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Homework Help: Modelling fuel economy and more economical options

  1. Nov 12, 2012 #1
    Consider a situation in which two motorists, Arwa and Bao, share the same driving route but own different sized vehicles. Arwa fills up her vehicle’s tank at a station along her normal route for US$ p(1) per litre. On the other hand, Bao drives an extra d kilometres out of his normal route to fill up his vehicle’s tank for US$ p(2) per litre where p(2) < p(1) .

    “Effective litres” is a method of comparing the cost of petrol bought under the two options
    described above. Effective litres, for a given vehicle, are those litres used when the vehicle travels its normal route (not including the extra distance travelled in option 2).

    1.) Use your variables and parameters to write algebraic expressions for E(1) and E(2) which
    represent the cost per effective litre under options 1 and 2 respectively.

    2.) Write a model that helps motorists decide on the more economical option for their

    3.) Use your model to find the farthest distance that Bao should drive to obtain a 2 % price

    4.) Investigate the relationship between d and p(2) when E(2) is kept constant (e.g. US$0.90,US$1.00, … etc.). Use technology to draw a family of curves for Arwa’s vehicle. Repeat for Bao’s vehicle.

    5.) For E(2) = US$0.90, provide Arwa with information on three different stations that yield
    this same cost per effective litre. Discuss how such information may be useful to Arwa.

    6.) For E(2) = US$1.00 and p(2) = US$0.80, compare the maximum distance that each motorist should drive and still save money.

    7.) Modify your model to account for the time taken to drive to and from an off-route station. Clearly justify any assumptions you make.


    My variables:
    p(1) = $1.00
    p(2) = $0.98
    d = 10km

    My parameters:
    Normal route = Dkm
    Fuel economy of Arwa = FE(A)km/l
    Fuel economy of Bao = FE(B)km/l

    E(1) = p(1) [since E(1) = (p(1)*D)/D = p(1)]

    E(2) = (p(2)[D+d])/D


    If E(2) < E(1), then E(2) is more economical.


    ([(p(1)*Total effective litres used in option 1) - (p(2)*Total effective litres used in option 2)]/(p(2)*Total effective litres used in option 2)) x 100% = 2. Then solve this for d.

    I'm not sure how to do 4,5, 6 and 7. For 4 I imagine you pick several values for E(2) and then them to plot a family of graphs? The others i have no clue!

    Sorry for the long post, ask if you need any more info. Don't worry if you don't have a definite solution, any pointers in the right direction would be just as welcome, thanks.
  2. jcsd
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