How is the definition of acceleration equation (v_avg = v₀ + v_f / 2) derived?

The equation you've referenced, v_avg = (v₀ + v_f) / 2, is actually the formula for average velocity under constant acceleration, not the definition of acceleration itself—though the two are intimately related. This formula derives from the geometric fact that constant acceleration produces a linear velocity-time graph, and the average of a linear function equals the mean of its endpoints. Mathematically, if velocity increases uniformly from v₀ to v_f, the average of all instantaneous velocities equals (v₀ + v_f) / 2, just as the average of any linearly increasing sequence is the average of its first and last terms.

To derive this formally, start with acceleration's definition a = (v_f - v₀) / t and the definition of average velocity v_avg = displacement / time. For constant acceleration, displacement follows s = v₀t + ½at². Dividing by t gives v_avg = v₀ + ½at. But from a = (v_f - v₀) / t, we get at = v_f - v₀, so ½at = ½(v_f - v₀). Substituting this into the v_avg equation: v_avg = v₀ + ½(v_f - v₀) = v₀ + ½v_f - ½v₀ = ½v₀ + ½v_f = (v₀ + v_f) / 2. This proves that average velocity equals the arithmetic mean of initial and final velocities when acceleration is constant.

This relationship breaks down for non-constant acceleration, where the velocity-time curve is non-linear and the average requires integration rather than simple averaging. For a car that accelerates rapidly at first then gradually levels off, the simple (v₀ + v_f) / 2 formula overestimates average velocity because more time is spent near v_f than proportional distribution would suggest. This is why the formula includes the constant-acceleration assumption—it's a special case of a more general calculus relationship, but it's the case that applies to most introductory physics problems where acceleration remains uniform throughout motion.