Fundamentals introduced the dot product's meaning as WORK:
  W = F · d = |F|·|d|·cos(θ)

The work-energy theorem connects total work done on an object
DIRECTLY to its change in kinetic energy, without needing to know
acceleration or time at all:
  W = ΔKE = ½mv_f² - ½mv_i²

This matters because it's a genuinely different tool than Fundamentals'
kinematics equations — useful specifically when you know FORCE and
DISTANCE but not time, or when force itself varies along the path
(where F = ma's instantaneous form is harder to apply directly than
integrating force over distance).

Given:  F = 5,000 N, d = 200 m, m = 2,000 kg, v_i = 0

Step 1 — compute work done:
  W = F · d = 5,000 × 200 = 1,000,000 J (1 MJ)

Step 2 — apply the work-energy theorem to solve for v_f:
  W = ½mv_f² - ½mv_i²
  1,000,000 = ½(2,000)v_f² - 0
  1,000,000 = 1,000 · v_f²
  v_f² = 1,000
  v_f = √1,000 = 31.62 m/s

Verified computationally: solving directly for v_f from the work-
energy relationship gives 31.62 m/s. Notice this required knowing
only force and distance -- not acceleration or time at all -- which
is precisely the situation where this theorem is the more direct
tool than Fundamentals' v = v0 + at.

Fundamentals' kinematics equations (v = v0 + at, x = x0 + v0t + ½at²)
ONLY work for constant acceleration -- a genuine simplification, since
real thrust decreases as propellant mass depletes, real drag changes
with speed, and real gravity changes with altitude.

For acceleration that varies with time, a(t), the exact relationship
is still v(t) = v0 + ∫a(t)dt -- but this integral doesn't always have
a clean closed-form answer. NUMERICAL INTEGRATION solves this by
breaking time into many small steps and approximating the integral
as a sum:
  v_new = v_old + a(t) · dt   (repeated for each small time step dt)

This is called EULER'S METHOD -- the simplest numerical integration
technique, and the direct computational version of "add up acceleration
over many tiny time slices" that a closed-form integral does exactly,
in the limit as dt approaches zero.

dt = 0.01
t, v = 0, 0
while t < 6:
    a = 40 - 2 * t
    v += a * dt
    t += dt
print(f"v(6) = {v:.2f} m/s")

Expected output: v(6) = 204.06 m/s

This case actually HAS a clean closed-form answer (verified
independently): a(t) = 40 - 2t integrates to v(t) = 40t - t²,
so v(6) = 40(6) - 6² = 240 - 36 = 204 m/s exactly -- the numerical
method's 204.06 m/s is very close (the small 0.06 m/s difference is
discretization error, from using small-but-not-infinitesimal steps
of dt = 0.01 s). This closed-form case is deliberately chosen so
the numerical method's accuracy can be checked directly against a
known-exact answer -- in a REAL scenario (genuinely irregular thrust
curves, drag that depends on a changing atmospheric density), no
closed-form answer would exist at all, and the numerical method would
be the only practical way to get an answer.