Velocity, acceleration and speed in space
Problem 9.397 · medium
A particle has position \( \displaystyle \mathbf r(t) = \left\langle t^{2}, t^{3}, t \right\rangle \). Find its velocity, acceleration and speed at \( \displaystyle t = 0 \).
- \[ \left[\begin{matrix}\frac{d}{d t} t^{2}\\\frac{d}{d t} t^{3}\\\frac{d}{d t} t\end{matrix}\right] = \left[\begin{matrix}2 t\\3 t^{2}\\1\end{matrix}\right] \]v = r′.✓ Proved
- \[ \left[\begin{matrix}\frac{d}{d t} 2 t\\\frac{d}{d t} 3 t^{2}\\\frac{d}{d t} 1\end{matrix}\right] = \left[\begin{matrix}2\\6 t\\0\end{matrix}\right] \]a = v′.✓ Proved
- \[ 1 \]Speed = ‖v(0)‖.✓ Proved
Answer \( \mathbf v = \left\langle 0, 0, 1 \right\rangle,\ \mathbf a = \left\langle 2, 0, 0 \right\rangle,\ \text{speed} = 1 \)
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: pass — The solution correctly computes the velocity and acceleration vectors by differentiation and evaluates them at t=0. The speed is correctly calculated as the magnitude of the velocity vector at t=0.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-08 — The solution correctly computes the velocity and acceleration vectors by differentiation and evaluates them at t=0. The speed is correctly calculated as the magnitude of the velocity vector at t=0.gpt-oss:20b: pass 2026-10-08gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution fails to explicitly evaluate the velocity and acceleration vectors at t=0 before stating the final answer. While the final numerical values are correct, the derivation skips the crucial step of substituting t=0 into the symbolic expressions for v(t) and a(t), making the logic incomplete.
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 2026-10-08 with SymPy 1.14.0.