 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ia/paper-4/3a/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

m \overset{x}{¨} = - V^{'} (x) .

\overset{x}{˙}

gives

\frac{d}{d t} (\frac{1}{2} m \overset{x}{˙}^{2} + V (x)) = \overset{x}{˙} (m \overset{x}{¨} + V^{'} (x)) = 0,

E = \frac{1}{2} m \overset{x}{˙}^{2} + V (x)

is conserved. On a segment where the direction of motion is fixed,

t - t_{0} = \pm \frac{m}{2} \int_{x_{0}}^{x} \frac{d u}{E - V ( u )} .

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