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 km/hv_{1}=10 \ \text{km/h} \ \\ v_{2}=40 \ \text{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 \ \text{km/h}



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