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Dynamics: Acceleration as a function of position

  • Thread starter kirab
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Homework Statement



The acceleration of a rocket travelling upward is given by a = (6 + 0.02s) m/(s^2), where s in in meters. Determine the rocket's velocity when s = 2 km = 2000 m and the time needed to reach this altitude. Initially, v = 0 and s = 0 when t = 0.

Homework Equations



Given: [tex] a = 6 + 0.02s, v_{0} = 0, s_{0} = 0 [/tex]
Find: [tex] v_{|s=2000}, t_{|s=2000} [/tex]


The Attempt at a Solution



Finding the velocity when the rocket is 2000 m from its starting position didn't give me many problems. I did the following:

[tex] a ds = v dv [/tex]

[tex] (6+0.02s)ds = v dv [/tex]
[tex]\int (6+0.02s)ds = \int v dv[/tex]

Note: I don't know how to do bounded integrals on Latex so the first integral goes from 0 to s, whereas the second integral goes from 0 to v.

[tex]\frac{v^2}{2} = 6s + 0.01s^2[/tex]

Which leads to:

[tex] v = \sqrt{12s+0.02s^2}[/tex]

and [tex] v(2000) = 322.6 m/s [/tex] which is correct.

As for the second part (finding the time t, when s = 2000 m again), this is where I need help:

[tex] v = \frac{ds}{dt} \Rightarrow dt = \frac{ds}{v}. [/tex] Meaning
[tex]\int dt = \int \frac{ds}{v} \Rightarrow [/tex]
[tex]\int dt = \int \frac {ds}{\sqrt{12s+0.02s^2}}[/tex]
(Integral goes from 0 to t, and from 0 to s, respectively.)


This is where I'm stuck. I looked in the Math appendix of my Dynamics book where I got this problem, and according to it, I should have a negative square root when the integral has been evaluated. However, I don't think that should be the case if I did this problem properly.

Any suggestions?

BTW, I'm sorry for the way the formulas are written and my half-use of Latex; I'm sure it could look more readable but it's my first time using Latex so I'm not 100% familiar with it.

Thanks in advance.
 
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Answers and Replies

  • #2
378
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you tried d^2s/dt^2 = a way?

integrate two times ...
 
  • #3
Redbelly98
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Not absolutely sure this will work, but how about completing the square of that quadratic expression? (0.02 s^2 + 12s)

That might get it to look like something in standard integral tables.
 
  • #4
378
2
Not absolutely sure this will work, but how about completing the square of that quadratic expression? (0.02 s^2 + 12s)

That might get it to look like something in standard integral tables.
Yes, it would work =P
http://www.integral-table.com/integral-table.pdf [Broken]
see (32)
might be a little mess ><
 
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  • #5
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you tried d^2s/dt^2 = a way?

integrate two times ...
I wouldn't be able to do that unless I know position as a function of time (in other words, s(t)).
But that's where I'm stuck. If I could solve that integral then I would have s(t). Or am I missing something here?

Not absolutely sure this will work, but how about completing the square of that quadratic expression? (0.02 s^2 + 12s)

That might get it to look like something in standard integral tables.
If I remember how to complete the square correctly, would [tex] 0.02s^2 + 12s = (0.02)((s+300)^2 - 90000) [/tex]? (Factoring it out yields 0.02s^2 + 12s so I think it's right).

However, I don't see the relationship between that equation and (32) on the integrals table. But I see a connection between [tex] 0.02^s + 12s [/tex] and (39), I'll try it out and post back.
 
  • #6
Redbelly98
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Integral (39) will work too.

Substituting x = (s+300), the completed-square version (that you did correctly) simplifies to the form in # 36.
 
  • #7
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Integrate a = dv/dt to get at = v-v(initial). Solve for t and plug in the numbers.

You should get t = 323 / (6 + 40) = 7 seconds.

You would normally be able to solve the integral you have in the end with basic trig substitution (calculus 2) but since you have an s in front of the 12 you can not use the arctangent. I did not bother working the integral so I am not sure if it will also produce 7 seconds. Although the answer is 7 seconds.
 
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  • #8
Redbelly98
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You might take note of the date this question was asked. It's a good bet that the OP is not actively trying to solve this one anymore.
 
  • #9
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That is a likely possibility, however, there are other students that may be in need of such information and look at older postings as well. This may not help this ONE student in particular, but it may help many others. Why would the older forum threads still be in existence if they were not open to be viewed for other students' benefit?
 
  • #10
Redbelly98
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Possible, yes. But there are so many such older postings, is it worth ones time to add hints to them because somebody else might not be able to solve it from the help that has already been given?

Do what you like, but I would leave it up to other students to come along and ask in those particular threads that they are curious about.

EDIT: Regardless of that, in the coming weeks this forum will become plenty busy with new homework help requests.
 
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