Solve a signed linear-resistance initial-value problem

Write a signed resistance law

Choose positive along a constant drive F. Linear resistance is−bv, so it opposes the signed velocity: if the object initially moves negatively, resistance points positively. Newton’s law is m dv/dt=F−bv. Define vt=F/b and tau=m/b; the equation becomes dv/dt=(vt−v)/tau. Mass controls the response time, while the steady velocity depends on the force-to-resistance ratio.

Separate variables and keep the initial condition

Away from the equilibrium solution, integrate dv/(vt−v)=dt/tau from initial velocity v0 at zero time. The logarithm gives v(t)=vt+(v0−vt)e^(−t/tau). The same expression includes v0=vt by direct substitution, even though dividing by vt−v would be invalid for that equilibrium path. Verify the result by differentiation rather than discarding this special case.

Recover position and acceleration

With x0=0, integration gives x=vt t+(v0−vt)tau[1−e^(−t/tau)]. Differentiation gives acceleration (vt−v0)e^(−t/tau)/tau. At time zero the velocity returns v0 and position returns zero. At late times velocity approaches vt, acceleration approaches zero, and position retains a linear term; position does not approach a finite asymptote under a nonzero persistent drive.

Compare three original starts

For m=2,F=4,b=1, vt=4 m/s and tau=2 s. From rest, v(3)=4(1−e⁻¹·⁵)≈3.10748 m/s and x(3)≈5.78504 m. Starting at 4 gives constant velocity. Starting at 8 gives a decreasing velocity approaching 4 from above. Thus terminal equilibrium is not a universal ceiling on a prescribed initial speed.

Connect a terminal fall without changing the logic

For a downward rest release with constant gravity force mg, the same linear model gives vt=mg/b and tau=m/b. Initial acceleration is g because initial drag is zero. As downward speed grows, resistance grows and net acceleration decreases. This linear law is a bounded model, not a universal description of every flow regime or falling object.

Transfer and independent evidence

For m=4,b=2,F=6, equilibrium velocity is 3 m/s and response time 2 s. Equal terminal velocities can arise from different force/resistance pairs; a time course adds information about m/b. A proposed physical test would need calibrated time, force, mass and a justified resistance regime. The workbench supplies those assumptions rather than claiming actual measurements.

Public numerical transfer practice

Try a new practice case

A mass 4 kg with b2 kg/s starts at rest under drive 6 N. Find terminal velocity in m/s.

Numerical result in the requested units Check this practice result

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Sources and evidence boundary

Original prose, models and cases, checked 30 October 2026 against the current official AP Physics C Mechanics CED and OpenStax University Physics. Values are supplied or generated, not measured apparatus records. No physical launch, suspension, moving machinery or practical safety validation is requested. Virtual captures and public answers do not certify independent mastery or laboratory participation.