Velocity, acceleration and speed in space
Problem 9.556 · medium
A particle has acceleration \( \displaystyle \mathbf a(t) = \left\langle \cos{\left(t \right)}, - \sin{\left(t \right)}, 1 \right\rangle \), initial velocity \( \displaystyle \left\langle 0, 3, -3 \right\rangle \) and initial position \( \displaystyle \left\langle -1, 1, -1 \right\rangle \). Find \( \displaystyle \mathbf r(t) \).
- \[ \left[\begin{matrix}\frac{d}{d t} \sin{\left(t \right)}\\\frac{d}{d t} \left(\cos{\left(t \right)} + 2\right)\\\frac{d}{d t} \left(t - 3\right)\end{matrix}\right] = \left[\begin{matrix}\cos{\left(t \right)}\\- \sin{\left(t \right)}\\1\end{matrix}\right] \]Integrating a and using v(0) gives v(t) = <sin(t), cos(t) + 2, t - 3>.✓ Proved
- \[ \left[\begin{matrix}\frac{d}{d t} \left(- \cos{\left(t \right)}\right)\\\frac{d}{d t} \left(2 t + \sin{\left(t \right)} + 1\right)\\\frac{d}{d t} \left(\frac{t^{2}}{2} - 3 t - 1\right)\end{matrix}\right] = \left[\begin{matrix}\sin{\left(t \right)}\\\cos{\left(t \right)} + 2\\t - 3\end{matrix}\right] \]Integrate again and use r(0).✓ Proved
- \[ \left[\begin{matrix}-1\\1\\-1\end{matrix}\right] \]r(0) is the given position.✓ Proved
Answer \( \mathbf r(t) = \left\langle - \cos{\left(t \right)}, 2 t + \sin{\left(t \right)} + 1, \frac{t^{2}}{2} - 3 t - 1 \right\rangle \)
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 | the motion integrated numerically from the initial conditions, compared at t = 1.5 |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution contains a sign error in the integration of the y-component of acceleration. The integral of -sin(t) is cos(t), but the solution uses -cos(t) (or implies a sign flip) leading to an incorrect velocity component cos(t)+2 instead of cos(t)+3, and subsequently an incorrect position component. Specifically, v_y(t) should be cos(t) + 3, not cos(t) + 2.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution contains a sign error in the integration of the y-component of acceleration. The integral of -sin(t) is cos(t), but the solution uses -cos(t) (or implies a sign flip) leading to an incorrect velocity component cos(t)+2 instead of cos(t)+3, and subsequently an incorrect position component. Specifically, v_y(t) should be cos(t) + 3, not cos(t) + 2.gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: fail (error) 2026-10-12 — The integration of the acceleration vector is incorrect. The integral of cos(t) is sin(t), not -cos(t) (as implied by the derivative check in step 1 where d/dt(-cos(t)) = sin(t) is used to justify v_x = sin(t), but the comment says 'Integrating a ... gives v(t) = <sin(t)...>'). Wait, let's look closer. Step 1 comment says: v(t) = <sin(t), cos(t) + 2, t - 3>. Let's check v(0): <0, 3, -3>. This matches the initial velocity. Let's check a(t) = v'(t): <cos(t), -sin(t), 1>. This matches the given acceleration. So v(t) is correct. Step 2 comment says: 'Integrate again and use r(0)'. The equation shows the derivative of the proposed r(t) equals v(t). Proposed r(t) = <-cos(t), 2t + sin(t) + 1, t^2/2 - 3t - 1>. Let's check r(0): <-1, 1, -1>. This matches the initial position. Let's check r'(t): <sin(t), 2 + cos(t), t - 3>. This matches v(t). So the math is actually correct. Why did I think it was wrong? Let's re-read the prompt's solution lines. Line 1: Derivative of sin(t) is cos(t). Derivative of cos(t)+2 is -sin(t). Derivative of t-3 is 1. The matrix on the right is <cos(t), -sin(t), 1>. This is a(t). The comment says 'Integrating a and using v(0) gives v(t) = <sin(t), cos(t) + 2, t - 3>'. This is correct. Line 2: Derivative of -cos(t) is sin(t). Derivative of 2t+sin(t)+1 is 2+cos(t). Derivative of t^2/2-3t-1 is t-3. The matrix on the right is <sin(t), cos(t)+2, t-3>. This is v(t). The comment says 'Integrate again and use r(0)'. This is correct. Line 3: Checks r(0). Correct. The solution is correct. The verdict should be pass.
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 2026-10-12 with SymPy 1.14.0.