5. The hare and the tortoise had a race over $100 \mathrm{~m}$, in which both maintained constant speeds. When the hare reached the finish line, it was $75 \mathrm{~m}$ in front of the tortoise. The hare immediately turned around and ran back towards the start line.
How far from the finish line did the hare and the tortoise meet?
A 54
B 60
C 64
D $66 \frac{2}{3}$
E 72
Solution
B
The tortoise is $75 \mathrm{~m}$ behind the hare when the hare has run the full $100 \mathrm{~m}$. Therefore the tortoise runs $25 \mathrm{~m}$ in the same time as the hare runs $100 \mathrm{~m}$. Hence the hare runs at four times the speed of the tortoise.

When the hare turns round, the hare and the tortoise are $75 \mathrm{~m}$ apart and running towards each other.

Since the hare is running at four times the speed of the tortoise, when they meet the hare has run $\frac{4}{5}$ ths of the $75 \mathrm{~m}$ they were apart when the hare turned around. The tortoise has run $\frac{1}{5}$ th of this distance.
Therefore they meet at a distance $\frac{4}{5} \times 75 \mathrm{~m}$, that is, $60 \mathrm{~m}$, from the finish line.
For INVESTIGATION
15.1 The distance-time graph below illustrates the paths of the hare and the tortoise.
The point $H$ corresponds to the position of the hare when the hare reaches the finish line and turns around. The point $T$ corresponds to the position of the tortoise at this time. The point $M$ corresponds to their common position when they meet. The points $F$ and $G$ are as shown.
Use the geometry of the diagram to work out how far from the finish line are the hare and tortoise when they meet.
[Hint: Extend the line $F M$ to meet the line through $G$ which is perpendicular to $F G$.]