Particle motion: position, velocity, acceleration
Problem 2.1612 · hard
A particle moves along a line with position \( \displaystyle s(t) = 2 t^{3} - 18 t^{2} + 30 t + 4 \) (meters, \( \displaystyle t \) in seconds). Find its velocity and acceleration, the times it is at rest, and the total distance it travels for \( \displaystyle 0 \le t \le 7 \).
- \[ \frac{d}{d t} \left(2 t^{3} - 18 t^{2} + 30 t + 4\right) = 6 t^{2} - 36 t + 30 \]Velocity is the derivative of position.✓ Proved
- \[ \frac{d}{d t} \left(6 t^{2} - 36 t + 30\right) = 12 t - 36 \]Acceleration is the derivative of velocity.✓ Proved
- \[ 6 t^{2} - 36 t + 30 = \left(t - 1\right) \left(6 t - 30\right) \]v = 0 at t = 1 and t = 5: 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, 1], [1, 5], [5, 7].Reviewed
- \[ 14 + 64 + 64 = 142 \]|s(1) − s(0)| + |s(5) − s(1)| + |s(7) − s(5)|, with s = 4, 18, -46, 18.✓ Proved
- For comparison, the displacement is s(7) − s(0) = 14 m.Reviewed
Answer \( v(t) = 6 t^{2} - 36 t + 30,\ a(t) = 12 t - 36,\ \text{at rest at } t = 1, 5,\ \text{distance} = 142\text{ m} \)
✓ Nihil obstat Lines: 4 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
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 and acceleration, finds the critical points, verifies the sign changes to justify splitting the integral for total distance, and computes the correct numerical result.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly identifies velocity and acceleration, finds the critical points, verifies the sign changes to justify splitting the integral for total distance, and computes the correct numerical result.gpt-oss:20b: pass 2026-10-05qwen3.6:27b-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=1 and t=5 by summing the absolute displacements of each segment.gpt-oss:20b: pass 2026-10-05
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.