Mechanics- problem solving

Question

The distance from point $A$ to point $B$ is $s$. In the motion from $A$ to $B$ and back, the speed for the first part of the motion is $v_1$ and the speed for the return part of the motion is $v_2$. The average speed for the entire motion is $v$. Deduce that it is impossible to average twice the speed of the first part of the motion, that is, it is impossible to have $v=2v_1$.

Answer 1

Let $t_1$ be the time needed to move from $A$ to $B$ and $t_2$ be the time to move from $B$ to $A$.

From the statement of the problem, we must assume that $s$ the distance from $A$ to $B$ is finite and positive, that is $0<s<+\infty$. So it is physically impossible to move from $A$ to $B$ (or back) instantaneously,so $t_1>0$ and $t_2>0$.

Since the statement of the problem says "the motion from A to B and back, the speed for the first part of the motion is $v_1$ and the speed for the return part of the motion is $v_2$", we have that it take a finite time to go from $A$ to $B$ (and back). So we have $t_1<+\infty$ and $t_2<+\infty$.

So, we have $0<s<+\infty$, $0<t_1<+\infty$ and $0<t_2<+\infty$.

Note that \begin{align*} v_1&= \frac{s}{t_1}\\ v_2&= \frac{s}{t_2} \end{align*} and $$ v = \frac{2s}{t_1+t_2} $$ So we have that $0<v_1<+\infty$, $0<v_2<+\infty$ and $0<v<+\infty$, and we also have $$ \frac{1}{v}=\frac{1}{2 } \left ( \frac{1}{v_1} + \frac{1}{v_2} \right) $$ If $v=2v_1$, then we must have $$\frac{1}{2 v_2}=0$$ which is not possible, since $0<v_2<+\infty$.