# Simple Question about Acceleration and Conservation of Energy

This was simply a question that came up during physics class:

What takes more energy?

$$1)$$ accelerating a vehicle from $$0$$ to $$30$$ mph, or

$$2)$$from $$30$$ to $$60$$ mph?

I thought both required the same energy, but I guess I was wrong, the energy required being $$\dfrac 12m(30)^2$$ for the first and $$\dfrac 12m(60^2-30^2)$$ for the second.

However, I'm still confused. What if you had an electric car (so mass is constant). It has a mass of 1 (keeping things simple.) If you have no external reference frame, and no idea how fast you're going, but know the energy level of the battery to be 450 units, won't you always accelerate by $$30$$ mph (as $$\dfrac 12(1)(30)^2=450$$)? (assume magical transmission + no friction or drag)

Now imagine it has $$900$$ units of energy in the battery. You accelerate to 30 mph. You now have $$450$$ units of energy in the battery (as $$\dfrac 12(1)(30)^2=450$$). Both a stationary person and a person in a chase vehicle at $$30$$ mph would agree on this, right?

Now there's $$450$$ units of energy left in the battery. You are in your car. Velocity is arbitrary, you can claim to be going $$0$$ mph for all you care. The energy source is in the same reference frame as you are. So can't you add another $$30$$mph to your velocity? Why or why not? Won't the same energy result in the same acceleration?

• "However, I'm still confused. What if you had an electric car (so mass is constant)." What make's you think that the car has to be electric in order for its mass to be constant? Commented Nov 6, 2019 at 0:21
• Well, a car consuming fuel would lose weight as it consumed energy, which would complicate things. I just say an electric car to make it clear that the consumption of fuel isn't a factor. Commented Nov 6, 2019 at 19:07
• OK but you probably should have said so. But I'm not sure it was necessary to consider mass anyway. It is the same mass going from 0 to 30 as it is going from 30 to 60, and you are not comparing an electric with a gasoline operated car. Commented Nov 6, 2019 at 19:19

You can't just switch reference frames here because of two things: 1) the car is accelerating, so it's not an inertial reference frame, and 2) the two reference frames aren't the same.

In the first case (car goes from 0 mph to 30 mph), in the initial state, the car is 0 mph, so an observer in the car will observe that he is stationary with respect to the rest of the world. In the second (car goes from 30 mph to 60 mph), in the initial state, an observer in the car will see the rest of the world going at -30 mph. As he accelerates to 60 mph, he's seeing the rest of the world accelerating to -60 mph. He can feel himself accelerating, so he knows it's not the rest of the world accelerating backwards. The kinetic energy equation still works, and there's no contradiction.

For the intermediate questions: yes, it takes more energy accelerating from 30 mph to 60 mph than from 0 mph to 30 mph. And yes, both a stationary person and a person in the vehicle would agree that it takes 450 units of energy to accelerate from 0 to 30 mph.

I thin you sort of have two different questions here. Given that it takes a different amount of energy to go from $$0$$ to $$v$$ than it does from $$v$$ to $$2v$$:

• Why is this, when we can just shift to a different reference frame and again see the car starting at 0?
• Where does the energy from the battery go if we shift to a different reference frame?

Yes, we can shift to a different frame, but it turns out that we can't consider the car in isolation. Since the car is changing speed by pushing on something (the ground), the speed of the ground becomes important. Rather than thinking about how fast the car is going we should be thinking about the difference in velocity between the car and the earth.

We could pick a frame where the car is moving at super high speed and has an enormous $$KE$$ in that frame. But without being able to "slow down" the car to zero in that frame, we can't extract the energy. What the battery is doing is creating a gradient (difference of speeds). And the energy stored in this gradient is not linear. The absolute speed of the car doesn't matter.

So if we have a constant amount of energy in the battery, why does the $$KE$$ for the car change by different amounts in different frames? Because the car is not the only partner. Again, the car and the ground are coupled together for this interaction, and both can change energy. If we assume no losses, then:

$$E_{battery} = \Delta KE_{car} + \Delta KE_{earth}$$

Earth is so big that we cannot measure the velocity change, but it is present and when multiplied by its mass (or moment of inertia), it can yield a macroscopic energy change. In the frame where the earth is at rest, the change in energy on the earth is nearly zero. It's easy to think of all the battery energy going into the car. In other frames, that's not true. In a frame where the car is moving forward at high speed, the earth would lose $$KE$$. In that frame the car would gain energy from the battery, plus any energy lost by the earth slowing.

Yes, more Work is needed to be done in order to accelerate the vehicle from $$30-60$$ mph than $$0-30$$mph: Change in Mechanical Energy (kinetic energy in this case), on the absence of friction forces, is equal to the amount of work needed to do this change.

Also, the idea of two observers on two different frames measuring the same Energy level of the battery at a given time is in contradiction with the -special- Relativity principles. If we call Event 1 that of observer 1 measuring the battery to have 450 units, and Event 2 that of observer 2 measuring also 450 units, then these two events cannot be simulatenously occurring, observer 1 (the one on the car) will see that observer 2 made the measurement a time slightly after him; observer 2 will see that observer 1 made the measurement some time slightly before him.

You had it right in your first paragraph, which is why many truckers get in trouble when they go too fast. Energy is proportional to the square of speed, speed is not arbitrary, it is referenced to 0 mph, standing still.

Energy comparison, proportionally, is easy, simply square your speed and compare.

30 × 30 = 900, 60 × 60 = 3600. 900/3600 = 1/4 = 25%

And there is your answer, it takes 4 times more energy input to reach 60 mph from 0 mph (and more to maintain it due to increased drag and friction).

The brakes get 4x hotter slowing you back down to 0 mph.