Athletic oval

On the 400 m athletic oval, two athletes started back to back from the same level. After the starting shot, they both ran away from each other at the same time. Athlete A ran at 15 km/h, and Athlete B ran at 17 km/h. A) in how many seconds did the athletes meet each other? B) how many meters did each athlete run before they met?

Correct result:

t =  45 s
s₁ =  187.5 m
s₂ =  212.5 m

Solution:

$$\begin{gathered} s=400\text{ m } \\ {v}_1=15\text{ km/h}\to \text{ m/s}=15\mathrm{/}3.6\text{ m/s}=4.16667\text{ m/s } \\ {v}_2=17\text{ km/h}\to \text{ m/s}=17\mathrm{/}3.6\text{ m/s}=4.72222\text{ m/s } \\ t={\displaystyle \frac{s}{v_1+v_2}}={\displaystyle \frac{400}{4.1667+4.7222}}=45\text{ s} \end{gathered}$$

$s_1=v_1\cdot t=4.1667\cdot 45={\displaystyle \frac{375}{2}}={\displaystyle \frac{375}{2}}\text{ m}=187.5\text{ m}$

$s_2=v_2\cdot t=4.7222\cdot 45={\displaystyle \frac{425}{2}}={\displaystyle \frac{425}{2}}\text{ m}=212.5\text{ m}$

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