Time, Speed & Distance

The language of races

Every race question uses one of four phrases. Decode them first, then it is plain arithmetic:

• "A beats B by x metres" — when A finishes the race (covers the full distance D), B is still x m from the finish, having run $D - x$.
• "A beats B by t seconds" — A finishes, and B needs t more seconds to finish.
• "A gives B a start of x metres" — B starts x m ahead of the common start line, so B runs only $D - x$.
• "A gives B a start of t seconds" — B starts running t seconds before A does. A "dead heat" means both finish together.

Beats by x metres → the speed ratio

When A finishes, B has covered $D - x$. Both ran for the same time, so $\frac{S_A}{S_B} = \frac{D}{D - x}$ In a 100 m race A beats B by 10 m → $S_A : S_B = 100 : 90 = 10:9$. That ratio answers most follow-ups directly (who wins a longer race, by how much, start needed for a dead heat, and so on).

Beats by t seconds / by both

• Beaten by t s: B covers D in $t_A + t$. If B's time is given, A's time is $t_B - t$; if A's time is given, B's is $t_A + t$.
• "A beats B by 20 m or 5 s" links them: B runs the final 20 m in 5 s → $S_B = \frac{20}{5} = 4$ m/s. Then A's time over D follows at once (200 m race: A runs 200 m while B runs 180 m, taking 45 s → $t_A = 45$ s, $S_A = \frac{40}{9}$ m/s).
• Race scores may quote the winner's time: A runs 100 m in 12 s while B takes 15 s → when A finishes, B has covered $\frac{100}{15} \times 12 = 80$ m → beaten by 20 m.

Starts and handicaps

A start of x m to B in a race of D: B runs $D - x$ while A runs D. If the result is a dead heat, $\frac{S_A}{S_B} = \frac{D}{D-x}$ exactly as in a "beats by x" race — a start problem and a beats problem are the same arithmetic read in opposite directions. If A gives a start and still WINS by y m, B effectively ran $D - x - y$: e.g. A gives B 30 m in a 300 m race and wins by 45 m → $S_A : S_B = 300 : 225 = 4:3$.

Games of 100 (chain rule)

"A can give B 20 points in a game of 100" means $S_A : S_B = 100 : 80$. Chain two such statements by multiplying: A gives B 20 (in 100) and B gives C 10 (in 100): $\frac{A}{B} = \frac{100}{80}$, $\frac{B}{C} = \frac{100}{90}$ → when A scores 100, C scores $80 \times \frac{90}{100} = 72$ → A can give C 28 points.

Chained races

"A beats B by 5 m in 100 m; B beats C by 8 m in 100 m" does NOT give A beats C by 13 m. Convert to ratios and multiply: $\frac{A}{B} = \frac{100}{95}$, $\frac{B}{C} = \frac{100}{92}$ → when A runs 100, C runs $95 \times \frac{92}{100} = 87.4$ → A beats C by 12.6 m.

Common traps

• Adding the two margins in a chained race (13 instead of 12.6 m).
• Forgetting that a beaten runner still has to run the remaining x m in "or t seconds" statements.
• Treating a start of x m as if B ran the full distance.
• Comparing speeds before both runners have run for the SAME time.

Quick revision

• Beats by x m → $S_A:S_B = D : (D-x)$.
• Beats by t s → $t_B = t_A + t$; "by x m or t s" → $S_B = \frac{x}{t}$.
• Start x m → B runs $D - x$; wins by y m after a start → B ran $D - x - y$.
• Chain margins through ratios, never by addition.