Motion With Variable Acceleration

When position is an explicit function of time, instantaneous velocity is the first time derivative of position and instantaneous acceleration is the second time derivative, since these derivatives capture the instantaneous rates of change rather than averages. Differentiating $x(t)=3t^3-2t^2+5$ once gives the velocity $v(t)=\frac{dx}{dt}=9t^2-4t$. Differentiating again gives the acceleration $a(t)=\frac{dv}{dt}=18t-4$. Evaluating at $t=2\text{ s}$ yields $a=18(2)-4=36-4=32{\text{ m/s}}^2$. The option 28 m/s^2 is wrong because it drops the constant term incorrectly or uses $18t-8$. The option 36 m/s^2 forgets to subtract the 4 from differentiating the quadratic term. The option 20 m/s^2 mistakenly evaluates the velocity expression's derivative coefficients wrongly. This applies the NCERT calculus definition of acceleration for non-uniform motion. It is essential to differentiate twice rather than plug into $v=u+at$, because that equation presumes a constant acceleration, an assumption that fails the moment acceleration depends on time as it does here. As a check, the acceleration is itself time-dependent and increases linearly with $t$, confirming that this motion is genuinely non-uniformly accelerated and lies outside the scope of the standard constant-acceleration kinematic equations.