Mathematics 9709/42 — October/November 2019

Walkthrough

The displacement is given as a function of time, so the velocity is the rate of change of displacement. Differentiate $s = t^3 - 6t^2 + 4t$ term by term:

$$v = \frac{ds}{dt} = 3t^2 - 12t + 4.$$

The acceleration is the rate of change of velocity, so differentiate $v$:

$$a = \frac{dv}{dt} = 6t - 12.$$

The question asks for the velocity at the instant when the acceleration is zero, so set $a = 0$:

$$6t - 12 = 0.$$

Solving gives $t = 2$. Finally, substitute $t = 2$ into the velocity expression:

$$v = 3(2)^2 - 12(2) + 4 = -8.$$

Therefore the velocity at that instant is $-8\text{ ms}^{-1}$.

Key Takeaways

• Velocity is the derivative of displacement with respect to time.
• Acceleration is the derivative of velocity with respect to time.
• When a condition such as "acceleration is zero" is given, first solve for the time, then substitute that time into the required expression.

Common Mistakes

• Differentiating only once and using the displacement expression when finding the acceleration.
• Setting $v = 0$ instead of $a = 0$.
• Making a sign error in the final arithmetic: $12 - 24 + 4 = -8$, not $8$.
• Omitting the differentiation steps; the mark scheme requires method marks for differentiating $s$ and $v$.

Things to Be Careful About

• The answer must include the correct units: $\text{ms}^{-1}$.
• The time found is $t = 2$, which is a valid positive time in this context.
• Show the substitution into $v$ clearly so that the final mark can be awarded.