## Task 3 - 070813

Three athletes started simultaneously and ran $100 \mathrm{~m}$. When the first one reached the finish line, the second one still had exactly $10 \mathrm{~m}$ to go. When the second one reached the finish line, the third one still had exactly $10 \mathrm{~m}$ to go.

How far was the third one from the finish line when the first one reached it? (Assume that each of the three athletes ran the entire distance at a constant speed.)
The formula for constant velocity can be applied: $v=s: t$ or rearranged: $t=s: v$.

Considering the time point 1, when the 1st runner (a) reaches the finish line, for the second (b) and third (c) it holds that: $t_{1}=s_{b_{1}}: v_{b_{1}}=s_{c_{1}}: v_{c_{1}}$,

where $s_{b_{1}}=100 \mathrm{~m}-10 \mathrm{~m}=90 \mathrm{~m}$ and $s_{c_{1}}$ is the distance to be found that $c$ had covered at the time point 1 when the first runner crossed the finish line.

At time point 2, when the second runner (b) reaches the finish line, it holds analogously: $t_{2}=s_{b_{2}}: v_{b_{2}}=s_{c_{2}}: v_{c_{2}}$, where $s_{b_{2}}=100 \mathrm{~m}$ and $s_{c_{2}}=100 \mathrm{~m}-10 \mathrm{~m}=90 \mathrm{~m}$, and $v_{b}:=v_{b_{1}}=v_{b_{2}}$ and $v_{c}:=v_{c_{1}}=v_{c_{2}}$ apply.

Now, the equations for the 1st and 2nd time points are combined:

$$
s_{c_{1}}=s_{b_{1}} \cdot \frac{v_{c}}{v_{b}}=90 \mathrm{~m} \cdot \frac{v_{c}}{v_{b}}=90 \mathrm{~m} \cdot \frac{s_{c_{2}}}{s_{b_{2}}}=90 \mathrm{~m} \cdot \frac{90 \mathrm{~m}}{100 \mathrm{~m}}=81 \mathrm{~m}
$$

Accordingly, the third runner was still $100 \mathrm{~m}-81 \mathrm{~m}=19 \mathrm{~m}$ away from the finish line when the first runner had crossed it.