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Rocket Equation-esque Problem

  1. Sep 17, 2015 #1
    1. The problem statement, all variables and given/known data

    A truck hauling a big tank of oil starts at position x=D (Chicago) and heads due west (–x direction) toward its destination at x = 0 (Des Moines). At Chicago, the total mass of the loaded truck is M and the mass of oil it is carrying is λM. (Thus M(1–λ) is the "tare" mass = the mass of the truck when it is empty.) The driver starts from rest at time t = 0 with his engines set to deliver a constant force of magnitude F throughout the trip.
    Unfortunately, the trucker's oil tank is leaking: it is losing oil at a constant rate-per-unit-distance of dm/dx=λM/D
    Ignore the small change in m from the truck's consumption of gasoline (it's tiny compared to the truck's mass).

    (a) Calculate m(x) in terms of x and the given constants D, M, λ, and/or F.

    (b) Calculate the truck's kinetic energy as a function of x and the given constants.

    (c) At what position x does the truck reach it's maximum speed and at what position does it reach it's maximum kinetic energy?

    2. Relevant equations
    dm/dx=λM/D

    F=dp/dt

    T=.5mv²
    3. The attempt at a solution
    I think I've got part a right as λ, M, and D are constants so:
    m(x)=λMx/D+(1-λ)M
    and I checked it where x=D the mass is M, so I think that is the right answer.

    As for parts b and c , I honestly have no idea where to begin solving for velocity. I understand to find T that I'll have to use my m(x) but I don't know what force equation to set up to solve for velocity.
     
  2. jcsd
  3. Sep 17, 2015 #2

    DEvens

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    That's a weird way of setting up the problem, having x decrease as you travel. Oh well. You seem to have that part ok.

    For the other parts: The problem statement does not explicitly tell you how the constant force of the engine changes the speed at which the truck moves. You will need to make some assumptions. One possible assumption: The friction force with the road is some function of the mass of the truck. So the lighter the truck the faster it goes for a given engine force. You could assume a simple linear dependence. The speed of the truck when it is full tells you the proportion. This is probably not particularly accurate compared to a real truck. But hey, how long could one really drive an oil truck spewing oil out the back on interstates between Chicago and Des Moines and not have the police pull you over?
     
  4. Sep 17, 2015 #3

    mfb

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    It will reach its maximum speed shortly after a police car discovers it?
    Yeah, that is weird. Also, I don't see any interpretation that would stop it from getting faster and faster until it reaches its destination.
     
  5. Sep 17, 2015 #4
    If you make the substitution dv/dt = (dv/dx)(dx/dt) = vdv/dx in the force eqn. you can equate an integral over dx to an integral over dv, i.e. ∫f(x)dx = ∫g(v)dv
     
  6. Sep 17, 2015 #5

    DEvens

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    Well, yeah. But that's the rocket equation. This is an oil tanker losing mass. One has to expect that the force the oil generates spraying out the back is not making a significant difference to the speed of the truck.
     
  7. Sep 17, 2015 #6
    I disagree. F0∫dx/m(x) = ∫vdv makes no reference to dm/dt or relative velocity of accreted mass.
     
  8. Sep 17, 2015 #7
    Using this I got an answer for kinetic energy that I'm pretty sure is right.
    T=(Fx+FD(1-λ)/λ)ln(DM(1-λ)+λMx)
    If anything, all the units work out right.
    Then I'm pretty sure to find maximum kinetic energy I need to take the derivative of T with respect to x and set it equal to zero and that should give me the position.
     
  9. Feb 9, 2017 #8
    I'm quite late to this discussion but I'll give advice anyways because other people might be searching for help on this problem. The equation T=(Fx+FD(1-λ)/λ)ln(DM(1-λ)+λMx) can't possibly be correct. At x = D, T must be equal to 0 and in this equation it would be T(x = 0) = (FD/λ)ln(DM).

    Since there is constant force the work is simply the usual F*dr integral but in the negative direction so you end up with w = F(D - x). This is equal to the total change in T which is .5[λMx/D+(1-λ)M]v^2

    If you like you can solve for v and get the derivative but the result is just that the velocity and kinetic energy are maximum at the finish point, x = 0. This is because, since the force is constant, the decreasing mass results in increasing acceleration the whole way.

    The problem is a poor one, one can guess the result easily without doing any math at all, except its such a strange scenario that its confusing. The result is also not interesting and so simple that it seems like its wrong.
     
  10. Feb 9, 2017 #9
    Sorry, I typed T(x = 0) which should be T(x = D).
     
  11. Feb 10, 2017 #10

    haruspex

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    I believe the leaked oil would have left with some of the KE it had acquired.
     
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