Hint
Choose and retain a positive direction; distinguish signed quantities from magnitudes. Differentiate the given velocity rule and use the requested time.
Worked solution
(a)
Differentiate the given velocity rule and use the requested time.
The derivative \(6 t - 24\) evaluates to \(\displaystyle \frac{18m}{s^{2}}\).
Answer: \(18\) \(\displaystyle \frac{m}{s^{2}}\)
(b)
Set the derivative of velocity equal to zero.
Solving \(6 t - 24 = 0\) gives \(\displaystyle t=4 s\).
Answer: \(4\) s
(c)
Check that velocity stays positive and then minimise its square-term form.
Velocity is \(3 \left(t - 4\right)^{2} + 8\), always positive; its square term vanishes at \(\displaystyle t=4\), leaving \(\displaystyle \frac{8m}{s}\).
Answer: \(8\) m/s
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternative methods and consequential reasoning. An unexplained result does not earn a method mark.
(a) · 2 marks
- 1 mark: Differentiate the given velocity rule and use the requested time.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: A zero acceleration does not imply zero velocity.
(b) · 2 marks
- 1 mark: Set the derivative of velocity equal to zero.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Instantaneous acceleration and interval-average acceleration need not agree.
(c) · 2 marks
- 1 mark: Check that velocity stays positive and then minimise its square-term form.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: A zero acceleration does not imply zero velocity.
Total: 6 marks.