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

Let $x(t)$ be the displacement of the particle at time $t$ from a fixed origin.

Choose a positive direction on the line. Then:

Speed is the magnitude of velocity: $\text{speed} = |v(t)|$. Velocity carries a sign; speed does not.

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 $\int_a^b |v(t)| \, dt$.

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

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

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

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

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

This is the net displacement, not the total distance, when the particle reverses direction.

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.

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