Won by less than a metre

The school sprint final was run on Friday afternoon over a hundred metres, with four runners on the track. The winner crossed the line in twelve and a half seconds.

Friday final, $100$ m, $4$ runners. Winner: $12.5$ s.

Distance $d$ m, winner's time $W$ s: $d=100,W=12.5$. $4$ runners.

The runner in second place finished in twelve point six two seconds, just twelve hundredths of a second later. The third runner took twelve point eight.

Second: $12.62$ s, so $12.62-12.5=0.12$ s later. Third: $12.8$ s.

Second $S$: $S=12.62$, and $S-W=0.12$. Third $12.8$.

The winner's average speed was eight metres every second. When the winner crossed the line, the runner in second place was still less than a metre behind, about ninety-five centimetres.

Winner's speed: $100÷12.5=8$ m/s. In $12.5$ s the runner in second place covered only $\frac{12.5}{12.62}\times 100\approx 99.05$ m, so was $100-99.05\approx 0.95$ m behind.

Speed $\frac{100}{12.5}=8$. At time $W$ the second had run $\frac{12.5}{12.62}\times 100\approx 99.05$, so behind by $100-99.05\approx 0.95$.

Had the second runner left the blocks twelve hundredths of a second sooner, the two would have tied for first place.

Leaving the blocks $0.12$ s sooner: $12.62-0.12=12.5$, a tie.

$S-0.12=W$: a tie.