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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 $v=2v_1$.

Andrei
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Shah21
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  • Welcome to MSE. What have you tried? What formulas did you use? Where are you stuck? Questions showing no effort tend to be voted down and/or closed – Andrei Oct 13 '20 at 16:55
  • try plugging some numbers for $v,v_1$ and $s$ Solve for the required times to complete the legs, and complete the complete circuit. – Doug M Oct 13 '20 at 18:09
  • So when i plug in some values for eg: v=50, v1=5, s=10, and by substititing these values in V= 2v1v2/(v1+v2), we get a finite value for v2. V2 cannot be equal to zero as it has a finite value, hence proves that its impossible to have v=2v1 – Shah21 Oct 14 '20 at 07:30
  • @Shah21 The reasoning does not follow exactly as you wrote. I have edited my answer to make it clearer. – Ramiro Oct 14 '20 at 12:11
  • @Shah21 I have edited my answer and made it very detailed. Please, let me know if you have any other question regarding my answer. – Ramiro Oct 14 '20 at 12:12
  • Thank you very much! – Shah21 Oct 14 '20 at 13:54
  • @Shah21 Since you are new here in Stack Exchange, I guess you may not know. If an answer provides relevant / helpful information regarding your question, you should upvote it. If the answer answers completely your question, you should accept it. You can upvote as many answers you want, but you can accept only one. You may change which answer is accepted, or simply un-accept the answer, at any time. It is common to upvote the answer you accept, but those are two separate operation (upvoting and accepting) – Ramiro Oct 14 '20 at 17:32
  • @Shah21 To upvote, click the triangle pointing upward above the number (of votes) in front of the question. To accept the answer, click on the check mark beside the answer to toggle it from greyed out to filled in. – Ramiro Oct 14 '20 at 17:34

1 Answers1

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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 $0<v_2<+\infty$.

Ramiro
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