The method and the results are on the calculus kinematics lab. Do the analysis yourself first, then check the answers below.
Quick answer: In calculus-based kinematics, velocity is the derivative of position and acceleration is the derivative of velocity. These answers go with the AP Physics C lab on calculus kinematics: the graph, the calculations, the uncertainties and the conclusion explained, for all 7 analysis questions.
Model answers use the sample table above. Your readings will differ a little. Also, the key skills are converting positions to velocities, estimating slopes and checking the area under the graph against the measured distance.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:M Unit 1
v = Δy / 0.05 s for each interval: 0.30 m/s (at 0.025 s), 0.76 m/s (at 0.075 s), 1.04 m/s (at 0.125 s), 1.20 m/s (at 0.175 s), 1.28 m/s (at 0.225 s), 1.36 m/s (at 0.275 s), 1.40 m/s (at 0.325 s), 1.42 m/s (at 0.375 s), 1.42 m/s (at 0.425 s), 1.44 m/s (at 0.475 s).
Question 2 Velocity–time graph
The graph is a curve that rises steeply and then levels off, so the slope, which is the acceleration, falls with time.
Question 3 Acceleration at the start and end
Start: (0.76 − 0.3) / 0.05 = 9.2 m/s², close to g. End: (1.44 − 1.42) / 0.05 = 0.4 m/s², close to zero within the reading error.
Question 4 Terminal velocity
The curve levels at about 1.44 m/s.
Question 5 Check by integration
Sum of v × 0.05 s = 0.581 m against the measured 0.581 m, so the two agree: both are the area under the v–t graph (the sum telescopes to the last position).
Question 6 Explain the shape
At release only weight acts, so a ≈ g. As speed rises, drag grows, the net force falls and the acceleration shrinks to zero when drag equals weight, giving a constant terminal velocity.
Question 7 Uncertainty
Judging the bottom edge of the stack against the rule gives about ±1 mm in y, which is the largest error in the small first intervals. Improvement: film closer, use a higher frame rate or add a bright marker.
