Net change and accumulation from a rate
Problem 4.673 · easy
A particle starts at position 1200 m with velocity \( \displaystyle v(t) = 2 t^{2} + 9 \) m/s. Where is it at \( \displaystyle t = 2 \)?
- The net change over [0, T] is the integral of the rate (the Fundamental Theorem of Calculus).
- \[ \int\limits_{0}^{2} \left(2 t^{2} + 9\right)\, dt = \frac{70}{3} \]The change over the interval.✓ Proved
- \[ \frac{3670}{3} \]Add the starting amount.✓ Proved
Answer \( \frac{3670}{3} \approx 1223\ \text{m} \)
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: fail (style) — [domain objection, downgraded to style] The solution incorrectly sets the lower bound of the integral to 0, implying the particle starts at t=0. However, the problem states the particle starts at position 1200 m, but does not specify that this occurs at t=0. More critically, even if we assume t=0 is the start time, the integral calculates the displacement from t=0 to t=2. The final position is Initial Position + Displacement. The solution adds 3670/3 (which is 1223.33) to... wait, 1200 + 70/3 = 3600/3 + 70/3 = 3670/3. The arithmetic is correct. The error is in the setup description: it says 'net change over [0, T]' but doesn't explicitly state that the starting time is t=0. If the particle starts at position 1200 at some time t_0 != 0, the answer is wrong. Standard convention implies t=0 is the start time unless specified otherwise. However, the sentence 'The net change over [0, T]' is a bit ambiguous if T is not defined as 2 in the text, though the equation uses 2. The main issue is that the sentence claims the integral is the net change over [0, T], but it doesn't define T. It should say 'over [0, 2]'. Also, it fails to explicitly state that the starting time is t=0. Without this assumption, the problem is unsolvable. Given standard calculus problem conventions, t=0 is the start. The calculation is correct. Is there a subtle error? 'Where is it at t=2?' implies we need position at t=2. If start is at t=0, pos(2) = pos(0) + int_0^2 v(t) dt. 1200 + 70/3 = 3670/3. The solution is mathematically correct under the standard assumption. Why would it fail? Let's look closer. 'The net change over [0, T] is the integral...'. T is not defined. Then it integrates from 0 to 2. It should define T=2. This is a style/minor error. But wait, is there a bigger error? No. The answer is correct. The notes are slightly sloppy. Let's look for 'misleading'. Does it teach something false? No. Is it 'style'? Yes, T is undefined. But usually 'pass' is for sound solutions. Is it sound? Yes, under standard conventions. Let's check the severity. If I mark it pass, I accept undefined variables. If I mark it fail, I demand rigor. The prompt asks to find what is wrong. The variable T is used in the sentence but not defined, then 2 is used in the equation. This is a disconnect. Also, it doesn't state the assumption that t=0 is the start time. This is a significant logical gap in the 'sentence' part. A student might think the integral from 0 to 2 is always the displacement from the start, regardless of when the start is. This is misleading. The displacement is from t_start to t_end. If t_start is not 0, the integral from 0 is wrong. The solution assumes t_start=0 without stating it. This is a 'misleading' or 'error' in the setup logic. I will call it 'error' because the setup is incomplete/incorrectly justified.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (style) 2026-10-08 — [domain objection, downgraded to style] The solution incorrectly sets the lower bound of the integral to 0, implying the particle starts at t=0. However, the problem states the particle starts at position 1200 m, but does not specify that this occurs at t=0. More critically, even if we assume t=0 is the start time, the integral calculates the displacement from t=0 to t=2. The final position is Initial Position + Displacement. The solution adds 3670/3 (which is 1223.33) to... wait, 1200 + 70/3 = 3600/3 + 70/3 = 3670/3. The arithmetic is correct. The error is in the setup description: it says 'net change over [0, T]' but doesn't explicitly state that the starting time is t=0. If the particle starts at position 1200 at some time t_0 != 0, the answer is wrong. Standard convention implies t=0 is the start time unless specified otherwise. However, the sentence 'The net change over [0, T]' is a bit ambiguous if T is not defined as 2 in the text, though the equation uses 2. The main issue is that the sentence claims the integral is the net change over [0, T], but it doesn't define T. It should say 'over [0, 2]'. Also, it fails to explicitly state that the starting time is t=0. Without this assumption, the problem is unsolvable. Given standard calculus problem conventions, t=0 is the start. The calculation is correct. Is there a subtle error? 'Where is it at t=2?' implies we need position at t=2. If start is at t=0, pos(2) = pos(0) + int_0^2 v(t) dt. 1200 + 70/3 = 3670/3. The solution is mathematically correct under the standard assumption. Why would it fail? Let's look closer. 'The net change over [0, T] is the integral...'. T is not defined. Then it integrates from 0 to 2. It should define T=2. This is a style/minor error. But wait, is there a bigger error? No. The answer is correct. The notes are slightly sloppy. Let's look for 'misleading'. Does it teach something false? No. Is it 'style'? Yes, T is undefined. But usually 'pass' is for sound solutions. Is it sound? Yes, under standard conventions. Let's check the severity. If I mark it pass, I accept undefined variables. If I mark it fail, I demand rigor. The prompt asks to find what is wrong. The variable T is used in the sentence but not defined, then 2 is used in the equation. This is a disconnect. Also, it doesn't state the assumption that t=0 is the start time. This is a significant logical gap in the 'sentence' part. A student might think the integral from 0 to 2 is always the displacement from the start, regardless of when the start is. This is misleading. The displacement is from t_start to t_end. If t_start is not 0, the integral from 0 is wrong. The solution assumes t_start=0 without stating it. This is a 'misleading' or 'error' in the setup logic. I will call it 'error' because the setup is incomplete/incorrectly justified.gpt-oss:20b: pass 2026-10-08gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: inconclusive 2026-10-08 — reviewer returned a non-object
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 2026-10-08 with SymPy 1.14.0.