∫Calc Practice

Particle motion: position, velocity, acceleration

Problem 2.1716 · hard

A particle moves along a line with position \( \displaystyle s(t) = 2 t^{3} - 30 t^{2} + 126 t + 10 \) (meters, \( \displaystyle t \) in seconds). Find its velocity and acceleration, the times it is at rest, and the total distance it travels for \( \displaystyle 0 \le t \le 9 \).
  1. \[ \frac{d}{d t} \left(2 t^{3} - 30 t^{2} + 126 t + 10\right) = 6 t^{2} - 60 t + 126 \]
    Velocity is the derivative of position.✓ Proved
  2. \[ \frac{d}{d t} \left(6 t^{2} - 60 t + 126\right) = 12 t - 60 \]
    Acceleration is the derivative of velocity.✓ Proved
  3. \[ 6 t^{2} - 60 t + 126 = \left(t - 3\right) \left(6 t - 42\right) \]
    v = 0 at t = 3 and t = 7: the particle is at rest.✓ Proved
  4. 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].
  5. \[ 64 + 64 + 162 = 290 \]
    |s(3) − s(0)| + |s(7) − s(3)| + |s(9) − s(7)|, with s = 10, 172, 108, 172.✓ Proved
  6. For comparison, the displacement is s(9) − s(0) = 162 m.
Answer \( v(t) = 6 t^{2} - 60 t + 126,\ a(t) = 12 t - 60,\ \text{at rest at } t = 3, 7,\ \text{distance} = 290\text{ m} \)

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
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4Not checked—a sentence; read, not computed
5✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
6Not checked—a sentence; read, not computed
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0integral of |v(t)| by numerical quadrature

Reviewers

  • gpt-oss:20b: fail (error) — The last leg of the distance calculation is incorrect: |s(9)-s(7)| = |162-108| = 54, not 162. The correct total distance is 162+64+54 = 280 m, not 290 m.
  • qwen3.6:27b-mlx: pass — The solution correctly identifies velocity and acceleration, finds the critical points where velocity is zero, and accurately calculates the total distance by summing the absolute changes in position over the intervals defined by these points.
Every verdict on record (4)
  • gpt-oss:20b: fail (error) 2026-10-06 — The last leg of the distance calculation is incorrect: |s(9)-s(7)| = |162-108| = 54, not 162. The correct total distance is 162+64+54 = 280 m, not 290 m.
  • qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly identifies velocity and acceleration, finds the critical points where velocity is zero, and accurately calculates the total distance by summing the absolute changes in position over the intervals defined by these points.
  • gpt-oss:20b: inconclusive 2026-10-06 — reviewer returned a non-object
  • qwen3.6:27b-mlx: pass 2026-10-06 — 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 direction of motion does not change.

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 2026-10-06 with SymPy 1.14.0.