Particle motion: position, velocity, acceleration
Problem 2.1618 · hard
A particle moves along a line with position \( \displaystyle s(t) = t^{3} - 15 t^{2} + 63 t + 9 \) (meters, \( \displaystyle t \) in seconds). Find its velocity and acceleration, the times it is at rest, and the total distance it travels for \( \displaystyle 0 \le t \le 9 \).
- \[ \frac{d}{d t} \left(t^{3} - 15 t^{2} + 63 t + 9\right) = 3 t^{2} - 30 t + 63 \]Velocity is the derivative of position.✓ Proved
- \[ \frac{d}{d t} \left(3 t^{2} - 30 t + 63\right) = 6 t - 30 \]Acceleration is the derivative of velocity.✓ Proved
- \[ 3 t^{2} - 30 t + 63 = \left(t - 3\right) \left(3 t - 21\right) \]v = 0 at t = 3 and t = 7: the particle is at rest.✓ Proved
- v changes sign at each of those times, so the particle turns around there. Distance adds the length of each leg: [0, 3], [3, 7], [7, 9].Reviewed
- \[ 32 + 32 + 81 = 145 \]|s(3) − s(0)| + |s(7) − s(3)| + |s(9) − s(7)|, with s = 9, 90, 58, 90.✓ Proved
- For comparison, the displacement is s(9) − s(0) = 81 m.Reviewed
Answer \( v(t) = 3 t^{2} - 30 t + 63,\ a(t) = 6 t - 30,\ \text{at rest at } t = 3, 7,\ \text{distance} = 145\text{ m} \)
Lines: 4 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 6 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | integral of |v(t)| by numerical quadrature |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly identifies velocity, acceleration, and rest times. The distance calculation is accurate, properly accounting for direction changes at t=3 and t=7.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly identifies velocity, acceleration, and rest times. The distance calculation is accurate, properly accounting for direction changes at t=3 and t=7.gpt-oss:20b: pass 2026-10-05qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly identifies velocity and acceleration, finds the correct times at rest, and accurately calculates the total distance by summing the absolute displacements over the intervals where the direction of motion does not change.gpt-oss:20b: fail (error) 2026-10-05 — The last distance segment is mis‑calculated: |s(9)−s(7)| = |81−58| = 23 m, not 81 m. The correct total distance is 81 + 32 + 23 = 136 m.
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/particle_motion, checked 2026-10-05 with SymPy 1.14.0.