Two friends

Two friends on a bicycle rode against each other from two places 49 km apart. The first set off at 8.00 at a speed of 20km/h, the second 12 minutes later at a speed of 25km/h. What time do they meet? How many km will each of them travel since then?

Result

t = 9:12
a =  24 km
b =  25 km

Solution:

$$\begin{gathered} s=49\text{ km } \\ {v}_1=20\text{ km/h } \\ {v}_2=25\text{ km/h } \\ {t}_1=8.00\text{ h } \\ {t}_2=8{\displaystyle \frac{12}{60}}={\displaystyle \frac{8\cdot 60+12}{60}}={\displaystyle \frac{492}{60}}={\displaystyle \frac{41}{5}}=8\frac{1}{5}=8.2\text{ h } \\ a+b=s \\ a=v_1\cdot (t-t_1) \\ b=v_2\cdot (t-t_2) \\ a+b=49 \\ a=20\cdot (t-8) \\ b=25\cdot (t-8.2) \\ a+b=49 \\ a-20t=-160 \\ b-25t=-205 \\ a=24 \\ b=25 \\ t={\displaystyle \frac{46}{5}}=9.2=9:12 \end{gathered}$$

$a=24=24\text{ km}$

$b=25=25\text{ km}$

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