Constant-acceleration equations

Constant acceleration means dv/dt = a, with a the same at every instant of the interval. Integrate from the clock reading t₀, where the velocity is v₀, to a later time t: ∫ dv from v₀ to v equals ∫ a dt from t₀ to t, so v − v₀ = a(t − t₀). The lower limits are where the initial condition enters; integrate without them and you get v = at + C with C unknown. The result is a straight v–t line of slope a, so equal time intervals bring equal changes in velocity. A cyclist rolling at 3.0 m s⁻¹ who accelerates at 0.60 m s⁻² for 8.0 s reaches 3.0 + 0.60 × 8.0 = 7.8 m s⁻¹. Most problems set t₀ = 0 and write v = v₀ + at, which is right only when the clock starts at the known state. A sensor that first sees the cyclist at t = 2.0 s needs t − 2.0 s in place of t.

Now v = dx/dt = v₀ + a(t − t₀). Integrating from t₀, where the position is x₀, gives x = x₀ + v₀(t − t₀) + ½a(t − t₀)². The graph says the same thing without symbols: the area under the v–t line over an interval Δt is a rectangle of height v₀ plus a triangle of base Δt and height aΔt, so Δx = v₀Δt + ½aΔt². For the cyclist that is 3.0 × 8.0 + ½ × 0.60 × 8.0² = 24.0 + 19.2 = 43.2 m. Two integrations need two initial conditions, and that is physics, not bookkeeping: d²x/dt² = a is satisfied by x₀ + v₀t + ½at² for every choice of x₀ and v₀, and only the initial conditions select the one your object follows. Two carts with the same a but different launch speeds obey the same differential equation and share nothing else.

Five quantities describe an interval: Δx, v₀, v, a and Δt, and each standard relation leaves exactly one out. Solve the velocity equation for Δt = (v − v₀)/a, substitute into the position equation, and time disappears: v² = v₀² + 2aΔx. The chain rule gets there directly, because a = dv/dt = v dv/dx, so ∫ v dv = ∫ a dx. Because v is linear in t, the average velocity over the interval is the mean of its end values, giving Δx = ½(v₀ + v)Δt, which skips a. Replace v₀ by v − aΔt and you get Δx = vΔt − ½aΔt², which skips v₀. Check the cyclist without the clock: v² = 3.0² + 2 × 0.60 × 43.2 = 9.00 + 51.84 = 60.84 m² s⁻², and √60.84 = 7.8 m s⁻¹. Choosing is mechanical: find the one quantity that is neither given nor wanted, and use the relation that omits it.

Every quantity except time is signed relative to an axis you declare before substituting anything. “Decelerating” is not a sign. It says the acceleration points opposite the velocity, which makes a negative only if v is positive. A cart moving in the −x direction at 4.0 m s⁻¹ and slowing at 0.50 m s⁻² has v₀ = −4.0 m s⁻¹ and a = +0.50 m s⁻². It stops where 0 = (−4.0)² + 2(0.50)Δx, so Δx = −16 m: 16 m on the negative side of its start, which is the way it was going. Enter a = −0.50 m s⁻² out of habit and the same equation returns Δx = +16 m, a stop in a direction the cart never moved. The time-free relation hides a second trap: v² cannot tell +v from −v. Its square root is a speed, and the sign comes from the motion you are describing.

Asking when an object reaches a position turns x(t) into a quadratic, and a quadratic answers twice. A cart launched up a smooth ramp at 6.0 m s⁻¹, with a = −2.0 m s⁻² along the ramp, is 5.0 m up when 6.0t − 1.0t² = 5.0, that is t² − 6.0t + 5.0 = 0, so t = 1.0 s or 5.0 s. Both are real events. The cart passes the 5.0 m mark going up, turns round at t = 3.0 s and x = 9.0 m, and passes the mark again coming down, because the same component of gravity acts the whole time. That is the test for every root: keep it only if the stated acceleration really holds from t₀ to that time. A negative root describes a moment before the initial condition was set. A root that falls after a force switches off describes a motion that never happens.

Nothing in the equations checks that a is constant. The forces have to. Three failures cover most cases. A force can switch off: brakes decelerate a car only while it moves, so the braking quadratic is wrong from the instant v reaches zero. A force can depend on speed: drag grows with v, so a coasting or falling body has an acceleration that changes as it goes. A force can depend on position: a stretched spring pulls hardest at full stretch. Release a block from 0.10 m on a spring with k/m = 100 s⁻², and at release |a| = 10 m s⁻². Pretend that stays constant and v² = 2 × 10 × 0.10 gives 1.41 m s⁻¹ at equilibrium; the true speed is √(k/m) × 0.10 m = 1.0 m s⁻¹, because the pull fades to zero on the way in. When a changes in steps, apply the equations to each step and hand one step's final state to the next as its initial condition. When it changes smoothly, integrate a(t) itself.