Home›Calculus 3›Velocity, acceleration and speed in space›Problem 9.234

Velocity, acceleration and speed in space

Problem 9.234 · medium

A particle has position \( \displaystyle \mathbf r(t) = \left\langle t, t^{2} - 1, 2 t \right\rangle \). Find its velocity, acceleration and speed at \( \displaystyle t = 1 \).

\[ \left[\begin{matrix}\frac{d}{d t} t\\\frac{d}{d t} \left(t^{2} - 1\right)\\\frac{d}{d t} 2 t\end{matrix}\right] = \left[\begin{matrix}1\\2 t\\2\end{matrix}\right] \]
v = r′.✓ Proved
\[ \left[\begin{matrix}\frac{d}{d t} 1\\\frac{d}{d t} 2 t\\\frac{d}{d t} 2\end{matrix}\right] = \left[\begin{matrix}0\\2\\0\end{matrix}\right] \]
a = v′.✓ Proved
\[ 3 \]
Speed = ‖v(1)‖.✓ Proved

Answer \( \mathbf v = \left\langle 1, 2, 2 \right\rangle,\ \mathbf a = \left\langle 0, 2, 0 \right\rangle,\ \text{speed} = 3 \)

Every line of this solution was proved by the computer algebra system SymPy. 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
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	difference quotients of the position

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The solution fails to explicitly evaluate the velocity vector at t=1 before calculating the speed, making the final equation '3 = 3' logically disconnected from the previous steps. It skips the crucial step of substituting t=1 into v(t) = <1, 2t, 2> to get v(1) = <1, 2, 2>.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution fails to explicitly evaluate the velocity vector at t=1 before calculating the speed, making the final equation '3 = 3' logically disconnected from the previous steps. It skips the crucial step of substituting t=1 into v(t) = <1, 2t, 2> to get v(1) = <1, 2, 2>.
gpt-oss:20b: pass 2026-10-04
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution fails to evaluate the velocity vector at t=1 before computing the speed, leaving the speed calculation as a tautology (3=3) rather than a derived result. Additionally, the acceleration is constant, so evaluating at t=1 is trivial but the lack of explicit substitution for velocity makes the logic incomplete.
gpt-oss:20b: pass 2026-10-04

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/motion_in_space, checked 2026-10-04 with SymPy 1.14.0.