Net change and accumulation from a rate
Problem 4.348 · easy
A particle starts at position 20 m with velocity \( \displaystyle v(t) = 3 t + 8 \) m/s. Where is it at \( \displaystyle t = 5 \)?
- The net change over [0, T] is the integral of the rate (the Fundamental Theorem of Calculus).
- \[ \int\limits_{0}^{5} \left(3 t + 8\right)\, dt = \frac{155}{2} \]The change over the interval.✓ Proved
- \[ \frac{195}{2} \]Add the starting amount.✓ Proved
Answer \( \frac{195}{2} \approx 97.5\ \text{m} \)
Lines: 2 proved, 1 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 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | numerical quadrature of the rate |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: inconclusive — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly integrates from t=0 to t=5, ignoring that the particle starts at t=0 with position 20m but the problem implies the motion s
Every verdict on record (4)
qwen3.6:27b-mlx: inconclusive 2026-10-04 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly integrates from t=0 to t=5, ignoring that the particle starts at t=0 with position 20m but the problem implies the motion sgpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: inconclusive 2026-10-04 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly integrates from t=0 to t=5, but the particle starts at t=0 with position 20m, implying the initial condition is at t=0. Howgpt-oss:20b: fail (misleading) 2026-10-04 — The solution skips the step of adding the initial position 20 m to the displacement 155/2 m. It simply states 195/2 = 195/2, which does not justify how the final position 195/2 m was obtained. This could mislead a student into thinking the integral alone gives the final position.
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/net_change, checked 2026-10-04 with SymPy 1.14.0.