Q.A cyclist starts from the centre $O$ of a circular park of radius $1\ \text{km}$. She first rides straight out along a radius to the edge $P$ of the park (with $O$ at the centre and the radius $OP$ pointing horizontally to the right). She then rides a quarter of the way around the circumference from $P$ to a point $Q$ on the edge, where $Q$ is positioned so that the radius $OQ$ is perpendicular to $OP$ (i.e. $Q$ lies directly above $O$). Finally she returns straight from $Q$ back to the centre $O$ along the radius $QO$, as shown below.

Concept

Displacement is the straight-line vector from the starting point to the finishing point. Average velocity $= \dfrac{\text{displacement}}{\text{time}}$ (a vector), while average speed $= \dfrac{\text{total path length}}{\text{time}}$ (a scalar). When start and end coincide, displacement is zero even though the distance travelled is not.

The three legs of the trip

• $OP$ (radius, out to the edge) $= R = 1\ \text{km}$.
• Arc $PQ$ (a quarter of the circumference) $= \dfrac{1}{4}(2\pi R) = \dfrac{1}{4}(2\pi \times 1) = \dfrac{\pi}{2}\ \text{km} \approx 1.571\ \text{km}$.
• $QO$ (radius, back to the centre) $= R = 1\ \text{km}$.

(a) Net displacement

The cyclist starts at $O$ and returns to $O$, so the initial and final positions are the same:

(b) Average velocity

(c) Average speed …

Concept: Displacement Is Defined by Endpoints Alone — Skip the Vector Sum Entirely

Method: Position-Vector Definition First (No Leg-by-Leg Vector Addition), Arc Length via $s=r\theta$

Both existing solutions implicitly treat the trip as three legs and could in principle add them as vectors to find displacement. This method skips that step altogether: displacement is defined as $\vec{r}_{\text{final}} - \vec{r}_{\text{initial}}$, so the moment we notice the cyclist's final position coincides with her initial position, the vector sum is already known to be zero — no need to ever write down the individual leg vectors for OP, arc $PQ$, or $QO$. For the speed half of the question, this method also uses the general arc-length formula $s = r\theta$ (radian measure) rather than "a quarter of the circumference," which generalizes immediately to any fraction of a circle without re-deriving a new fraction each time.

Step 1 — Displacement, straight from the definition

By definition, for any journey, however winding:

The cyclist's journey is $O \to P \to Q \to O$ — she starts at $O$ and the last leg explicitly returns her to $O$. So $\vec{r}_{\text{final}} = \vec{r}_{\text{initial}} = \vec{r}_O$, and:

This holds regardless of the shape of the middle leg (arc $PQ$) or how far out the radius $R$ is — any closed trip has zero displacement, by the definition alone.

Step 2 — Average velocity follows immediately

Step 3 — Average speed needs the actual path length, via $s=r\theta$

Unlike displacement, speed genuinely depends on the shape walked, so here the individual legs must be added — but as scalar lengths, not vectors.

• $OP = R = 1\ \text{km}$ (radius).
• Arc $PQ$: the radius sweeps through $\theta = 90^\circ = \dfrac{\pi}{2}\ \text{rad}$ (from $OP$ to $OQ$, a right angle since $OQ\perp OP$). Using $s=r\theta$ directly in radians:

• $QO = R = 1\ \text{km}$.

Total path length: …