AP · AP C:M · extended response
Calculus extension: a reversal hidden inside the interval
AP Physics C: Mechanics · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 4
- Answer
- Complete
Selected position samples on a cubic trajectory
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Position / m
Horizontal axis (x): Time / s
Read the exact data as a table
| Time / s | Position / m |
|---|---|
| 0 | 0 |
| 1 | 4 |
| 2 | 2 |
| 3 | 0 |
| 4 | 4 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Choose your AP course · School year 2026–27 · May 2027 exam
Choose a course to check this question’s reviewed parts and required depth. Its original AP label and topic codes do not establish suitability for the current syllabus.
Choose reviewed AP practice →A 2.0 kg cart has one-dimensional position x(t) = t³−6t²+9t metres for 0≤t≤4 s, with t in seconds. The supplied position samples are evidence points on this cubic, not straight-line motion between them. No other translational kinetic-energy stores are included. Supplied data (Time / s; Position / m): (0, 0), (1, 4), (2, 2), (3, 0), (4, 4).
- (a)
Differentiate to find velocity, acceleration and net force. Locate reversals and compute total distance and displacement.
3 marks - (b)
Calculate net impulse from 0 to 4 s and check net work from 0 to 2 s with kinetic energy. Explain why zero impulse need not mean no force.
3 marks
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(a)
Differentiate to find velocity, acceleration and net force. Locate reversals and compute total distance and displacement.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Differentiate x(t): v = dx/dt = 3(t−1)(t−3), and a = dv/dt = 6t−12. Newton's law gives F = ma = 12t−24 N.
- 2
Velocity changes sign at 1 and 3 s. Position values at 0,1,3,4 s are 0,4,0,4 m, so path length is 4+4+4 = 12 m.
- 3
Net displacement is x(4)−x(0) = 4 m. The intermediate t = 2 s sample does not create an additional reversal; smooth cubic behaviour must be used.
(b)
Calculate net impulse from 0 to 4 s and check net work from 0 to 2 s with kinetic energy. Explain why zero impulse need not mean no force.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Integrating force gives J = ∫₀⁴(12t−24)dt = [6t²−24t]₀⁴ = 0 N s, matching m[v(4)−v(0)] = 2(9−9).
- 2
At t = 0, v = 9 m s⁻¹ and K = 81 J. At t = 2, v = −3 m s⁻¹ and K = 9 J, so net work is −72 J.
- 3
A direct work integral is ∫F(t)v(t)dt, not ∫F(t)dt. Positive and negative force impulses can cancel even though the force is nonzero during most of the interval.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- Differentiate x(t): v = dx/dt = 3(t−1)(t−3), and a = dv/dt = 6t−12. Newton's law gives F = ma = 12t−24 N.
- Velocity changes sign at 1 and 3 s. Position values at 0,1,3,4 s are 0,4,0,4 m, so path length is 4+4+4 = 12 m.
- Net displacement is x(4)−x(0) = 4 m. The intermediate t = 2 s sample does not create an additional reversal; smooth cubic behaviour must be used.
- Integrating force gives J = ∫₀⁴(12t−24)dt = [6t²−24t]₀⁴ = 0 N s, matching m[v(4)−v(0)] = 2(9−9).
- At t = 0, v = 9 m s⁻¹ and K = 81 J. At t = 2, v = −3 m s⁻¹ and K = 9 J, so net work is −72 J.
- A direct work integral is ∫F(t)v(t)dt, not ∫F(t)dt. Positive and negative force impulses can cancel even though the force is nonzero during most of the interval.
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