"The magnitude of average velocity is equal to average speed". This statement is not always correct and not always incorrect. Explain with an example.

(N/A) Consider a car moving along a straight line as shown in the figure. Let the car move from $O$ to $P$ in $18 \, s$ and then return from $P$ to $Q$ in $6 \, s$.
Case $1$: Motion from $O$ to $P$
Total distance = $OP = 360 \, m$
Total displacement = $360 \, m - 0 \, m = 360 \, m$
Time taken = $18 \, s$
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{360}{18} = 20 \, ms^{-1}$
Average velocity = $\frac{\text{Total displacement}}{\text{Total time}} = \frac{360}{18} = 20 \, ms^{-1}$
Here, the magnitude of average velocity is equal to the average speed.
Case $2$: Motion from $O$ to $P$ and then to $Q$
Total distance = $OP + PQ = 360 + (360 - 240) = 360 + 120 = 480 \, m$
Total displacement = $OQ = 240 \, m - 0 \, m = 240 \, m$
Total time = $18 \, s + 6 \, s = 24 \, s$
Average speed = $\frac{480}{24} = 20 \, ms^{-1}$
Average velocity = $\frac{240}{24} = 10 \, ms^{-1}$
Here, the magnitude of average velocity $(10 \, ms^{-1})$ is not equal to the average speed $(20 \, ms^{-1})$.
Thus, the statement is not always correct and not always incorrect.