Kinetic energy, work and the marginal cost of velocity

Factual Questions
1

Perhaps this is bleedingly obvious to those with some familiarity with physics but I wanted to check if I have something right.

Say I want a car to go at 25mph and going at 25mph requires 1 unit of work.

If I want the car to go at 50mph, I will need 4 units of work, correct?

If I want the car to go at 100mph, I will need 16 units of work, still correct?

Does this not imply then that the marginal cost of velocity from 25mph to 50mph has been 3 additional units of work? This means that while the first 25mph cost 1 unit of work, the second 25mph cost 3 which is to say that the marginal cost goes up at the square of 1.73?

Say I wanted to know exactly how much additioanl work is required to go from 25mph to 26mph or from 25mph to 30mph, what would be the formula that would tell me the marginal cost of the velocity increase?

2

Assuming you’re speaking only in terms of ideal physics without resistance the energy (or work, same difference) required to reach a certain velocity from zero goes up with the square of the velocity. Choosing appropriate units E=1/2 m v^2
To get the marginal cost of velocity increase you derive that:

marginal cost of velocity increase = mv

So the marginal cost of velocity increase is proportional to the velocity.

m’s a constant so I’ll ignore that (or pretend the mass is 1 and my units “special”)

v = 100, E = 5000
v = 101, E = 5100.5
increase 100.5
v = 102, E = 5202
increase 101.5
v = 103, E = 5304.5
increase 102.5

v = 1000, E = 500 000
v = 1001, E = 501 000.5
increase 1000.5
v = 1002, E = 502 002
increase 1001.5

The marginal cost of velocity increase doesn’t tell you exactly how much additional work is required to go from 25-26 though, and definitely not from 25 - 30. It’s an approximation, and with the increase being 4% (or worse, 20%) of the current velocity, not that good a one. For the exact answer you have to do what you did initially. Calculate the energy at each velocity and subtract.

3

This is what calculus was invented for. As stated above, the kinetic energy is (mv^2)/2. The rate of change of energy with respect to velocity (the “marginal cost of velocity” in the OP’s parlance) is just the derivative of (mv^2)/2 with respect to v, i.e. mv, as shown by naita.

4

0.5 x m x (v[sub]2[/sub][sup]2[/sup] - v[sub]1[/sub][sup]2[/sup])

5

As noted, this ignores all other forces. In reality, friction and drag would be significant factors, becoming more significant the higher the speed. And drag scales like the square of the speed.

6

I’m hoping the OP has some follow-up questions, because some of those will have interesting answers, but I’ll just confuse things if I give them now.

7

Chronos,

I was mainly thinking in terms of what that mean for the additional amount of propellant and barrel length required for a projectile to gain velocity.

If you can think of follow-up questions which aren’t coming to me but which would have interesting answers, please go ahead.