Derive terminal motion and its approach time separately
A horizontal point starts from rest under constant drive F=10 N and linear resistance−b v relative to a stationary medium. Newton's law gives m dv/dt=F−b v. The coefficient b has units kg/s. A terminal state solves force balance, but a force-balance value alone says nothing about how quickly it is approached.
Learning outcome: Compare masses at fixed drag, then drag coefficients at fixed mass. Retain force balance, finite-time force residual and integrated displacement separately.
Starting recap
Differentiate/integrate with explicit initial conditions; identify each force recipient and the inertial frame. Preserve force dimensions, geometric constraints and the distinction between model and solver error.
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At m=1,b=1 and t=2, velocity10(1−e⁻²)≈8.64665 m/s. The same terminal10 m/s is approached faster than by the2 kg point.
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Doubling m at fixed drive F and linear coefficient b changes:
Experiment
Compare masses at fixed drag, then drag coefficients at fixed mass. Retain force balance, finite-time force residual and integrated displacement separately.
Derive the transient and terminal limits separately
Worked teaching model · synthetic values · Horizontal virtual point starts from rest with fixed drive10 N and drag−b v relative to a stationary medium. Linear drag is a stipulated bounded regime, not a universal fluid law; evaluation2 s.
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Forces on the selected recipient
- Solid: Drive; east 10 N, north 0 N.
- Dashed: Drag at2 s; east -6.321206 N, north 0 N.
Both components use the same pixels per newton. Every arrow acts on the named recipient; third-law partners on other bodies are not added here. The net force is the sum, not an extra interaction.
Approach to terminal velocity
Unlike graph quantities rescale independently; read units and values, not screen angles. 1: linear-resistance solution (solid). Joined curves show the supplied model, not additional observations.
Equivalent numerical coordinates for Approach to terminal velocity
linear-resistance solution
| time (s) | velocity (m/s) |
|---|---|
| 0 | 0 |
| 0.05 | 0.2469009 |
| 0.1 | 0.4877058 |
| 0.15 | 0.7225651 |
| 0.2 | 0.9516258 |
| 0.25 | 1.175031 |
| 0.3 | 1.39292 |
| 0.35 | 1.60543 |
| 0.4 | 1.812692 |
| 0.45 | 2.014838 |
| 0.5 | 2.211992 |
| 0.55 | 2.404279 |
| 0.6 | 2.591818 |
| 0.65 | 2.774726 |
| 0.7 | 2.953119 |
| 0.75 | 3.127107 |
| 0.8 | 3.2968 |
| 0.85 | 3.462302 |
| 0.9 | 3.623718 |
| 0.95 | 3.781149 |
| 1 | 3.934693 |
| 1.05 | 4.084446 |
| 1.1 | 4.230502 |
| 1.15 | 4.372951 |
| 1.2 | 4.511884 |
| 1.25 | 4.647386 |
| 1.3 | 4.779542 |
| 1.35 | 4.908436 |
| 1.4 | 5.034147 |
| 1.45 | 5.156754 |
| 1.5 | 5.276334 |
| 1.55 | 5.392962 |
| 1.6 | 5.50671 |
| 1.65 | 5.61765 |
| 1.7 | 5.725851 |
| 1.75 | 5.83138 |
| 1.8 | 5.934303 |
| 1.85 | 6.034686 |
| 1.9 | 6.13259 |
| 1.95 | 6.228076 |
| 2 | 6.321206 |
| 2.05 | 6.412035 |
| 2.1 | 6.500623 |
| 2.15 | 6.587022 |
| 2.2 | 6.671289 |
| 2.25 | 6.753475 |
| 2.3 | 6.833632 |
| 2.35 | 6.91181 |
| 2.4 | 6.988058 |
| 2.45 | 7.062423 |
| 2.5 | 7.134952 |
| 2.55 | 7.20569 |
| 2.6 | 7.274682 |
| 2.65 | 7.34197 |
| 2.7 | 7.407597 |
| 2.75 | 7.471604 |
| 2.8 | 7.53403 |
| 2.85 | 7.594915 |
| 2.9 | 7.654297 |
| 2.95 | 7.712213 |
| 3 | 7.768698 |
| 3.05 | 7.823789 |
| 3.1 | 7.87752 |
| 3.15 | 7.929924 |
| 3.2 | 7.981035 |
| 3.25 | 8.030883 |
| 3.3 | 8.079501 |
| 3.35 | 8.126918 |
| 3.4 | 8.173165 |
| 3.45 | 8.218269 |
| 3.5 | 8.262261 |
| 3.55 | 8.305166 |
| 3.6 | 8.347011 |
| 3.65 | 8.387824 |
| 3.7 | 8.427628 |
| 3.75 | 8.46645 |
| 3.8 | 8.504314 |
| 3.85 | 8.541242 |
| 3.9 | 8.577259 |
| 3.95 | 8.612387 |
| 4 | 8.646647 |
At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.
| Quantity | Value | Unit |
|---|---|---|
| Relaxation time | 2 | s |
| Terminal velocity | 10 | m/s |
| Velocity at2 s | 6.321206 | m/s |
| Position at2 s | 7.357589 | m |
| Net force at2 s | 3.678794 | N |
Mass changes the relaxation time while fixed drive and drag coefficient set the terminal velocity.
Solve dv/dt+(b/m)v=F/m with integrating factor exp(bt/m). From rest v=(F/b)(1−exp(−bt/m)); integrating gives x=(F/b)[t−(m/b)(1−exp(−bt/m))]. Terminal force balance does not imply that the point is already terminal after any finite chosen time. Linear and quadratic drag coefficients have different dimensions and regime assumptions.
Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.
Mass changes the relaxation time while fixed drive and drag coefficient set the terminal velocity.
Solve dv/dt+(b/m)v=F/m with integrating factor exp(bt/m). From rest v=(F/b)(1−exp(−bt/m)); integrating gives x=(F/b)[t−(m/b)(1−exp(−bt/m))]. Terminal force balance does not imply that the point is already terminal after any finite chosen time. Linear and quadratic drag coefficients have different dimensions and regime assumptions.
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Why is an asymptotic terminal value not a finite-time initial condition?
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