Motion

If you go at speed 3.7 km/h, you come to the station 42 minutes after leaving the train. If you go by bike to the station at speed 27 km/h, you come to the station 56 minutes before its departure.
How far is the train station?

Correct result:

s =  7.003 km

Solution:

$$\begin{gathered} v_1=3.7\text{ km/h } \\ {v}_2=27\text{ km/h } \\ {t}_1=(x+42)\mathrm{/}60 \\ {t}_2=(x-56)\mathrm{/}60 \\ s=v_1\cdot t_1=v_2\cdot t_2 \\ {v}_1(x+42)=v_2(x-56) \\ x=(v_1\cdot 42+v_2\cdot 56)\mathrm{/}(v_2-v_1)=(3.7\cdot 42+27\cdot 56)\mathrm{/}(27-3.7)\doteq 71.5622\text{ min } \\ {t}_1=(x+42)\mathrm{/}60=(71.5622+42)\mathrm{/}60={\displaystyle \frac{441}{233}}\doteq 1.8927\text{ h } \\ {t}_2=(x-56)\mathrm{/}60=(71.5622-56)\mathrm{/}60\doteq 0.2594\text{ h } \\ s=s_1=s_2 \\ {s}_1=v_1\cdot t_1=3.7\cdot 1.8927\doteq 7.003\text{ km } \\ s=v_2\cdot t_2=27\cdot 0.2594=7.003\text{ km} \end{gathered}$$

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