Net change and accumulation from a rate
Problem 4.969 · easy
A particle starts at position 500 m with velocity \( \displaystyle v(t) = 3 t + 10 \) m/s. Where is it at \( \displaystyle t = 6 \)?
- The net change over [0, T] is the integral of the rate (the Fundamental Theorem of Calculus).Reviewed
- \[ \int\limits_{0}^{6} \left(3 t + 10\right)\, dt = 114 \]The change over the interval.✓ Proved
- \[ 614 \]Add the starting amount.✓ Proved
Answer \( 614 \approx 614\ \text{m} \)
Lines: 2 proved, 1 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 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | 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: pass — The solution correctly applies the Fundamental Theorem of Calculus to find the displacement and adds it to the initial position. The logic is sound and the calculations are correct.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: pass 2026-10-12 — The solution correctly applies the Fundamental Theorem of Calculus to find the displacement and adds it to the initial position. The logic is sound and the calculations are correct.qwen3.6:27b-mlx: fail (error) 2026-10-12 — The integral limits are incorrect. The problem states the particle starts at position 500 m, but does not specify that this occurs at t=0. However, standard convention implies t=0 is the start unless specified otherwise. Wait, let's re-read carefully. 'A particle starts at position 500 m... Where is it at t=6?'. This implies the motion starts at t=0 with position 500. So the integral from 0 to 6 is correct for the displacement. Let's check the math. Integral of 3t+10 from 0 to 6 is [1.5t^2 + 10t] from 0 to 6 = 1.5(36) + 60 = 54 + 60 = 114. Final position = 500 + 114 = 614. The solution is actually correct. Why would it fail? Let me look closer at the sentence. 'The net change over [0, T] is the integral of the rate'. This is correct. The equation checks out. The final answer is correct. Is there a subtle error? Maybe the prompt implies the start time isn't 0? 'starts at position 500 m' usually implies t=0. If the start time was different, it would be stated. Let's look at the 'Stated answer'. 614. The solution derives 614. It seems correct. Wait, is there a trap? 'v(t) = 3t + 10'. If t=0, v=10. If the particle 'starts' at 500m, does it mean at t=0? Yes. Is the integral calculation correct? 114. 500+114=614. Everything seems fine. Let me check the JSON format requirements again. 'Reply with JSON only'. Okay. Is there any ambiguity? 'Where is it at t=6?'. If the start was at t=1, the answer would be different. But 'starts' implies the beginning of the observation, typically t=0. Let's assume standard physics problem conventions. The solution is sound. Wait, looking at the provided solution text again. Line 2 says 'Integral(3*t + 10, (t, 0, 6)) = 114'. This is correct. Line 3 says '614 = 614'. This is correct. The sentence in line 1 is correct. I will mark it as pass.gpt-oss:20b: pass 2026-10-12
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 2026-10-12 with SymPy 1.14.0.