Hint
Choose and retain a positive direction; distinguish signed quantities from magnitudes. Solve \(\displaystyle v=0\) and verify the sign changes within the permitted interval.
Worked solution
(a)
Solve \(\displaystyle v=0\) and verify the sign changes within the permitted interval.
The velocity is zero at \(\displaystyle t=4\) and changes from negative to positive.
Answer: \(4\) s
(b)
Evaluate the first signed definite integral.
Integrating from 0 to 4 gives \(-8\) m.
Answer: \(-8\) m
(c)
Integrate over the second sign-consistent interval and use a nonnegative path length.
Velocity is positive on the second half; its triangular area is \(8\) m.
Answer: \(8\) 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: Solve \(\displaystyle v=0\) and verify the sign changes within the permitted 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 take only the absolute value of the final net displacement.
(b) · 2 marks
- 1 mark: Evaluate the first signed definite integral.
- 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: Find and test all velocity zeros within the time domain.
(c) · 2 marks
- 1 mark: Integrate over the second sign-consistent interval and use a nonnegative path length.
- 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 take only the absolute value of the final net displacement.
Total: 6 marks.