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Ratio of Power between Two Cars

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  1. Apr 27, 2016 #1
    1. The problem statement, all variables and given/known data
    A Grand Prix race car accelerates to twice the speed of the car in sample problem 1, in the same amount of time. Calculate the ratio of the power needed by the Grand Prix car to the power needed by the car in sample problem 1.
    Given: initial speed of car=0m/s,

    final speed of car in sample problem 1=22.7m/s, so speed of Grand Prix car will be 2(22.7)=55.6m/s,

    time=7.7s

    h=0, so there's no gravitational potential energy, only kinetic energy.

    power in sample problem 1=55kW=55000W

    2. Relevant equations

    W=ΔEg=1/2mvf^2-1/2mvi^2,

    After finding W, use P=W/t to solve for power and then write what the ratio between the power is
    3. The attempt at a solution
    I don't know how to solve this because we're not given the mass of the Grand Prix car. I can't think of a way to solve this problem.
     
  2. jcsd
  3. Apr 27, 2016 #2

    rcgldr

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    My guess is that you're supposed to assume the mass of both cars is the same, and the question only asks for the ratio of power, so the common mass term would drop out of the ratio.

    Other assumptions are also involved. For example, is the acceleration constant, so that the maximum power only occurs at the end of the time period, or is the power constant, which requires infinite friction at the start, or is it some combination of the two.

    What is the source of the relevant equations? There are alternative equations that could be used.
     
    Last edited: Apr 27, 2016
  4. Apr 27, 2016 #3
    I tried doing that, but I got the wrong answer. The answer is 1:4
     
  5. Apr 27, 2016 #4

    Simon Bridge

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    Considering the iplied level ... you are probably finding the ratio of average power, properly given:
    $$\bar P =\frac{\Delta E}{\Delta t}$$
    ... adjust your strategy: it is best practise do the algebra before you use numbers ... so work entirely in variables. Make sure you use a different variable for the things that are different.
    i.e. ##\bar P_1## would be the average power of the 1st car (the one in sample problem 1 say) and ##v_1## would be it's final speed.
    This will allow you to eliminate the variables that have the same value without having to know what they are.

    I take it you are given the mass of the car in sample problem 1, but not problem 2 ... you may assume that the grand prix car has the same mass (or near enough) or you can look up the regulation weight of a grand prix car and use that. Make sure you list your assumptions.

    I have a feeling that you are expected to use a shortcut here ... but since you didn't spot it, you should go through the long way.
     
  6. Apr 27, 2016 #5

    Simon Bridge

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    That answer suggests the shortcut I mentioned above... it is correct for assuming the cars are equal mass.

    Since you tried that, it means you did something wrong in your maths.
     
  7. Apr 27, 2016 #6
    I tried assuming that the masses of both cars are the same (1100 kg), but I got the wrong answer. The correct answer is 1:4
     
  8. Apr 27, 2016 #7

    Simon Bridge

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    See post #5 (above yours). Our posts crossed.
    1:4 is the correct answer if the cars have the same mass.
     
  9. Apr 27, 2016 #8
    I did Em=Eg=.5(1000)(55.6)^2. Em=1700248--plug that in for W: P=W/t. P=1700248/7.7=220811.4268.

    That would make the ratio between the cars 220811.4268:55000, which is wrong
     
  10. Apr 27, 2016 #9

    Simon Bridge

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    So the car in problem 1 reached 27.8m/s then? You definitely did something wrong ... maybe the probelm 1 power calc is wrong or there is a calculator button-press error somewhere?

    Do what I suggested and do the algebra first.
    If car 1 reaches speed ##v_1## in time ##T## then car 2 reaches speed ##v_2=2v_1## in time ##T## and the power ratio is given by:

    $$\frac{\bar P_2}{\bar P_1} = \frac{\frac{1}{2T}mv_2^2}{\frac{1}{2T}mv_1^2} = \cdots$$ ... and simplify.
     
    Last edited: Apr 27, 2016
  11. Apr 27, 2016 #10

    Simon Bridge

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    But: 0.5*1000*(55.6)^2 = 1545680 ... check your arithmetic.
     
  12. Apr 27, 2016 #11
    I tried this and I got the right answer because I ended up recalculating the power for the sample problem, which gave me a non-rounded value. The problem was that the sample problem in the book only writes the rounded value and that was what I used. Thanks !
     
  13. Apr 27, 2016 #12

    Simon Bridge

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    Continuing the algebra from post #9 ... the ##\frac{1}{2T}m## terms divide out leaving:
    $$\frac{\bar P_2}{\bar P_1} = \frac{v_2^2}{v_1^2} = \frac{(2v_1)^2}{v_1^2} = 4$$ so the ratio is: ##\bar P_1:\bar P_2 = 1:4##

    The shortcut is to realise that, since power depends on the square of the speed, doubling the speed will quadruple the power.
    Now you don't need to do so much maths in future. ;)

    Aside: the regulation minimum weight for a grand prix car is 750kg ... most actual cars weight less than this and the teams add additional weights to meet the regs, so ##m_2 = 3m_1/4## is reasonable... this means that the correct ratio should be 1:3 instead of 1:4... can you see how that is calculated without having to crunch all the numbers?
     
    Last edited: Apr 27, 2016
  14. Apr 27, 2016 #13
    Yeah, that makes sense. If you keep all other values the same but double the velocity, it'll result in a quadrupled power
     
  15. Apr 27, 2016 #14

    rcgldr

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    Formula 1 2016 season minimum "weight" limit is 702kg with driver, but not including the 100 kg max fuel at the start of a race (Formula 1 cars do not refuel during a race). As a further aside, a few years ago FIA banned the usage of depleted uranium for ballast (based on a cost issue, about $2750 USA per kg, so for a team with 3 cars at up to 75kg of ballast per car, that would be $618,750 USA).

    For those in the USA, grand prix is also used to refer to Indy Racing League cars, where the weight limit is 691kg, but without driver or fuel.
     
  16. Apr 27, 2016 #15

    Simon Bridge

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