- #36
brochesspro
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Can you open this link?brochesspro said:
What is the speed of the bicycle?
- Thread starter brochesspro
- Start date
In summary, you seem to have pasted in an image containing your work rather than typing in your work directly. This is a common mistake and can make it difficult to understand what you are doing.
- #37
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You need to post some Latex or, at least, an image of your work.brochesspro said:
- #38
brochesspro
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It has an image, if you can open it.
- #39
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You need to post your solution in this thread; not a link to an external site.brochesspro said:
- #40
brochesspro
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Ok, but is that a rule?PeroK said:
- #41
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Yes.brochesspro said:
- #42
brochesspro
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- #43
brochesspro
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Then what is the "Insert link" button used for?
- #44
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There are a few mistakes there. ##t_2 = 3s## is correct. The calculation of ##x_{t_2}## has gone wrong somewhere. You added an extra ##12s## for some reason(?).brochesspro said:
- #45
brochesspro
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I added the 12 seconds by mistake. I should have added 1 second for the phase where the speed is constant.
- #46
brochesspro
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I too feel I have gone wrong in that part.PeroK said:
- #47
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The way you did it, ##t_2## is the final time. You don't need to add anything.brochesspro said:
I don't see the error immediately - but I used ##T## to keep things simple.brochesspro said:
- #48
brochesspro
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Oh, I see, so I did a mistake in something basic.
- #49
brochesspro
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I think I neglected the constant ##C## while integrating for the first time for finding ##x_{t_2}##.
- #50
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You have ##a(t) = -6(t - 1)## (let's do the maths without units). So: $$v(t) = -3t^2 + 6t + k$$ where ##k## is the constant of integration. We know that ##v(1) = 12## (note that we do not have ##v(0) = 12##). So:brochesspro said:
$$k = 12 + 3 - 6 = 9$$ and $$v(t) = -3t^2 + 6t + 9$$ That gives ##v(3) = 0##. Integrating again:
$$x(t) = -t^3 +3t^2 +9t + l$$
I think that's where you went wrong, by losing the ##9t## term.
- #51
brochesspro
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Never mind, I saw your post.
- #52
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The constant of integration in the equation for ##x(t)## doesn't matter, as you are calculating the distance traveled between ##t = 1## and ##t =3##.brochesspro said:
- #53
brochesspro
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After integration, I got ##x_{t_2}-x_{t_1}=16## after solving and after substituting I got ##x_{t_2}=28m## and thus the speed of bicycle is ##9.3 m/s## as per the question.
- #54
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The bicycle is another ##17m## ahead to begin with. It must be Mark Cavendish after all!brochesspro said:
- #55
brochesspro
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Oh yeah, so I get the average velocity as ##15 m/s##.
- #56
brochesspro
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Thanks a lot.
So, how do I close this thread?
So, how do I close this thread?
- #57
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It doesn't need to get closed. It can stay as it is.brochesspro said:
- #58
brochesspro
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Oh, I see. Thanks again.