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.

Result

v1 =  17.9 km/h
v2 =  4.5 km/h

Solution:

37=v135+7160+v27160 37=v135+71+81060+v271+860  2220=106v1+71v2 2220=104v1+79v2  v1=17.9  km/h 37= v_1 \cdot \dfrac{ 35+71} { 60} + v_2\cdot \dfrac{ 71} {60} \ \\ 37 = v_1 \cdot \dfrac{ 35+71+8-10} { 60} + v_2 \cdot \dfrac{ 71+8 }{60} \ \\ \ \\ 2220 = 106 v_1 + 71 v_2 \ \\ 2220 = 104 v_1 + 79 v_2 \ \\ \ \\ v_1 = 17.9 \ \text{ km/h }
v2=4.5  km/h v_2 = 4.5 \ \text{ km/h }



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