Constant Acceleration

From force to position

1. Newton's second law

F₀ = m dv/dt,   a₀ = F₀/m

2. Integrate once and apply the initial velocity

v(t) = a₀t + C₁

At t = t₀, we require v = v₀. Therefore C₁ = v₀ − a₀t₀.

v(t) = v₀ + a₀(t − t₀)

3. Integrate again and apply the initial position

x(t) = ½a₀t² + C₁t + C₂

Requiring x(t₀) = x₀ gives C₂ = x₀ − v₀t₀ + ½a₀t₀².

x(t) = x₀ + v₀τ + ½a₀τ²,   τ = t − t₀

The constant position is where we start. The linear term is the motion we already had. The quadratic term is the additional displacement caused by the force. Its factor ½ comes from integrating the linear velocity increment.

A checkpoint at clock time 5 s

With +4 N, only 3 s have elapsed: v = 9 m/s and x = 19 m. With −2 N, the particle is momentarily at rest at x = 5.5 m, but its acceleration remains −1 m/s². It then reverses. After 6 s elapsed it returns to 1 m: zero displacement, yet 9 m traveled.

In the coasting case, zero net force leaves velocity unchanged. The equal and opposite third-law force acts on the force-producing agent, not on this particle.

Full LaTeX derivation includes the double definite integral, the two constants, dimensions, work–energy checks and the computational method.