Velocity, acceleration and speed in space
Problem 9.393 · medium
A particle has position \( \displaystyle \mathbf r(t) = \left\langle 3 t, 4 t^{2}, 2 t^{3} \right\rangle \). Find its velocity, acceleration and speed at \( \displaystyle t = 0 \).
- \[ \left[\begin{matrix}\frac{d}{d t} 3 t\\\frac{d}{d t} 4 t^{2}\\\frac{d}{d t} 2 t^{3}\end{matrix}\right] = \left[\begin{matrix}3\\8 t\\6 t^{2}\end{matrix}\right] \]v = r′.✓ Proved
- \[ \left[\begin{matrix}\frac{d}{d t} 3\\\frac{d}{d t} 8 t\\\frac{d}{d t} 6 t^{2}\end{matrix}\right] = \left[\begin{matrix}0\\8\\12 t\end{matrix}\right] \]a = v′.✓ Proved
- \[ 3 \]Speed = ‖v(0)‖.✓ Proved
Answer \( \mathbf v = \left\langle 3, 0, 0 \right\rangle,\ \mathbf a = \left\langle 0, 8, 0 \right\rangle,\ \text{speed} = 3 \)
✓ Nihil obstat 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. Reviewers found nothing wrong with the explanation.
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 identified 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 identified 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: 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.
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.