Rates, accumulation, and sampled motion

• Estimate an accumulated quantity from sampled rates and explain how interval size and bias affect the result.

• algebra-trig

A rate says how quickly one quantity changes with another. Speed is distance per time; current is charge per time; power is energy per time. If a toy car travels at a steady 0.30 m/s for 4 s, distance is rate × duration = 1.2 m. When the rate varies, split time into intervals and add each interval's contribution. That sum is the practical idea behind integration.

Formally, if position x changes with velocity v, then v = dx/dt and the change in position is Δx = ∫v dt. With sampled data, a rectangular estimate is Δx ≈ Σ v_k Δt. A trapezoidal estimate uses the average of consecutive samples: Δx ≈ Σ((v_k + v_(k+1))/2)Δt. The same structure estimates energy from power readings or battery charge from current. Units provide a strong check: (m/s)×s becomes m.

Sampling is a design decision. Long intervals save storage but can miss rapid changes, such as a printer carriage accelerating and stopping. Short intervals reveal detail but do not cure a biased sensor. A velocity bias of only 0.01 m/s accumulates 0.6 m of position error in 60 s. For a desk robot, this is why wheel-speed integration needs periodic external correction rather than blind trust.

A common misconception is that integration means merely adding raw samples. Samples must be multiplied by their time widths; otherwise changing the sample rate changes the answer. Another failure is assuming the rate stayed constant between widely spaced readings. State the interpolation assumption, inspect gaps, and compare rectangular and trapezoidal results. Accumulation turns small persistent errors into large drift, so a plausible final number is not evidence that each interval was sound.

Problem

A toy car's measured speeds at 0, 1, 2, and 3 s are 0, 0.4, 0.6, and 0.2 m/s. Estimate distance over 3 s with trapezoids.

Steps

1. From 0–1 s: ((0 + 0.4)/2) × 1 s = 0.20 m.
2. From 1–2 s: ((0.4 + 0.6)/2) × 1 s = 0.50 m.
3. From 2–3 s: ((0.6 + 0.2)/2) × 1 s = 0.40 m.
4. Sum the interval distances: 0.20 + 0.50 + 0.40 = 1.10 m.

Result

The sampled trapezoidal estimate is 1.10 m.

Sanity check

The average speed implied by the result is 1.10 m / 3 s = 0.367 m/s, between the observed minimum and maximum.

A constant current of 0.25 A flows for 8 s. How much charge passes?

A. 2 C B. 0.03125 C C. 8.25 C D. 32 C

Your answer: ______________________________

Reasoning:

Goal

Compare sampled-speed integration with a directly measured travel distance.

Equipment

• unpowered toy car
• 1 m tape measure
• masking tape
• phone stopwatch or video timer
• spreadsheet or calculator

Procedure

1. Mark a straight 800 mm path and predict a gentle hand-push duration.
2. Record the car crossing 200 mm marks and compute one average speed per interval.
3. Accumulate interval distance as speed × measured interval duration and compare with the tape-measured total.
4. Deliberately omit one interval duration and add the speed numbers alone; diagnose the units and sample-rate dependence of the faulty result.
5. Repeat with 100 mm marks and compare sensitivity to timing error.

What to observe

• Expected: correctly paired rate and duration reconstruct approximately 800 mm; record the actual discrepancy and likely timing uncertainty.
• Expected fault clue: the raw sum of speeds has units of m/s, not metres, and changes when more samples are taken.

Transfer challenge

Given five one-minute power readings for a screen, describe how you would estimate watt-hours and what behavior might be missed between readings.

Cautions

• Use a level floor or table with a stop block so the car cannot roll into people or fall.
• Do not place yourself in a walkway while timing.

Learner: ____________________ Date: ____________________

Evidence artifact or file: ____________________________________________

Notes: