# Energy and acceleration

1. Aug 25, 2014

### brake4country

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. Aug 25, 2014

### 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.

3. Aug 25, 2014

### brake4country

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!

4. Aug 25, 2014

### Staff: Mentor

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

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, $\Delta K \neq \frac{1}{2} m (\Delta v)^2$.