Hint
Choose and retain a positive direction; distinguish signed quantities from magnitudes. Subtract the leftward distance from the rightward distance.
Worked solution
(a)
Subtract the leftward distance from the rightward distance.
Add the signed changes 15 and -9 to obtain 6 m.
Answer: \(6\) m
(b)
Add the lengths of both stages without a direction sign.
Both path lengths contribute positively: 24 m.
Answer: \(24\) m
(c)
Divide the signed displacement by the complete journey time.
The total displacement divided by 9 s is \(\displaystyle \frac{2}{3}\) m/s.
Answer: \(\displaystyle \frac{2}{3}\) 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: Subtract the leftward distance from the rightward distance.
- 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: Adding signed displacements does not calculate total distance.
(b) · 2 marks
- 1 mark: Add the lengths of both stages without a direction sign.
- 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: Divide by the complete elapsed time, including both stages.
(c) · 2 marks
- 1 mark: Divide the signed displacement by the complete journey 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: Adding signed displacements does not calculate total distance.
Total: 6 marks.