The law of motion of a point is s = 3t² + 2t + 1, where s

The law of motion of a point is $s = 3t^{2} + 2t + 1$, where $s$ is the distance in metres and $t$ the time in seconds. Find the velocity of the point at $t = 2$.

Step 1: Differentiate to obtain the velocity function

The instantaneous velocity is the derivative of the distance with respect to time, $v(t) = s'(t)$. Applying the power rule term by term: \[v(t) = \frac{ds}{dt} = \frac{d}{dt}\bigl(3t^{2}\bigr) + \frac{d}{dt}(2t) + \frac{d}{dt}(1) = 6t + 2 + 0.\]

Step 2: Evaluate at $t = 2$

\[ v(2) = 6\cdot 2 + 2 = 12 + 2 = 14. \]

Step 3: Verification from the definition

The average velocity over the interval $[2,\,2+\Delta t]$ is

\[\frac{\Delta s}{\Delta t} = \frac{3(2+\Delta t)^{2}+2(2+\Delta t)+1 - \bigl(12+4+1\bigr)}{\Delta t} = \frac{14\,\Delta t + 3(\Delta t)^{2}}{\Delta t} = 14 + 3\,\Delta t\]

\[\lim_{\Delta t \to 0}\bigl(14 + 3\,\Delta t\bigr) = 14\ \text{m/s} \;\;\checkmark\]