Motion

Cyclist started at 9:00 from point S to point T. After 10 minutes, followed him at the same speed the second cyclist. Walker, which went from T to S, started at 9:35. After 71 minutes he met the first cyclist and after next 8 minutes the second cyclist.
Distance between the two cities is 37 km.

Determine the (same) speed of cyclists and walker.

Correct result:

v₁ =  17.9 km/h
v₂ =  4.5 km/h

Solution:

$$\begin{gathered} 37=v_1\cdot {\displaystyle \frac{35+71}{60}}+v_2\cdot {\displaystyle \frac{71}{60}} \\ 37=v_1\cdot {\displaystyle \frac{35+71+8-10}{60}}+v_2\cdot {\displaystyle \frac{71+8}{60}} \\ 2220=106v_1+71v_2 \\ 2220=104v_1+79v_2 \\ {v}_1=17.9\text{ km/h} \end{gathered}$$

$v_2=4.5\text{ km/h}$

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