Hint
Both models use the same time origin and units. Equate both position components using the same t.
Worked solution
(a)
Equating coordinates gives the same nonnegative time in both equations; substituting gives the common position.
Answer: \(3\); \(\left[\begin{matrix}16\\12\end{matrix}\right]\)
(b)
Substitute the common time into either position model.
\(16\)
Answer: \(16\)
(c)
Verify the second component as well.
\(12\)
Answer: \(12\)
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) · 3 marks
- 1 mark: Equate both coordinate positions at the same nonnegative time.
- 1 mark: Verify that time in both coordinates and evaluate the common position.
- 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: Equal path coordinates reached at different times are not a collision.
(b) · 2 marks
- 1 mark: Substitute the common time into either position model.
- 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: Velocity is a vector; speed is its magnitude.
(c) · 2 marks
- 1 mark: Verify the second component as well.
- 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: Equal path coordinates reached at different times are not a collision.
Total: 7 marks.