Particle motion: position, velocity, acceleration
Problem 2.1614 · hard
A particle moves along a line with position \( \displaystyle s(t) = t^{3} - 9 t^{2} + 15 t \) (meters, \( \displaystyle t \) in seconds). Find its velocity and acceleration, the times it is at rest, and the total distance it travels for \( \displaystyle 0 \le t \le 6 \).
- \[ \frac{d}{d t} \left(t^{3} - 9 t^{2} + 15 t\right) = 3 t^{2} - 18 t + 15 \]Velocity is the derivative of position.✓ Proved
- \[ \frac{d}{d t} \left(3 t^{2} - 18 t + 15\right) = 6 t - 18 \]Acceleration is the derivative of velocity.✓ Proved
- \[ 3 t^{2} - 18 t + 15 = \left(t - 1\right) \left(3 t - 15\right) \]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, 6].
- \[ 7 + 7 + 32 = 46 \]|s(1) − s(0)| + |s(5) − s(1)| + |s(6) − s(5)|, with s = 0, 7, -25, -18.✓ Proved
- For comparison, the displacement is s(6) − s(0) = -18 m.
Answer \( v(t) = 3 t^{2} - 18 t + 15,\ a(t) = 6 t - 18,\ \text{at rest at } t = 1, 5,\ \text{distance} = 46\text{ m} \)
Lines: 4 proved, 2 not checked. 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 | Not checked | — | a sentence; read, not computed |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 6 | Not checked | — | 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: fail (misleading) — The solution incorrectly claims the particle turns around at t=1. The velocity remains negative from t=0 to t=5, so there is no sign change at t=1; only at t=5 does the particle reverse direction.qwen3.6:27b-mlx: pass
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-05gpt-oss:20b: fail (misleading) 2026-10-05 — The solution incorrectly claims the particle turns around at t=1. The velocity remains negative from t=0 to t=5, so there is no sign change at t=1; only at t=5 does the particle reverse direction.qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly derives velocity and acceleration, identifies the times at rest, and accurately calculates the total distance by summing the absolute displacements over the intervals where the velocity does not change sign.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.