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Homework Help: Tanker in water with drag

  1. Nov 7, 2009 #1
    1. The problem statement, all variables and given/known data
    The tanker has a weight 4000 MN and is traveling forward at speed 1m/s in still water when the engines are shut off. If the drag resistance of water is proportional to the speed of tanker at any instant and can be approximated by F=0.65v MN, determine the time needed for the tanker's speed to become 0.5m/s?

    2. Relevant equations
    F=0.65v, F=ma
    a = dv/dt

    3. The attempt at a solution
    [tex]F=ma \Rightarrow a = \frac {F}{m} \Rightarrow a = \frac {-0.65v}{\frac {4000}{9.81} } \Rightarrow
    a = \frac {dv}{dt} \Rightarrow dt = \frac {dv}{a} \Rightarrow \int dt= \int \frac {dv}{a} = \int \frac {dv}{\frac {-0.65v}{\frac {4000}{9.81}}} \Rightarrow t = 434 s [/tex]

    4. Extra
    The correct answer should be 44.3s. The solution manual adds an extra "g", making
    [tex] a = \frac {-0.65v * 9.81} { \frac {4000}{9.81} } [/tex]
    I don't understand why there's an extra "g"
    Last edited: Nov 7, 2009
  2. jcsd
  3. Nov 7, 2009 #2


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    Your solution is correct.

  4. Nov 7, 2009 #3
    I get 434 seconds, but the textbook's answer is 44.3 seconds... Is there something to this?
    The way I did it seems reasonable, but it's incorrect (as the book is concerned); I'm also curious as to why the book adds an extra "g" (mentioned on my question)
  5. Nov 7, 2009 #4


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    There can be mistakes in a textbook. There can not be two g-s. Even if the tanker kept its initial deceleration, 313 s would be needed to slow to half its initial speed. But the deceleration decreases with speed, so this time should be longer.
    It is also useful to check dimensions if you are in doubt.
    The dimension of your "a" is really in m/s^2, that of the book is m^2/s^4.

  6. Nov 7, 2009 #5
    I am not sure but if you are integrating dt with respect to velocity then the result is not time but displacement
  7. Nov 7, 2009 #6
    Hmm. But doesn't explain why in the solution manual there's an extra "g"

    I'm kinda confused with the unit checking part (with the integrals). According to the solution manual, they set "a" as
    [tex] a(v) = - \frac {kN* \frac {s}{m} * \frac {m}{s^2}}{kg} * v[/tex]
    I'm puzzled about the unit "v" (last variable), is it unit-less or is it m/s as it usually is?

    If assuming "v" is unit-less, then a(v)'s unit will then become
    [tex] a(v) = - \frac {kN* \frac {s}{m} * \frac {m}{s^2}}{kg} = - \frac {kg * \frac {m}{s^2} * \frac {s}{m} * \frac {m}{s^2}}{kg} = - \frac {m}{s^3}[/tex]
    which isn't a unit of acceleration

    and even if "v" is m/s in the equation, a(v) is
    [tex] a(v) = - \frac {kN* \frac {s}{m} * \frac {m}{s^2}}{kg} = - \frac {kg * \frac {m}{s^2} * \frac {s}{m} * \frac {m}{s^2}}{kg} * \frac {m}{s} = - \frac {m^2}{s^4}[/tex]

    Is there anything wrong with my derviation?
    Last edited: Nov 7, 2009
  8. Nov 8, 2009 #7


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    The unit of v must be m/s, as it is speed. The manual got "a" in m^2/s^4, which is obviously wrong. You are right, trust in yourself! :).

  9. Nov 8, 2009 #8
    Haha, ok ok :)
    I guess this problem is solved then, thanks everyone :)
  10. Nov 8, 2009 #9
    I dont know if you can solve for dt but even if you could I have a problem with your solution: You found t=434s. With my way I find t=418s..
    I dont know how to write the maths in Latex so it will be a little messy:

    F=0.65u => ma=0.65u. Now lets assume that at t0=0 the engines stop. We are looking for t1
    So if we integrate this expression with respect to t and from t0=0 to t1 we get

    mu(t1)=0.65x(t1) => x(t1)=2000/0.65*9.81= 313.65m that is the distance it travelled during the deceleration.

    Now assuming you are correct we can find the average Force acted on the body:

    u(t)=u(t0) - at solving for t=t1 we find the avg acceleration: a=-0.0011 m/s^2 and the avg force: F=-0.4698 MN

    Now that we know the F and the distance traveled we can find the Work done by the force:

    Wf= F*d => Wf=-147.03MJ

    However, we can find the work done by the force from Wf=ΔK => Wf= 1/2m(u(t1)^2-u(t0)^2) => Wf=1/2m(0.5^2 - 1^2)= 1/2m(-0.75)=152.90MJ which is not what we found before..

    I am not sure whether I am right but I think that proves you are wrong ( I can be wrong because I am not sure about that thing with the avg force ).

    However, from x=313m if you solve:

    1) u(t)=u(t0)-at (assume t=t1 and solve for a)
    2) y= u(t0)t-1/2at^2 (substitute the acceleration with what you found above)

    The time will be 418s which results to a Work done 152.90MJ.

    IF I am right ( which I dont really think so) I think that your mistake is that you solved for dt. I am not sure if this can happen.

    First Post here : ) Please correct me If I am wrong..
  11. Nov 8, 2009 #10


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    The work with average force is a good approximation sometimes, but it is an approximation. Silentwf presented the exact solution. I write it in a bit more comprehensive form.

    a = \frac {F}{m}= \frac {-0.65v}{4000/g}

    as a=dv/dt, we have a diffrential equation for v:

    \frac {dv}{dt}= \frac {-0.65gv}{4000}

    The standard method of solution for such type of differential equations is to rearrange them so that identical variables are at one side, and integrate:

    \int_{v1}^{v2}{\frac {dv}{v}}=\frac {-0.65g}{4000} \int_{t1}^{t2}{ dt}

    The result is

    ln{(\frac {v_2}{v_1})}=\frac {-0.65g}{4000} (t_2-t_1)

    As v2/v1 =0.5, t2-t1 = 434.8 s.

  12. Nov 8, 2009 #11
    I only have very basic knowledge of integrals ( 1st year in Physics ). I cant understand why its an approximation in this situation. I base the force on actual numbers. Also if you are right I cannot see why the Work done is different from what I find. In my method I try to find the force that if it substitutes the real force can have the same result. I cannot see why this time is an approximation.

    Just trying to get the whole thing..
  13. Nov 8, 2009 #12


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    How do you calculate average acceleration?

  14. Nov 8, 2009 #13
    Echild i thonk you have a mistake.The units are wrong in the equation you wrote when you integrate.In the one side of the equation where you integrate velocity with respect to velocity you have m^2/s^2.Whereas in the second where you integrate time with respect to time you have m.
  15. Nov 8, 2009 #14
    I found the distance moved during the deceleration: from the definite integral of ma=0.65u with respect to time and from t0=0 to t1 (which is acurate).

    Then knowing the distance and both the initial and final speeds you can safely state that you can find the acceleration which could be responsible for the exact type of motion. ( hope you understand what I mean.. dont know how to state it better )

    That is from:
    u(t)= u(t0) - a*t and
    y=u(t0)*t - 1/2a*t^2

    This will not result to approximate numbers because it is based on accurate calculations. The only assumption is that we assume a stable acceleration and therefore a stable force ( even if we know it s not ) that would have the same results as the force given. The results are not based on any other assumption. Also the units dont seem very ok with your integrals..
  16. Nov 8, 2009 #15


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    your equations are valid when the acceleration is constant. It is not true in this problem.

  17. Nov 8, 2009 #16
    I know that. *But* the point of this is not to find the Actual acceleration of the tanker.
    The point is to find an acceleration that can substitute during the whole motion the true acceleration of the tanker. Thats why I use them. Its like trying to create an equal situation which can be described by these equations and has the same results as the actual problem
  18. Nov 9, 2009 #17


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    Yes, but your method does not give the same result as solving the original problem directly which is quite easy if you know a bit about differential equations.

    You have to use calculus in Physics, and you will learn it soon.

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