Demonstration 1 of 4
Two ledgers for one trip
The traveler moves 5 units. Why do they end only 1 unit from the start?
The left panel shows each block's signed contribution. The right panel follows position: up 3, then back. The odometer adds every movement as positive.
\[+3 + (-2) = +1\ \text{unit}\]
\[|+3| + |-2| = 3 + 2 = 5\ \text{units}\]
Velocity is signed: + means the positive direction, - the negative one. The bars | | give speed, the size without the sign. s(t) is position in units.
Predict first. Make the second velocity -3. What are the displacement and the distance?
Choose an example
Constructed example: the chapter's +3 then -2 trip from s(0) = 10, plus other second velocities.
Calculated values
- Displacement
- +3 + (-2) = +1 unit
- Distance
- |+3| + |-2| = 3 + 2 = 5 units
- Ending position
- s(2) = 10 + 1 = 11 units
Displacement keeps signs: +3 + (-2) = +1. Distance adds sizes: 3 + 2 = 5. Ending position: 10 + 1 = 11. The traveler moved 5 units but ended 1 from the start, so adding 5 to 10 would put them in the wrong place.
Use the idea
To find where something ends up, add signed changes to the start. To find how much it moved, add sizes. Never add distance to the start.
Where the conclusion applies
Constant velocity on each block and one straight line of motion. A coordinate direction is chosen in advance; flipping it flips every sign but not the distance.
Check your understanding: Starting at 10 with velocities +3 then -5, where does the traveler end, and how far did they go?
Chapter 32 source: section "The two-block trip keeps both ledgers". Demonstration C32-D01.
Demonstration 2 of 4
Accumulation stores the rate in its slope
If A(x) is the area so far, how fast does A grow as x moves?
The new strip is almost a rectangle of height f(x) and width h. Dividing its area by h leaves a number near f(x); the extra is exactly h for this straight line.
\[A'(x) = f(x)\]
f(t) = 2t + 1 is a continuous rate. A(x) is the area under it from a = 0 to x. h is a small step to the right, and the gold strip is the area it adds.
Predict first. At x = 2, shrink h from 1 to 0.1. What number does strip / h approach?
Choose an example
Constructed example: the chapter's continuous rate 2t + 1, accumulated from 0.
Calculated values
- A(x), area from 0 to x
- 2 x (1 + 5) / 2 = 6
- Added strip A(x + h) - A(x)
- 8.75 - 6 = 2.75
- Strip / h
- 2.75 / 0.5 = 5.5
- f(x) = 2x + 1
- 5
- Gap from f(x)
- 0.5
A(2) = 2 x (1 + 5) / 2 = 6 and A(2.5) = 8.75, so the strip adds 8.75 - 6 = 2.75. Divide by h: 2.75 / 0.5 = 5.5, close to f(2) = 2 x 2 + 1 = 5. The gap is 0.5, exactly h; as h shrinks, the rate of accumulation becomes f(x).
Use the idea
When you know a running total, its rate of growth at a moment is the rate being accumulated at that moment.
Where the conclusion applies
A continuous rate on an interval starting at 0. Here the gap is exactly h because the rate is a straight line; for a curve the gap is not that tidy but still shrinks.
Check your understanding: At x = 1 with h = 0.5, what does the strip add, and what is strip / h?
Chapter 32 source: section "Accumulating a continuous rate creates a function". Demonstration C32-D02.
Demonstration 3 of 4
Check, then subtract endpoints
How does F(t) = t^2 + t turn a velocity into a displacement?
The shaded area on the left and the gold rise on the right are the same number. Shifting F up moves both endpoint values together, so their difference stays put.
\[F(t) = t^2 + t\]
\[F'(t) = 2t + 1 = v(t)\]
v(t) = 2t + 1 is velocity in length units per time unit. F is an antiderivative: its derivative is v. The displacement from 0 to the upper end is F(end) - F(0).
Predict first. Add 5 to F. Does the displacement from 0 to 3 change?
Choose an example
Constructed example: the chapter's v(t) = 2t + 1 on 0 to 3, plus shorter intervals and a shifted F.
Calculated values
- Derivative check
- d(t^2 + t)/dt = 2t + 1
- F(upper end) - F(0)
- (3^2 + 3) - (0^2 + 0) = 12
- Graph check
- (1 + 7)/2 x 3 = 12
- Displacement
- 12 length units
- Distance
- 12 length units (v stays positive)
F(3) - F(0) = (3^2 + 3) - (0^2 + 0) = 12 - 0 = 12 length units. Graph check: the rate rises from 1 to 7, average (1 + 7)/2 = 4, times 3 = 12.
Use the idea
Before using an antiderivative, differentiate it and confirm it returns the rate. Then subtract and check the answer against the graph.
Where the conclusion applies
A continuous velocity on a closed interval. Because v stays positive here, distance equals displacement; that would fail if v changed sign.
Check your understanding: What is the displacement from t = 0 to t = 2?
Chapter 32 source: section "Work the rate, then audit the units". Demonstration C32-D03.
Demonstration 4 of 4
A velocity that crosses zero
When velocity changes sign at t = 2, why do displacement and distance part ways?
The signed pieces add with their signs for displacement. For distance, the negative piece has its sign flipped before adding, so nothing cancels.
\[v(t)=2t-4\]
\[F(t)=t^2-4t\]
v(t) is velocity in meters per second, negative before 2 seconds and positive after. F(t) = t^2 - 4t is an antiderivative. Pink pieces are negative, teal are positive.
Predict first. Stop the clock at t = 4. What is the displacement, and what is the distance?
Choose an example
Constructed example: the chapter's v(t) = 2t - 4 on 0 to 5 seconds, plus earlier stopping times.
Calculated values
- F(t) = t^2 - 4t at the end
- F(5) = 5^2 - 4 x 5 = 5
- Displacement
- 5 meters
- Distance
- 13 meters
- Ending position from 10 meters
- 10 + 5 = 15 meters
Displacement: F(5) - F(0) = 5 - 0 = 5 meters. Distance: -[F(2) - F(0)] + [F(5) - F(2)] = 4 + 9 = 13 meters. Split at t = 2, where velocity turns from negative to positive, before adding sizes.
Use the idea
When a rate can change sign, find where it does, split the interval there, and add the sizes of the pieces.
Where the conclusion applies
A continuous velocity on one line, assuming a start at 10 meters, as the chapter supposes. A zero of v is only a candidate for a reversal; the sign on each side decides.
Check your understanding: Stopping at t = 3, what are the displacement and the distance?
Chapter 32 source: section "Worked example: a continuous velocity crosses zero". Demonstration C32-D04.