Integrate a force ramp with the original initial conditions
A force law gives acceleration, but an acceleration equation still needs initial conditions. The original2 kg point has x0=0 and v0=1 m/s, with net force F(t)=F0+k t. The coefficient k has units N/s; it is a slope in time, not a spring constant.
Learning outcome: Integrate two ramps, verify each derivative and initial condition, and compare an endpoint-force shortcut with the actual time integral.
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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Try
With F0=−2 N and k=0, acceleration is−1 m/s², velocity at2 s is−1 m/s and position returns to0. The zero displacement does not mean the point never moved.
Open the experimentPredict
For F0=0,k=2 N/s over0–2 s, average net force is:
Experiment
Integrate two ramps, verify each derivative and initial condition, and compare an endpoint-force shortcut with the actual time integral.
Integrate a time-dependent net force
Worked teaching model · synthetic values · Virtual2 kg point has x0=0 and v0=1 m/s, with prescribed net force F(t)=F0+k t over0–2 s. Motion can reverse; no mechanical device or control feasibility is implied.
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Integrated velocity under ramped net force
Unlike graph quantities rescale independently; read units and values, not screen angles. 1: v0+integral F/m (solid). Joined curves show the supplied model, not additional observations.
Equivalent numerical coordinates for Integrated velocity under ramped net force
v0+integral F/m
| time (s) | signed velocity (m/s) |
|---|---|
| 0 | 1 |
| 0.025 | 1.000312 |
| 0.05 | 1.00125 |
| 0.075 | 1.002813 |
| 0.1 | 1.005 |
| 0.125 | 1.007813 |
| 0.15 | 1.01125 |
| 0.175 | 1.015313 |
| 0.2 | 1.02 |
| 0.225 | 1.025313 |
| 0.25 | 1.03125 |
| 0.275 | 1.037813 |
| 0.3 | 1.045 |
| 0.325 | 1.052812 |
| 0.35 | 1.06125 |
| 0.375 | 1.070313 |
| 0.4 | 1.08 |
| 0.425 | 1.090313 |
| 0.45 | 1.10125 |
| 0.475 | 1.112812 |
| 0.5 | 1.125 |
| 0.525 | 1.137812 |
| 0.55 | 1.15125 |
| 0.575 | 1.165312 |
| 0.6 | 1.18 |
| 0.625 | 1.195313 |
| 0.65 | 1.21125 |
| 0.675 | 1.227812 |
| 0.7 | 1.245 |
| 0.725 | 1.262812 |
| 0.75 | 1.28125 |
| 0.775 | 1.300312 |
| 0.8 | 1.32 |
| 0.825 | 1.340313 |
| 0.85 | 1.36125 |
| 0.875 | 1.382813 |
| 0.9 | 1.405 |
| 0.925 | 1.427812 |
| 0.95 | 1.45125 |
| 0.975 | 1.475313 |
| 1 | 1.5 |
| 1.025 | 1.525313 |
| 1.05 | 1.55125 |
| 1.075 | 1.577812 |
| 1.1 | 1.605 |
| 1.125 | 1.632813 |
| 1.15 | 1.66125 |
| 1.175 | 1.690313 |
| 1.2 | 1.72 |
| 1.225 | 1.750313 |
| 1.25 | 1.78125 |
| 1.275 | 1.812812 |
| 1.3 | 1.845 |
| 1.325 | 1.877813 |
| 1.35 | 1.91125 |
| 1.375 | 1.945313 |
| 1.4 | 1.98 |
| 1.425 | 2.015313 |
| 1.45 | 2.05125 |
| 1.475 | 2.087813 |
| 1.5 | 2.125 |
| 1.525 | 2.162812 |
| 1.55 | 2.20125 |
| 1.575 | 2.240312 |
| 1.6 | 2.28 |
| 1.625 | 2.320313 |
| 1.65 | 2.36125 |
| 1.675 | 2.402813 |
| 1.7 | 2.445 |
| 1.725 | 2.487813 |
| 1.75 | 2.53125 |
| 1.775 | 2.575312 |
| 1.8 | 2.62 |
| 1.825 | 2.665312 |
| 1.85 | 2.71125 |
| 1.875 | 2.757813 |
| 1.9 | 2.805 |
| 1.925 | 2.852813 |
| 1.95 | 2.90125 |
| 1.975 | 2.950312 |
| 2 | 3 |
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 |
|---|---|---|
| Initial acceleration | 0 | m/s² |
| Acceleration at2 s | 2 | m/s² |
| Velocity at2 s | 3 | m/s |
| Position at2 s | 3.333333 | m |
| Average net force over0–2 s | 2 | N |
The force law supplies acceleration; initial conditions still select the trajectory.
For m2, integrate (F0+k t)/2: v=1+F0t/2+k t²/4 and x=t+F0t²/4+k t³/12. The endpoint force is not the average force over a ramp. A negative velocity is allowed and does not turn the prescribed time-dependent force into a friction law.
Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.
The force law supplies acceleration; initial conditions still select the trajectory.
For m2, integrate (F0+k t)/2: v=1+F0t/2+k t²/4 and x=t+F0t²/4+k t³/12. The endpoint force is not the average force over a ramp. A negative velocity is allowed and does not turn the prescribed time-dependent force into a friction law.
Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.
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How would a force tied to velocity differ from this prescribed time ramp?
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