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 $v(0)$).

Sign conventions

Choose a positive direction on the line. Then:

• IMATH_6 means the particle is on the positive side of the origin.
• IMATH_7 means the particle is moving in the positive direction.
• IMATH_8 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 $\int_a^b |v(t)| \, dt$.

Worked example

A particle moves so that its velocity is $v(t) = t^2 - 4 t + 3$ m/s for $t \geq 0$. Given $x(0) = 2$, find when the particle is at rest, the displacement at $t = 5$, and the total distance travelled from $t = 0$ to $t = 5$.

At rest. $v(t) = (t - 1)(t - 3) = 0$ gives $t = 1$ s and $t = 3$ s.

Displacement. $x(t) = \int v(t) \, dt = \frac{t^3}{3} - 2 t^2 + 3 t + C$. With $x(0) = 2$, $C = 2$.

$x(5) = \frac{125}{3} - 50 + 15 + 2 = \frac{125}{3} - 33 = \frac{125 - 99}{3} = \frac{26}{3} \approx 8.67$ m.

Total distance. The particle reverses at $t = 1$ and $t = 3$. Compute $x(0) = 2$, $x(1) = \frac{1}{3} - 2 + 3 + 2 = \frac{10}{3}$, $x(3) = 9 - 18 + 9 + 2 = 2$, $x(5) = \frac{26}{3}$.

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

Common traps

Treating "at rest" as "at the origin". At rest means $v = 0$. At the origin means $x = 0$. These are unrelated in general.

Forgetting the constant of integration. When you integrate velocity to recover displacement, you must use the initial position to fix $C$.

Computing distance as $|x(\text{final}) - x(\text{initial})|$. This is the net displacement, not the total distance, when the particle reverses direction.

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.

Forgetting units. Displacement is in metres, velocity in m/s, acceleration in m/s$^2$. Markers deduct for missing or wrong units.

In one sentence

In straight-line motion, differentiate displacement to get velocity, differentiate velocity to get acceleration, and integrate (with initial conditions) to reverse those steps.