A toy train on a straight track has the position x = 2t³, with x in metres and t in seconds…

Step 1. Average velocity. $x(1\ \text{s}) = 2\ \text{m}$ and $x(3\ \text{s}) = 54\ \text{m}$, so $v_{\text{av}} = \dfrac{54\ \text{m} - 2\ \text{m}}{2\ \text{s}} = 26\ \text{m/s}$.

Step 2. Velocity at 2 s. $v = \dfrac{dx}{dt} = 6t^2 = 24\ \text{m/s}$.

Step 3. When is v equal to 26 m/s? $6\ \text{m/s}^3 \times t^2 = 26\ \text{m/s}$, so $t = \sqrt{4.33\ \text{s}^2} = 2.08\ \text{s}$.