Quick questions on Motion along a straight line: displacement, velocity and acceleration via calculus

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 $x(0) = 2$.

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.

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.