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

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.

Going the other way:

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 $x$.