Direct route

From two different places A and B connected by a direct route, Adam (from city A) and Bohus (from city B) started at a constant speed. As Adam continued to go from A to B, Bohus turned around at the time of their meeting and at the same speed he returned to city B. He came there two hours earlier than Adam. How long did they come to meet when you know that Bohus went two times faster than Adam.

Result

t =  2 h

Solution:

 v2=2 v1 s1=s/3 s2=2/3 s s=s1+s2 t1=(s1+s2)/v1=s/v1 t2=(s2+s2)/v2=(2/3s+s/3s)/(2v1)=2/3 s/v1 t2=2/3 t1 t1=2+t2 t1=2/(12/3)=6 h t2=t12=62=4 h t=t1/3=6/3=2=2  h  \ \\ v_{ 2 } = 2 \ v_{ 1 } \ \\ s_{ 1 } = s/3 \ \\ s_{ 2 } = 2/3 \ s \ \\ s = s_{ 1 }+s_{ 2 } \ \\ t_{ 1 } = (s_{ 1 }+s_{ 2 })/v_{ 1 } = s/v_{ 1 } \ \\ t_{ 2 } = (s_{ 2 }+s_{ 2 })/v_{ 2 } = (2/3s+s/3s)/(2v_{ 1 }) = 2/3 \ s/v_{ 1 } \ \\ t_{ 2 } = 2/3 \ t_{ 1 } \ \\ t_{ 1 } = 2 + t_{ 2 } \ \\ t_{ 1 } = 2/(1-2/3) = 6 \ h \ \\ t_{ 2 } = t_{ 1 }-2 = 6-2 = 4 \ h \ \\ t = t_{ 1 }/3 = 6/3 = 2 = 2 \ \text{ h }



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