function of displacement, the whole method is "choose the right form, integrate over $x$, take the square root". The panels below run that method on the worked example below, $a = -\dfrac{4}{x^2}$ with $v = 3$ at $x = 1$, so you can see each curve the calculation produces.

Stage 1, choose the form. First read off what $a$ depends on. The decision flow turns that into the form to use: a function of time integrates with $\dfrac{dv}{dt}$, a function of velocity uses $v\dfrac{dv}{dx}$ or $\dfrac{dv}{dt}$, and a function of displacement uses $\dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$, highlighted, because it hands you $v^2$ directly.

Stage 2, plot what you are given. Here $a = -\dfrac{4}{x^2}$, a retarding acceleration: negative throughout, strongest near $x = 1$ and fading towards zero as $x$ grows. Because it depends on $x$, the highlighted form applies.

Stage 3, integrate to $v^2$. Integrating $\dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right) = -\dfrac{4}{x^2}$ over $x$ and applying $v = 3$ at $x = 1$ gives $v^2 = \dfrac{8}{x} + 1$. The retarding acceleration removes kinetic energy as $x$ grows, so $v^2$ falls from $9$ towards $1$.

Stage 4, square-root to $v$. Finally take the positive square root, $v = \sqrt{\dfrac{8}{x} + 1}$, the velocity as a function of position. The sign of the root comes from the initial condition: the body starts with $v = 3 > 0$, so the positive root is correct. Substituting $x = 1$ returns $v = 3$, confirming the answer.

Acceleration as a function of displacement

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

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

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:

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

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.

Deciding the form from the question

The decision tree is short and worth memorising. If acceleration is given as a function of time, integrate $\dfrac{dv}{dt}$ directly. If acceleration is a function of displacement, use $\dfrac{d}{dx}\left(\frac{1}{2}v^2\right)$, because integrating it with respect to $x$ delivers $v^2$ in one step. If acceleration is a function of velocity, choose $\dfrac{dv}{dt}$ when the question wants time and $v\dfrac{dv}{dx}$ when it wants displacement. Picking the wrong form leaves variables tangled and the integral non-separable, so reading the question to see what the answer should depend on is the most important early move.

The energy interpretation

The form $a = \dfrac{d}{dx}\left(\frac{1}{2}v^2\right)$ is more than a trick; it is the work-energy principle in disguise. The quantity $\frac{1}{2}v^2$ is the kinetic energy per unit mass, and integrating acceleration (force per unit mass) over distance gives the change in that energy. This is why a displacement-dependent acceleration integrates so cleanly to $v^2$: you are summing the work done. Recognising the energy structure helps you anticipate the shape of the answer and provides a physical check on the algebra.

Self-checking with initial conditions

Every integration introduces a constant, and every constant must be pinned down by a stated initial condition. A reliable final step is to substitute the initial values back into your result: if the model does not reproduce the given starting velocity at the starting position, a constant is wrong. This single check, shown in the worked example, catches most sign and arithmetic slips before they cost marks.

How exam questions ask about velocity and acceleration functions

The phrasing tells you which acceleration form to reach for:

• "The acceleration is $a = f(x)$. Find $v^2$ / the speed at $x = \ldots$" Use $\dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right) = f(x)$ and integrate over $x$; this is the most common type and the one the velocity-displacement relation for SHM is built on.
• "The acceleration is $a = g(v)$. Find $v$ as a function of time" / "the time to reach a given speed." Use $\dfrac{dv}{dt} = g(v)$ and separate, integrating $\int \dfrac{dv}{g(v)} = t + C$.
• "The acceleration is $a = g(v)$. Find $v$ as a function of distance" / "the distance to come to rest." Use $v\dfrac{dv}{dx} = g(v)$ instead, integrating $\int \dfrac{v\,dv}{g(v)} = x + C$. Time never appears.
• "Show that $v = \ldots$" A "show that" almost always wants the separation and integration written out, then the initial condition applied to fix the constant; quoting the final formula earns little.
• "Hence find $x$ as a function of $v$" / "$t$ as a function of $v$." Rearrange the relation you derived; you usually do not need to invert it to $x(t)$ unless the question explicitly asks.
• "A particle is released from rest ..." The initial condition is $v = 0$ at the given position; use it to fix the constant.
• "Find where the particle comes to rest / its limiting speed." Set $v = 0$ (rest) or let $x \to \infty$ (limiting behaviour) in your $v^2(x)$ relation.