How do I prove that kinetic energy = 1/2 mv^2?

Answer 1

For an object in motion along the x-axis, the object is at the origin x₀ at time t = 0 and its velocity at t = 0 is v₀ . Acceleration is a. Average velocity is v_avg = (v_{0 +} v)/2

Substitute v = (v₀ + at) into x = v_avgt or (v₀ + v)t/2 to obtain x = (v₀ + v₀ + at)t/2 or x = (2v₀ + at)t/2 equal to v₀t+ at²/2.

Next, from v = (v₀ + at), write t = (v - v₀)/a and put this expression for t into x = (v₀ + v)t/2 .

Write x = (v₀ + v)(v - v₀)/2a or x = (v² - v₀²)/2a which gives v² = v₀² + 2ax.

Since work or energy is a force pushing an object of mass (m) through a distance (x) and force is mass (m) times acceleration (a), it follows that ke (kinetic energy) = (m)(a)(x) (mass times acceleration times distance). From v² = v₀² + 2ax, obtain x = (v²- v₀²)/2a and write ke = ma(v²- v₀²)/2a or m(v² - v₀²)/2. For v₀ = 0, this gives ke = mv²/2 or 1/2 of mv².