Theorem

Over a fixed distance, time varies inversely with speed

Arithmetic • Time, Speed and Distance

Formula

t_{2} = \frac{s _{1}}{s _{2}} t_{1} (d fixed)

For a fixed distance

d

, if the speed is multiplied by

m > 0

then the time is multiplied by

\frac{1}{m}

; equivalently, speeds in the ratio

s_{1} : s_{2}

give times in the ratio

s_{2} : s_{1}

How to use it

1Convert a percentage change in speed into a fraction first: ' $20%$ faster' is $m = \frac{6}{5}$ , so the time becomes $\frac{5}{6}$ of what it was.
2The time saved is then $(1 - \frac{1}{m})$ of the original time, which recovers the original time from a stated saving.
3Never apply the speed's percentage to the time directly: a $20%$ rise in speed is a $16 \frac{2}{3} %$ fall in time, not a $20%$ one.

A cyclist covering a fixed route increases speed by

25%

and arrives

10

minutes earlier. How long did the original journey take?

Step-by-step

1
$m = \frac{5}{4} \Rightarrow t^{'} = \frac{4}{5} t$
Why: The distance is unchanged, so the time is divided by the speed factor.
2
$t - \frac{4}{5} t = \frac{t}{5} = 10 \Rightarrow t = 50$
Why: The saving is the missing fifth of the original time.

Answer

50

minutes

Watch out

Holds only while the distance is the same for both journeys; if the route changes as well, speed and time must be combined through

d = s t

rather than by ratio.

Also known as

time inversely proportional to speed, speed up by x percent time saved, time ratio is the inverse of the speed ratio