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)/60=4412331.8927 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 0.25947.003=7.003  km v_{ 1 } = 3.7 \ km/h \ \\ v_{ 2 } = 27 \ 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 \ min \ \\ t_{ 1 } = (x + 42)/60 = (71.5622 + 42)/60 = \dfrac{ 441 }{ 233 } \doteq 1.8927 \ h \ \\ t_{ 2 } = (x - 56)/60 = (71.5622 - 56)/60 \doteq 0.2594 \ h \ \\ \ \\ s = s_{ 1 } = s_{ 2 } \ \\ s_{ 1 } = v_{ 1 } \cdot \ t_{ 1 } = 3.7 \cdot \ 1.8927 \doteq 7.003 \ km \ \\ s = v_{ 2 } \cdot \ t_{ 2 } = 27 \cdot \ 0.2594 \doteq 7.003 = 7.003 \ \text{ km }



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