Uphill and downhill

The cyclist moves uphill at a constant speed of v1 = 10 km/h. When he reaches the top of the hill, he turns and passes the same track downhill at a speed of v2 = 40 km/h. What is the average speed of a cyclist?

Result

v =  16 km/h

Solution:

v1=10 km/h v2=40 km/h   s=v1 t1=v2 t2  t2=v1v2 t1  v=s+st1+t2=v1 t1+v2 t2t1+t2  v=v1 t1+v2 v1v2 t1t1+v1v2 t1  v=v1+v2 v1v21+v1v2  v=v1+v11+v1v2=10+101+1040=16=16  km/h v_{ 1 } = 10 \ km/h \ \\ v_{ 2 } = 40 \ km/h \ \\ \ \\ \ \\ s = v_{ 1 } \ t_{ 1 } = v_{ 2 } \ t_{ 2 } \ \\ \ \\ t_{ 2 } = \dfrac{ v_{ 1 } }{ v_{ 2 } } \cdot \ t_{ 1 } \ \\ \ \\ v = \dfrac{ s+s }{ t_{ 1 }+t_{ 2 } } = \dfrac{ v_{ 1 } \cdot \ t_{ 1 }+v_{ 2 } \cdot \ t_{ 2 } }{ t_{ 1 }+t_{ 2 } } \ \\ \ \\ v = \dfrac{ v_{ 1 } \cdot \ t_{ 1 }+v_{ 2 } \cdot \ \dfrac{ v_{ 1 } }{ v_{ 2 } } \cdot \ t_{ 1 } }{ t_{ 1 }+\dfrac{ v_{ 1 } }{ v_{ 2 } } \cdot \ t_{ 1 } } \ \\ \ \\ v = \dfrac{ v_{ 1 }+v_{ 2 } \cdot \ \dfrac{ v_{ 1 } }{ v_{ 2 } } }{ 1+\dfrac{ v_{ 1 } }{ v_{ 2 } } } \ \\ \ \\ v = \dfrac{ v_{ 1 }+v_{ 1 } }{ 1+\dfrac{ v_{ 1 } }{ v_{ 2 } } } = \dfrac{ 10+10 }{ 1+\dfrac{ 10 }{ 40 } } = 16 = 16 \ \text{ km/h }



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