Solve a nonlinear falling branch by separating variables
Choose downward positive for a1 kg point released from rest. A quadratic resisting force is−k v|v|, not always−k v² regardless of direction. On this falling branch v remains nonnegative, so dv/dt=g−k v²/m. Here g=10 and k has units kg/m.
Learning outcome: Compare two k values and two times, check velocity and displacement derivatives, and show the initial acceleration and terminal limit.
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.
All lessons are open directly. Your saved experiments remain when you leave and return.
Try
At k=.1, terminal speed is10 m/s. At t=1, v=10tanh1≈7.61594 m/s and x=10lncosh1≈4.33781 m, not10 m.
Open the experimentPredict
For this downward-from-rest branch, terminal speed is:
Experiment
Compare two k values and two times, check velocity and displacement derivatives, and show the initial acceleration and terminal limit.
Integrate a quadratic-drag falling branch
Worked teaching model · synthetic values · Virtual1 kg point falls from rest with downward positive, g10 and resisting force−k v|v|. The supplied solution applies to its v≥0 branch. No ground, buoyancy, changing density or actual fall experiment.
Reset keeps your prediction, captures and practice history.
Falling branch with quadratic resistance
Unlike graph quantities rescale independently; read units and values, not screen angles. 1: vt tanh(gt/vt) (solid). Joined curves show the supplied model, not additional observations.
Equivalent numerical coordinates for Falling branch with quadratic resistance
vt tanh(gt/vt)
| time (s) | downward velocity (m/s) |
|---|---|
| 0 | 0 |
| 0.025 | 0.2498959 |
| 0.05 | 0.4991683 |
| 0.075 | 0.7472001 |
| 0.1 | 0.9933862 |
| 0.125 | 1.23714 |
| 0.15 | 1.477898 |
| 0.175 | 1.715125 |
| 0.2 | 1.94832 |
| 0.225 | 2.177017 |
| 0.25 | 2.400791 |
| 0.275 | 2.619258 |
| 0.3 | 2.83208 |
| 0.325 | 3.038961 |
| 0.35 | 3.239651 |
| 0.375 | 3.433946 |
| 0.4 | 3.621685 |
| 0.425 | 3.802748 |
| 0.45 | 3.977056 |
| 0.475 | 4.144571 |
| 0.5 | 4.305286 |
| 0.525 | 4.45923 |
| 0.55 | 4.606463 |
| 0.575 | 4.747068 |
| 0.6 | 4.881156 |
| 0.625 | 5.008858 |
| 0.65 | 5.130322 |
| 0.675 | 5.245713 |
| 0.7 | 5.355208 |
| 0.725 | 5.458995 |
| 0.75 | 5.557268 |
| 0.775 | 5.65023 |
| 0.8 | 5.738086 |
| 0.825 | 5.821043 |
| 0.85 | 5.89931 |
| 0.875 | 5.973094 |
| 0.9 | 6.0426 |
| 0.925 | 6.10803 |
| 0.95 | 6.169585 |
| 0.975 | 6.227456 |
| 1 | 6.281835 |
| 1.025 | 6.332902 |
| 1.05 | 6.380836 |
| 1.075 | 6.425807 |
| 1.1 | 6.46798 |
| 1.125 | 6.507511 |
| 1.15 | 6.544552 |
| 1.175 | 6.579246 |
| 1.2 | 6.611732 |
| 1.225 | 6.642139 |
| 1.25 | 6.670592 |
| 1.275 | 6.697209 |
| 1.3 | 6.722102 |
| 1.325 | 6.745377 |
| 1.35 | 6.767133 |
| 1.375 | 6.787467 |
| 1.4 | 6.806465 |
| 1.425 | 6.824214 |
| 1.45 | 6.840792 |
| 1.475 | 6.856274 |
| 1.5 | 6.87073 |
| 1.525 | 6.884226 |
| 1.55 | 6.896824 |
| 1.575 | 6.908583 |
| 1.6 | 6.919557 |
| 1.625 | 6.929797 |
| 1.65 | 6.939351 |
| 1.675 | 6.948265 |
| 1.7 | 6.956581 |
| 1.725 | 6.964338 |
| 1.75 | 6.971573 |
| 1.775 | 6.978321 |
| 1.8 | 6.984614 |
| 1.825 | 6.990482 |
| 1.85 | 6.995954 |
| 1.875 | 7.001057 |
| 1.9 | 7.005814 |
| 1.925 | 7.01025 |
| 1.95 | 7.014385 |
| 1.975 | 7.018241 |
| 2 | 7.021834 |
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 |
|---|---|---|
| Terminal downward speed | 7.071068 | m/s |
| Downward speed | 7.021834 | m/s |
| Downward displacement | 10.69384 | m |
| Downward acceleration | 0.1387683 | m/s² |
| Drag magnitude | 9.861232 | N |
Separating variables gives a nonlinear approach to the same force-balance limit.
For this downward branch dv/dt=g−k v²/m=g[1−(v/vt)²]. Integrate dv/[1−(v/vt)²]=g dt from rest to obtain v=vt tanh(gt/vt); another integral gives x=(vt²/g)lncosh(gt/vt). The complete drag vector opposes velocity, so upward motion would not use the same positive-branch scalar equation without checking its sign.
Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.
Separating variables gives a nonlinear approach to the same force-balance limit.
For this downward branch dv/dt=g−k v²/m=g[1−(v/vt)²]. Integrate dv/[1−(v/vt)²]=g dt from rest to obtain v=vt tanh(gt/vt); another integral gives x=(vt²/g)lncosh(gt/vt). The complete drag vector opposes velocity, so upward motion would not use the same positive-branch scalar equation without checking its sign.
Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.
Understand
Apply
Public practice checks a specific claim. It does not establish independent mastery.
Why must the drag direction be revisited for an upward-moving point?
Continue
Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.
Skipping retains experiments and awards no mastery.
Save a science backup or import one
Science work is stored locally in this browser, separately from mathematics and AI. It does not sync automatically. Use this science-only backup; this checkout's older all-subject export does not include science.
A legacy file is available only when the complete snapshot fits the 36 MB limit. For larger snapshots, prepare a complete file set below; never discard notebooks to make a smaller file.
Matching imported notebook drafts replace their current drafts. Omitted lessons and known first attempts remain. Export your current work first if you want to retain both drafts.
Complete file sets and course backups
A file set can hold more work than one science file. Keep its manifest and every part together. Each file is limited to 36 MB. This format needs this version of LearnAI or a compatible newer version; older versions can use only the legacy single-file backup above.
The set includes every saved science course and global continuation at this moment. Later edits need a new export. A conflicting tab exports its own visible work, which may differ from saved work in another tab.
0 file(s) selected. File order and filenames do not determine identity. Missing, changed or conflicting parts block the entire restore; no good part is applied by itself.
Science storage: loading. No account or remote computation is required.