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Snapshot of the question

In this problem, P.A. Tipler writes

this is not the average of running and jogging speed because she ran for 10 s but jogged for 30 s.

Why must the time interval for running and jogging speeds be equal for this to be the average of those speds?

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    $\begingroup$ Hi and welcome to physics.SE. Please use the preview window (right below the area in which you are typing your post) to make sure your question looks like it is supposed to look. I have fixed the picture and removed a lot of superfluous and ungrammatical sentences. I think the latter half of your question was supposed to be a question about a different exercise, but there was no second picture, so I removed it. In any case, please ask only one question per post. $\endgroup$
    – ACuriousMind
    Sep 22, 2016 at 16:56
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    $\begingroup$ Rollback to Revision 2? $\endgroup$
    – Qmechanic
    Sep 25, 2016 at 18:45
  • $\begingroup$ I just wanted to add the remaining portion of this question which was edited ,that's why i rolled back . $\endgroup$
    – Who I Am
    Sep 25, 2016 at 18:59

3 Answers 3

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It's implicit in the way the author is using the word 'average'. The author is implicitly defining 'average' to mean 'average over time', rather than 'average over distance'. By this definition, the time intervals must be the same in order for the numerical average (of 10 m/s and -1.67 m/s) to be equal to average velocity.

In fairness, this is the way most people use the word 'average': if you were in a car going 50 mph for 1 min and then 60 mph for 1 min, most people would say your average speed was 55 mph. However, given the nature of the problem, the definition of 'average' should probably have been made explicit.

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You can't average values that have different units/denominators, because those values mean different things. Same concept applies when you say that the average of 10 feet and 20 meters isn't 15. Units of 10s and 20s are no different.

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  • $\begingroup$ Can you please give a simple and more clear example regarding this $\endgroup$
    – Who I Am
    Sep 28, 2016 at 8:01
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i think according to definition of average velocity; V(avg)=(u+v)/2 u=100/10 =10m/s and v=-50/30 =-1.67m/s. :- V(avg)=(10+-1.67)/2 =4.17m/s. Because the definition of average velocity is not (average displacement over total time)

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    $\begingroup$ Welcome to Physics SE! Please don't post formulas as pictures or plain text, but use MathJax instead. MathJax is easy for people on all devices to read, and can show up clearer on different screen sizes and resolutions. Look at this Math SE meta post for a quick tutorial. $\endgroup$
    – user191954
    Oct 6, 2018 at 9:13

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