The Fastest Path

The question

Two slides run from the same start to the same finish. One is dead straight — the shortest possible path. The other dips steeply and travels FARTHER. Which ball arrives first?

The straight path loses, every single time. Distance is not the only currency — speed is the other, and v = √(2g·drop) says speed comes from how far you have fallen. The curve buys speed early and spends it over the whole journey. The perfect such curve is called the brachistochrone, and finding it was the challenge that launched a whole branch of mathematics.

The key idea

The fastest descent path between two points is not the straight line but a curve — the brachistochrone (Greek: 'shortest time'). It is a cycloid: the curve traced by a point on the rim of a rolling wheel.

Time along a path is the sum of (distance ÷ speed) over every little segment, $t = \int ds/v$. The straight line minimises the distance but keeps $v$ small for the longest stretch; a steep early dive makes $v$ large almost immediately. The optimum trade is the cycloid. **All forms:** $v = \sqrt{2gh} \Rightarrow h = v^2/2g$. **Limiting case:** make the finish directly below the start and the brachistochrone collapses into the straight vertical drop — the curve only matters when the journey has horizontal distance to cover. **Connect it:** this is pure conservation of energy ($mgh = \tfrac{1}{2}mv^2$, mass cancels) plus one new idea — that WHEN you receive your speed matters. The same trade-off explains why skiers dive into a tuck early and why roller-coasters drop steepest first.

The formula

• ·v = speed (m/s) after dropping a height h (m) from rest
• ·with g = 9.8 m/s². The path taken does not appear in the formula at all — only the drop does.

What to remember

• ·Speed from a drop is path-independent: v = √(2gh) — energy conservation with m cancelled.
• ·Arrival time is path-dependent: t = ∫ds/v rewards getting fast early.
• ·The optimal curve is the brachistochrone — a cycloid, not a straight line.
• ·Mass never matters on frictionless slides — every result here survives doubling it.