However, when the object is already faster, it travels a longer distance per unit time, and $dE = \vec F\cdot d\vec x$ where $d\vec x = \vec v\cdot dt$. The force $\vec F = m\vec a$ is the same, regardless of the velocity, for a fixed acceleration.
One needs to spend more energy to accelerate a faster object by $\delta v$ because of your derivation based on kinetic energies (which go like the velocity squared, and that's more quickly increasing than proportionality $v$) or, equivalently, because the work per unit time (power) is Then spaceship will achieve max possibly velocity, and working engine will do nothing at all (i.e. On this logic, if we turn on engine forever, spaceship will lose acceleration with time, and in time converging to infinity acceleration will become 0. What is difference for engine, moving spaceship with velocity 0, or 20, or some another? $\Delta v$ will depend on velocity at moment of engine turning on (because $\Delta W$ depends on it and A is const). Question: why $W_3-W_2 > W_2-W_1$? I intuitively feel that if we turn engine at any moment on some time t = const (so, engine will make some work A = const), spaceship velocity will change to some $\Delta v$ = const. Then, it turns on its engine, and changes velocity from 0 to 20 (delta v = 20). At beginning, it has velocity 0, and kinetic energy 0. Imagine spaceship in vacuum with mass = 1.
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