Reconstruct motion with two integration constants

Integrate a rate with its initial value

If a(t)=dv/dt, then v(t)=v0+∫ from 0 to t a(u)du. The dummy variable u separates the integration variable from its upper endpoint. Acceleration determines changes in velocity, while v0 selects one member of the family. Integrating again gives x(t)=x0+∫ from 0 to t v(u)du; x0 is a different required datum.

Derive the prescribed history

Here a(t)=2+jt, with 2 in m/s² and j in m/s³. Integration gives v=v0+2t+jt²/2. A second integration gives x=−1+v0t+t²+jt³/6. Check by differentiating: dx/dt returns v and dv/dt returns a. At zero time the expressions return the declared x0 and v0.

Work and challenge a numerical case

For j=1,v0=2,t=3, v=2+6+4.5=12.5 m/s and x=−1+6+9+4.5=18.5 m. The displacement is 19.5 m, not 18.5, because the initial position is −1. The final acceleration is 5 m/s²; treating it as constant over all 3 s would incorrectly predict v=17 m/s.

Recover a special case

When j=0, v=v0+2t and x=x0+v0t+(2/2)t². These are the usual constant-acceleration equations derived from the integral, not a separate rule for all motion. Negative jerk can make acceleration change sign; the integrals remain valid without assuming monotonic position or velocity.

Connect graph areas and average rates

Before checking, predict the velocity change from the area under a=2+t over 0–3 s. The rectangle contributes 2×3=6 m/s and the triangle contributes (1/2)×3×3=4.5 m/s, totaling 10.5 m/s. Average acceleration is Δv/Δt=3.5 m/s², different from the endpoint instantaneous acceleration 5 m/s². The signed area under the derived velocity curve is displacement 19.5 m, not final position 18.5 m. A negative part of a velocity graph subtracts displacement while still adding distance.

For constant acceleration a, eliminate time using v−v0=at and Δx=(v+v0)t/2 to obtain v²−v0²=2aΔx. If a=0, the same relation reduces to v²=v0² while direction is fixed by v=v0. This third constant-acceleration relation is not valid with a changing acceleration merely replaced by its final value. As a changed-case check, constant a=3 m/s² and initial velocity 2 m/s over displacement 2 m imply final speed 4 m/s on the forward branch; the sign requires the motion context, not just a square root.

Transfer and verification

For a=3+2t and v0=−1, v=−1+3t+t², giving 9 m/s at 2 s. A position answer still needs x0. In retained work, write both initial conditions and differentiate the proposed solution before comparing its numerical output. A fitted acceleration model can be internally consistent yet fail outside its measured or stipulated interval.

Public numerical transfer practice

Try a new practice case

For a(t)=3+2t m/s², v(0)=−1 m/s, find v(2).

Numerical result in the requested units Check this practice result

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Original explanations and cases, checked 30 October 2026 against the official AP Physics C Mechanics framework, OpenStax calculus kinematics and vector kinematics. All numerical records are supplied or generated, not actual laboratory measurements. No physical launch, drop or moving apparatus is requested. Public feedback is practice, not independent certification.