Velocity, acceleration and speed in space
Problem 9.240 · 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 \( \displaystyle t = 0 \).
- \[ \left[\begin{matrix}\frac{d}{d t} t\\\frac{d}{d t} \left(t^{2} - 1\right)\\\frac{d}{d t} 2 t\end{matrix}\right] = \left[\begin{matrix}1\\2 t\\2\end{matrix}\right] \]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
- \[ \sqrt{5} \]Speed = ‖v(0)‖.✓ Proved
Answer \( \mathbf v = \left\langle 1, 0, 2 \right\rangle,\ \mathbf a = \left\langle 0, 2, 0 \right\rangle,\ \text{speed} = \sqrt{5} \)
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: passqwen3.6:27b-mlx: fail (error) — The solution fails to explicitly evaluate the velocity vector at t=0 before computing its magnitude, making the final step logically disconnected from the previous symbolic results. While the final answer is correct, the derivation skips the crucial step of substituting t=0 into v(t) = <1, 2t, 2> to get <1, 0, 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=0 before computing its magnitude, making the final step logically disconnected from the previous symbolic results. While the final answer is correct, the derivation skips the crucial step of substituting t=0 into v(t) = <1, 2t, 2> to get <1, 0, 2>.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly computes the velocity and acceleration vectors by differentiating the position vector, and correctly calculates the speed as the magnitude of the velocity vector at t=0.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.