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Math Help - Velocity problem!

  1. #1
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    Velocity problem!

    In this problem I have a rocket that is accelerated from rest by thrust that varies with time according to: F=1000t, if 0<t<2.
    I need to find velocity equation for the rocket if mass remains constant and the drag force is equal to ten times its velocity. Mass of the rocket is 1600 lb.
    What I did so far:

    m(dv/dt)=F-kv

    After trying to separate variables I end up with:

    dv + (kv/m)dt = (1000t/m)dt

    As you can see I am not able to separate variables, and it does not look like diff. equation of the first order.

    Can you help me?
    Thank you.
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  2. #2
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    Is there any gravity? Is the rocket working against gravity? If so, are you assuming that the gravitational force is constant, or does the inverse square thing?
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    Is there any gravity? Is the rocket working against gravity? If so, are you assuming that the gravitational force is constant, or does the inverse square thing?
    There is no gravity. Information in my first post is all I have.
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  4. #4
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    It is a first-order linear equation for v(t). You can use the integrating factor method. What do you get?
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  5. #5
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    From this: dv + (kv/m)dt = (1000t/m)dt
    My P=k/m , Q=1000t/m
    I need to find e^integral Pdt.
    Do I need to substitute given values before integrating. m=1600 lb, and k=10, because I can not integrate k/m.
    Thank you.
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  6. #6
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    You don't need to substitute the given values before-hand. I'll bet you can integrate k/m. Isn't k a constant? And isn't m a constant? Therefore ...
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  7. #7
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    m(dv/dt)=F-kv

    dv/dt+(k/m)v=F/m (This fits into standard form of diff. equation of first order, where P=k/m, and Q=F/m)

    ∫Pdt=(k/m)t

    e^∫Pdt=e^(k/m)t

    ve^(k/m)t = ∫(F/m) e^(k/m)t dt

    After integrating right side, solving for v, and substituting in given values I get:

    v=100t+Ce^-t/160

    Does this make sense?
    Thank you.
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  8. #8
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    No, you have integrated as if F were a constant, then put in F= 1000t.

    F= 1000t is a function of t so that integral is 1000\int t e^{kt/m}dt which you can do by parts.
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  9. #9
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    Quote Originally Posted by HallsofIvy View Post
    No, you have integrated as if F were a constant, then put in F= 1000t.

    F= 1000t is a function of t so that integral is 1000\int t e^{kt/m}dt which you can do by parts.
    I see what you are saying.
    But, my Q is F/m, so would integral be 1000/1600\int t e^{kt/m}dt
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  10. #10
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    Quote Originally Posted by HallsofIvy View Post
    No, you have integrated as if F were a constant, then put in F= 1000t.

    F= 1000t is a function of t so that integral is 1000\int t e^{kt/m}dt which you can do by parts.
    Instead of integrating by parts I used Table of integrals to integrate.
    I used this form \int u e^{au} du = {e^{au} (au-1)}/a^2
    I came up with:
    v=100t-16000+Ce^{-t/160}
    How does it look?
    Thank you
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  11. #11
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    Plug your solution back into the DE and see if it satisfies it. This is a standard step to perform when solving any DE. It's usually quite straight-forward to do (differentiation is generally easier than integration), and it prevents mistakes. So, what do you get?
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