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Dynamics: Acceleration as a function of position

  1. May 17, 2008 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

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


    3. 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.
     
    Last edited: May 17, 2008
  2. jcsd
  3. May 17, 2008 #2
    you tried d^2s/dt^2 = a way?

    integrate two times ...
     
  4. May 17, 2008 #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.
     
  5. May 17, 2008 #4
    Yes, it would work =P
    http://www.integral-table.com/integral-table.pdf
    see (32)
    might be a little mess ><
     
  6. May 18, 2008 #5
    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?

    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.
     
  7. May 18, 2008 #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.
     
  8. Aug 29, 2009 #7
    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.
     
    Last edited: Aug 30, 2009
  9. Aug 30, 2009 #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.
     
  10. Aug 30, 2009 #9
    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?
     
  11. Aug 30, 2009 #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.
     
    Last edited: Aug 30, 2009
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