Kinematics: Total Distance from a Velocity Function

The examiner’s trap in kinematics is the phrase “total distance”. The particle reverses direction partway through, so integrating velocity straight across the interval gives displacement, and understates the distance. You must find where it turns and add the legs separately.

A particle moves in a straight line so that, at time $t$ seconds after passing a fixed point $O$, its velocity is $v = 3t^2 - 12t + 9\ \text{m s}^{-1}$. (i) Find the times at which the particle is instantaneously at rest. [2] (ii) Find the total distance travelled in the first $3$ seconds. [5]

The working

(i) “At rest” means $v = 0$: $3t^2 - 12t + 9 = 0 \;\Rightarrow\; t^2 - 4t + 3 = 0 \;\Rightarrow\; (t - 1)(t - 3) = 0$ $t = 1\ \text{s} \quad \text{and} \quad t = 3\ \text{s} \quad \text{(M1, A1)}$

The particle is momentarily at rest at $t = 1$ and $t = 3$, so it changes direction at $t = 1$.

(ii) Displacement is the integral of velocity: $s = \int (3t^2 - 12t + 9)\,dt = t^3 - 6t^2 + 9t \quad (+\,C,\ \text{take } C = 0 \text{ since } s = 0 \text{ at } t = 0) \quad \text{(M1)}$

Evaluate the position at the key times: $s(0) = 0, \qquad s(1) = 1 - 6 + 9 = 4, \qquad s(3) = 27 - 54 + 27 = 0 \quad \text{(A1)}$

The particle moves from $0$ to $4$ (first $1$ s), then back from $4$ to $0$ (next $2$ s). Add the two legs as distances: $\text{Leg 1: } |4 - 0| = 4\ \text{m}, \qquad \text{Leg 2: } |0 - 4| = 4\ \text{m} \quad \text{(M1)}$ $\text{Total distance} = 4 + 4 = 8\ \text{m} \quad \text{(A1)}$

Note the displacement over the $3$ s is $s(3) - s(0) = 0$, the particle ends where it started, which is exactly why “distance” and “displacement” give different answers here.

Where the marks are won and lost

• Part (i) is not decoration: the rest times are the split points for part (ii). Skipping them almost guarantees the distance error.
• Integrating $v$ from $0$ to $3$ in one go gives $s(3) - s(0) = 0$. A student who writes “distance $= 0$ m” has computed displacement and lost three marks.
• Each leg’s contribution is a modulus. The backward leg is $-4$ as a signed displacement but $4$ as a distance.

Common mistakes

• Reporting the displacement ($0$ m) as the distance.
• Forgetting the constant of integration reasoning ($s = 0$ at $t = 0$).
• Splitting at the wrong time, the turning point is where $v = 0$, i.e. $t = 1$ inside the interval (the $t = 3$ endpoint is the interval’s end).

Full method: Kinematics notes. Topic home: Calculus pillar.