Hint
Choose and retain a positive direction; distinguish signed quantities from magnitudes. Locate the velocity zero within the stated interval.
Worked solution
(a)
Locate the velocity zero within the stated interval.
The equation \(\displaystyle 12-4t=0\) gives \(\displaystyle t=3 s\).
Answer: \(3\) s
(b)
Use the signed velocity rule after the reversal.
Substitute the full end time to obtain \(\displaystyle - \frac{20m}{s}\), directed left.
Answer: \(-20\) m/s
(c)
Add the velocity graph areas with their signs.
Positive area \(18\) minus the below-axis magnitude \(50\) gives \(-32\) m.
Answer: \(-32\) m
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: Locate the velocity zero within the stated interval.
- 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: Do not cancel areas when calculating total distance.
(b) · 2 marks
- 1 mark: Use the signed velocity rule after the reversal.
- 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: Locate the zero crossing before dividing the motion into directional stages.
(c) · 2 marks
- 1 mark: Add the velocity graph areas with their signs.
- 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: Do not cancel areas when calculating total distance.
Total: 6 marks.