How Much More Energy to Accelerate a Car from 30 to 60 km/h?

[brake4country]

Homework Statement

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:

Homework Equations

K = 1/2 mv^2

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?

 

[DrClaude]

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?

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.

 

[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!

 

[DrClaude]

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!

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