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Energy and acceleration

  1. Aug 25, 2014 #1
    1. The problem statement, all variables and given/known data
    The chemical potential energy in gasoline is converted to kinetic energy in cars. If a car accelerates from zero to 60 km/h, compared to the energy necessary to increase the velocity of the car from zero to 30 km/h the energy necessary to increase the velocity of the car from 30 to 60 km/h is:


    2. Relevant equations
    K = 1/2 mv^2

    3. The attempt at a solution

    I approached this problem in a conceptual manner. If we are to compare the energy from 0 to 60, it wouldn't matter if we go 0 to 30 or 30 to 60 since acceleration is not factored in. Thus it would be the same. However, the correct answer is 3 times as great. How is this possible?
     
  2. jcsd
  3. Aug 25, 2014 #2

    DrClaude

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    Staff: Mentor

    You have to compare the energy required to bring the car from 0 km/h to 30 km/h, versus the energy required to take it from 30 km/h to 60 km/h.
     
  4. Aug 25, 2014 #3
    So, that would be K = 1/2 m (30)^2 = 450 m; and then K = 1/2 m (30)^2 = 450 m. This makes no sense!
     
  5. Aug 25, 2014 #4

    DrClaude

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    Staff: Mentor

    Of course, if you're calculating the same thing twice, you get the same number :wink:

    I guess that you are putting in the second equation (60 - 30)2 for the velocity, but that is not correct. What you need to calculate is the difference in kinetic energy,
    $$
    K(v = 30\ \mathrm{km/h}) - K(v = 0\ \mathrm{km/h})
    $$
    compared to $$
    K(v = 60\ \mathrm{km/h}) - K(v = 30\ \mathrm{km/h})
    $$

    To be clear, ##\Delta K \neq \frac{1}{2} m (\Delta v)^2##.
     
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