Hint
Split a distance calculation at every direction change. Set the velocity factors equal to zero.
Worked solution
(a)
Integrate velocity and choose C to satisfy the stated initial position.
Answer: \(s(t)=t^{3} - 15 t^{2} + 63 t + 11\)
(b)
The rest times are \(3\) and \(7\). Signed integration gives displacement. Integrate with signs +,−,+ over the three successive intervals to obtain distance.
Answer: \(130\) m displacement; \(194\) m total distance.
(c)
Integrate signed velocity from 0 to T.
\(130\)
Answer: \(130\)
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: Integrate velocity and use the given initial position to determine the constant.
- 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: Total distance is not the absolute net displacement when direction changes.
(b) · 3 marks
- 1 mark: Integrate signed velocity for displacement.
- 1 mark: Split at both reversals and integrate its absolute value for distance.
- 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: An area below the time axis contributes negative displacement.
(c) · 2 marks
- 1 mark: Integrate signed velocity from 0 to T.
- 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: Total distance is not the absolute net displacement when direction changes.
Total: 7 marks.