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?

Result

s =  7.003 km

Solution:

v1=3.7 km/h v2=27 km/h t1=(x+42)/60 t2=(x56)/60 s=v1 t1=v2 t2 v1(x+42)=v2(x56) x=(v1 42+v2 56)/(v2v1)=(3.7 42+27 56)/(273.7)71.5622 min t1=(x+42)/60=(71.5622+42)/604412331.8927 h t2=(x56)/60=(71.562256)/600.2594 h  s=s1=s2 s1=v1 t1=3.7 1.89277.003 km s=v2 t2=27 0.25947.0037.003 kmv_{1}=3.7 \ \text{km/h} \ \\ v_{2}=27 \ \text{km/h} \ \\ t_{1}=(x + 42)/60 \ \\ t_{2}=(x - 56)/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)/(v_{2}-v_{1})=(3.7 \cdot \ 42+27 \cdot \ 56)/(27-3.7) \doteq 71.5622 \ \text{min} \ \\ t_{1}=(x + 42)/60=(71.5622 + 42)/60 \doteq \dfrac{ 441 }{ 233 } \doteq 1.8927 \ \text{h} \ \\ t_{2}=(x - 56)/60=(71.5622 - 56)/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 \doteq 7.003 \doteq 7.003 \ \text{km}



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