Hint
Restrict t to the stated physical interval. Differentiate the position and set \(\displaystyle t=0\).
Worked solution
(a)
Differentiate position once for velocity and twice for acceleration, retaining their units.
Answer: \(v=\left(t - 5\right) \left(2 t - 4\right)\), \(a=4 t - 14\). Velocity is in m/s and acceleration in \(\displaystyle \frac{m}{s^{2}}\).
(b)
The velocity is an upward quadratic with two simple roots. Its sign changes at each root, so both rest events are reversals.
Answer: \(2\) s and \(5\) s; velocity is negative between these times and positive outside them for \(\displaystyle t\ge 0\).
(c)
Solve \(\displaystyle v(t)=0\).
\(2\)
Answer: \(2\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternatives and consequential reasoning; an unexplained final answer does not earn a method mark.
(a) · 2 marks
- 1 mark: Differentiate position once for signed velocity and again for acceleration, with units.
- 1 mark: Give all results requested in this part, with the correct signs, exact values, units, branch and boundary inclusion where applicable.
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 mean rest.
(b) · 2 marks
- 1 mark: Solve for all physical rest times and use velocity signs to describe each intervening direction.
- 1 mark: Give all results requested in this part, with the correct signs, exact values, units, branch and boundary inclusion where applicable.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Do not discard negative velocity.
(c) · 2 marks
- 1 mark: Solve \(\displaystyle v(t)=0\).
- 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 mean rest.
Total: 6 marks.