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I have this question:

A particle moving in a straight line covers half the distance with speed 3m/s. The other half of the distance is covered in two equal time intervals with speed 4.5m/s and 7.5m/s respectively. The average speed of the particle during this motion is what? (4m/s - Ans)

I did so by taking the total distance as x, half the distance as x/2 and the other half which was further divided in two equal parts as x/4 and x/4. Then, as I know the speeds, I calculated the time which was: x/6,x/18 and x/30 respectively. Now since it's ' average speed = total distance/total time I did, x/[x/6+x/18+x/30] which came out to be 23/90 or 3.9 m/s

The real solution showed the answer to be 4.0m/s and I didn't get the method. So, can 3.9m/s be rounded off as 4.0m/s?

Correct me :)

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closed as off-topic by user191954, Emilio Pisanty, Aaron Stevens, ZeroTheHero, stafusa Oct 9 '18 at 11:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Community, Emilio Pisanty, Aaron Stevens, ZeroTheHero, stafusa
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions. Can you try making a question about some concepts that you'd need to solve this problem? $\endgroup$ – user191954 Oct 8 '18 at 15:25
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I did so by taking the total distance as x, half the distance as x/2 and the other half which was further divided in two equal parts as x/4 and x/4

Why? For the 2nd half of the journey, the particle spends equal time at each velocity and does not cover equal distance at each velocity.

From the problem statement:

The other half of the distance is covered in two equal time intervals with speed 4.5m/s and 7.5m/s respectively.

This isn't a conceptual problem; I believe you misread / misinterpreted the problem statement.

Now, solve the problem properly and put the solution into your question or into an answer.

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    $\begingroup$ It's both misread as well as misinterpreted =_= Thank you so much for pointing that out. Have a great day! $\endgroup$ – yena shah Oct 9 '18 at 0:56
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I like this problem, because it forces you to think about when you add things linearly, and when you add them harmonically. It also forces you to think about what you need to know, and what you don't need to know.

What does average velocity over some time interval,$T$, mean? Generally:

$$ \bar v = \frac{\int_0^T{v(t)dt}}{\int_0^T{ dt}}=\frac 1 T \int_0^T{v(t)dt}$$

Notice that $x$ does not appear in this equation explicitly. If you apply that to the 2nd half the trip, you'll see that you don't need to know $x$ or $T$ to compute the average velocity of they second half of the trip. This is good news, because you don't know $x$, nor do you know $T$.

So you do that, and get an average velocity, $v_2$. At this point you know for half the distance you went $v_1$ (given), and for the other half of distance you went $v_2$ (computed).

How do you compute the time-averaged $v(t)$ when you only know $v(x)$? Here you have to weight the time spent at $x$ moving at $v(x)$:

$$ \bar v = \frac{\int_0^X{dx}}{\int_0^X\frac{dx}{v(x)}}$$

This is a harmonic average. Note that you don't need to know $X$ to evaluate this, all you need to know is $v(x)$, relative to $X$.

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  • $\begingroup$ Do you think this answers the confusion in the OP or can be used to solve the problem at the level of the OP? $\endgroup$ – nasu Oct 8 '18 at 22:43
  • $\begingroup$ @nasu idk. Every time I think I am helping and am explicit, my answer gets put on hiatus. As far as I can tell, the confusion of the OP is exactly not-understanding the difference between the 2 types of averaging. $\endgroup$ – JEB Oct 9 '18 at 15:15
  • $\begingroup$ Yes, but you can use the definition of average velocity at the level of this question (total distance over total time) to explain this specific problem. This is the definition used in high-school physics (and introductory college level). $\endgroup$ – nasu Oct 9 '18 at 20:43

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