1. Richard is cycling at a speed $v \mathrm{~km} / \mathrm{h}$ when he looks at his cycle computer to see how long it will take him to get home at his current speed. It shows a time $t$ hours. He cycles at this speed for 40 minutes, then instantaneously slows down by $1 \mathrm{~km} / \mathrm{h}$ and checks his cycle computer; the predicted time to get home at his new speed is still $t$ hours.
After cycling at this new speed for 45 minutes, he instantaneously slows down by another $1 \mathrm{~km} / \mathrm{h}$ and checks his cycle computer; the predicted time to get home at this speed is again $t$ hours.
How far from home was he when he first looked at the computer?
Solution
Let the initial speed be $v$ and the distance from home he starts be $d$.
Then the distance from home where he started can be found in three ways from the three bits of information given.
$$
\begin{array}{l}
d=v t \\
d=\frac{2}{3} v+(v-1) t \\
d=\frac{2}{3} v+\frac{3}{4}(v-1)+t(v-2)
\end{array}
$$

Equating the first two expressions gives $t=\frac{2}{3} \nu$.
Equating the first and third and substituting in gives $\frac{2}{3} v^{2}=\frac{2}{3} v+\frac{3}{4}(v-1)+\frac{2}{3} v(v-2)$.
Solving this gives $v=9$. Substituting in gives $t=6$. The distance he started from home is therefore $54 \mathrm{~km}$.