Two trains meet

From A started at 7:15, the express train at a speed of 85 km/h to B. From B, the passenger train started at 8:30 in the direction of A and at a speed of 55 km/h. The distance A and B are 386 1/4 km. At what time and distance do the two trains meet from B?

Result

t = 10:30 hh:mm
b =  110 km

Step-by-step explanation:

$$\begin{gathered} v_1=85\text{km/h} \\ {t}_1=7:15=7hr15min=7+\frac{15}{60}=7.25hr=7.25 \\ {v}_2=55\text{km/h} \\ {t}_2=8:30=8hr30min=8+\frac{30}{60}=8.5hr=8.5 \\ s=386\frac{1}{4}=386+\frac{1}{4}=\frac{386\cdot 4+1}{4}=\frac{1544+1}{4}=\frac{1545}{4}=386.25\text{lm} \\ a+b=s \\ a=v_1\cdot (t-t_1) \\ b=v_2\cdot (t-t_2) \\ a+b=386.25 \\ a=85\cdot (t-7.25) \\ b=55\cdot (t-8.5) \\ a+b=386.25 \\ a-85t=-616.25 \\ b-55t=-467.5 \\ Row2-Row1\to Row2 \\ a+b=386.25 \\ -b-85t=-1002.5 \\ b-55t=-467.5 \\ Row3+Row2\to Row3 \\ a+b=386.25 \\ -b-85t=-1002.5 \\ -140t=-1470 \\ t=\frac{-1470}{-140}=10.5 \\ b=\frac{-1002.5+85t}{-1}=\frac{-1002.5+85\cdot 10.5}{-1}=110 \\ a=386.25-b=386.25-110=276.25 \\ a=\frac{1105}{4}=276.25 \\ b=110 \\ t=\frac{21}{2}=10.5=10:30\text{hh}:\text{mm} \end{gathered}$$

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