Answer:
(c) [tex]\frac{2v_0(v_1+v_2)}{v_1+v_2+2v_0}[/tex]
Explanation:
Average speed is calculated as (total distance)/(total time). We have three segments in the journey, indexed by 0, 1, and 2:
[tex]v = \frac{s}{t}=\frac{v_0t_0+v_1t_1+v_2t_2}{t_0+t_1+t_2}\\t_1=t_2\implies\\v=\frac{v_0t_0+v_1t_1+v_2t_1}{t_0+t_1+t_1}=\frac{v_0t_0+v_1t_1+v_2t_1}{\frac{v_0t_0}{v_0}+\frac{v_1t_1}{v_1}+\frac{v_2t_1}{v_2}}[/tex]
We also know that the distance of the first segment is the same as one of segment 2 and 3 together:
[tex]v_0t_0=v_1t_1+v_2t_1\\v_0\frac{t_0}{t_1}=v_1+v_2[/tex]
Going back to the average speed expression, divide by t_1:
[tex]\frac{v_0t_0+v_1t_1+v_2t_1}{\frac{v_0t_0}{v_0}+\frac{v_1t_1}{v_1}+\frac{v_2t_1}{v_2}}=\frac{\frac{v_0t_0}{t_1}+v_1+v_2}{\frac{v_0t_0}{v_0t_1}+2}[/tex]
and combine the two equations:
[tex]\frac{\frac{v_0t_0}{t_1}+v_1+v_2}{\frac{v_0t_0}{v_0t_1}+2}=\frac{2(v_1+v_2)}{\frac{v_1+v_2}{v_0}+2} = \frac{2v_0(v_1+v_2)}{v_1+v_2+2v_0}[/tex]
The last form matches your choice (c).
