{ "format": "recallweave-deck/1", "title": "Rates and accumulation: motion, signed area and traveled distance", "attribution": "Original teaching text, questions and worked examples prepared for RecallWeave in October 2026. Mathematical references: Gilbert Strang and Edwin Herman, OpenStax Calculus Volume 1, sections 3.1 (defining the derivative), 5.3 (fundamental theorem) and 5.4 (net change): https://openstax.org/books/calculus-volume-1/ . References checked 2026-10-08.", "license": "Original teaching text and examples. This contribution makes no additional reuse-license grant. Linked teaching sources retain their own terms; no textbook prose, exercise or figure is copied.", "concepts": [ "Rate and units", "Signed accumulation", "Local and interval rates", "Reading a continuous curve" ], "items": [ { "id": "ra-01", "concept": "Rate and units", "prerequisites": [], "prompt": "An object moves with constant velocity +3 m/s for 2 s. What is its displacement during those two seconds?", "options": [ "1.5 m", "6 m", "6 m/s", "3 m" ], "answer": 1, "explanation": "Displacement is velocity multiplied by elapsed time for this constant-rate interval: (+3 m/s) × (2 s) = +6 m. The seconds cancel, leaving metres. Dividing rate by time or retaining m/s confuses a rate with an accumulated change.", "transfer": "If the same velocity lasts 5 s, find the displacement and explain its sign." }, { "id": "ra-02", "concept": "Rate and units", "prerequisites": [], "prompt": "During an interval, velocity remains negative while its graph slopes upward toward zero. What happens to the motion during that interval?", "options": [ "It moves in the positive direction and speeds up.", "It moves in the negative direction and speeds up.", "It stays at rest while its acceleration increases.", "It moves in the negative direction and slows down." ], "answer": 3, "explanation": "The negative velocity fixes the negative direction. As that value approaches zero, its magnitude decreases, so speed decreases. An upward slope means positive acceleration, not necessarily positive motion. The statement applies while the velocity remains negative.", "transfer": "Sketch a velocity interval that is negative and slopes downward. Describe its direction and change in speed." }, { "id": "ra-03", "concept": "Rate and units", "prerequisites": [], "prompt": "A motion starts at position x(0) = 10 m. Its signed velocity area from 0 s to 3 s is −3 m. What is x(3)?", "options": [ "7 m", "−3 m", "13 m", "−7 m" ], "answer": 0, "explanation": "The signed area gives the change in position, not the absolute position. Add it to the initial value: 10 m + (−3 m) = 7 m. A negative change moves the endpoint in the negative direction without requiring the endpoint itself to be negative.", "transfer": "Explain what information is missing if someone reports only the velocity graph but asks for absolute position." }, { "id": "ra-04", "concept": "Signed accumulation", "prerequisites": [ "Rate and units" ], "prompt": "Velocity increases linearly from 0 m/s at 0 s to 6 m/s at 4 s. What is the displacement over this interval?", "options": [ "24 m", "6 m", "12 m", "3 m" ], "answer": 2, "explanation": "The region under this line is a triangle of base 4 s and height 6 m/s. Its signed area is (1/2) × 4 × 6 = 12 m. Equivalently, the interval's average velocity is (0 + 6)/2 = 3 m/s, and 3 × 4 = 12 m. The endpoint velocity alone does not describe the whole interval.", "transfer": "Use a line from 2 m/s to 6 m/s over the same four seconds. Find its trapezoid area." }, { "id": "ra-05", "concept": "Signed accumulation", "prerequisites": [ "Rate and units" ], "prompt": "Velocity is the straight line from +3 m/s at 0 s to −1 m/s at 4 s. What is the net displacement from 0 s to 4 s?", "options": [ "5 m", "8 m", "−4 m", "4 m" ], "answer": 3, "explanation": "For a linear rate, the signed trapezoid area is duration times the mean of the endpoint velocities: 4 × (3 + (−1))/2 = 4 m. The line crosses zero at 3 s. Its positive area is 9/2 m and its negative area is −1/2 m; adding them also gives 4 m. Net displacement retains direction.", "transfer": "Explain why replacing the negative endpoint with +1 changes the quantity being calculated." }, { "id": "ra-06", "concept": "Signed accumulation", "prerequisites": [ "Rate and units" ], "prompt": "For the same straight-line velocity from +3 m/s at 0 s to −1 m/s at 4 s, what total distance is traveled?", "options": [ "5 m", "4 m", "8 m", "0 m" ], "answer": 0, "explanation": "Distance adds the magnitudes of the two pieces. The zero crossing is at 3 s: the first triangle contributes (1/2) × 3 × 3 = 9/2 m, and the second contributes (1/2) × 1 × 1 = 1/2 m. The total is 5 m. Taking the absolute value of the net 4 m would lose the reversal.", "transfer": "Find the distance for a line from +1 m/s to −2 m/s over one second. Locate the zero crossing first." }, { "id": "ra-07", "concept": "Local and interval rates", "prerequisites": [ "Signed accumulation" ], "prompt": "A continuous velocity graph has vertices (time s, velocity m/s): (0,0), (2,4), (4,0), (6,−4), (8,0). What is the average velocity over all eight seconds?", "options": [ "2 m/s", "−2 m/s", "0 m/s", "4 m/s" ], "answer": 2, "explanation": "The positive triangle contributes +8 m and the negative triangle contributes −8 m. Net displacement is zero, so average velocity is 0 m / 8 s = 0 m/s. Total distance is 16 m, giving an average speed of 2 m/s. Average velocity and average speed answer different questions.", "transfer": "At the end of the fourth second, compute both the average velocity and the average speed so far." }, { "id": "ra-08", "concept": "Local and interval rates", "prerequisites": [ "Signed accumulation" ], "prompt": "For the straight-line velocity from +3 m/s at 0 s to −1 m/s at 4 s, what are the instantaneous velocity at 2 s and the average velocity over [0,2]?", "options": [ "2 m/s; 1 m/s", "1 m/s; 2 m/s", "1 m/s; 1 m/s", "2 m/s; 2 m/s" ], "answer": 1, "explanation": "The line has slope −1 m/s², so v(2) = 3 − 2 = 1 m/s. The signed area over the first two seconds is 2 × (3 + 1)/2 = 4 m; dividing by 2 s gives an average velocity of 2 m/s. One is a value at a moment, the other summarizes an interval.", "transfer": "Choose a constant-velocity curve and explain why these two velocities would agree there." }, { "id": "ra-09", "concept": "Local and interval rates", "prerequisites": [ "Signed accumulation" ], "prompt": "Two motions have the same velocity at every time. The second starts 10 m farther along the same axis. Which comparison must hold over the same interval of positive duration?", "options": [ "Their displacements and average velocities are equal.", "Their displacements differ by 10 m and their average velocities are equal.", "Their displacements are equal and their average velocities differ by 10 m/s.", "Their displacements and average velocities both differ." ], "answer": 0, "explanation": "A change in initial position adds a constant to the position curve. Its derivative is unchanged, and subtracting final minus initial position removes that constant. Both motions therefore have the same displacement and, for a nonzero interval, the same average velocity. Their positions remain 10 m apart.", "transfer": "Would their traveled distances also agree over that interval? Explain using the shared speed." }, { "id": "ra-10", "concept": "Reading a continuous curve", "prerequisites": [ "Signed accumulation", "Local and interval rates" ], "prompt": "Velocity decreases linearly from +2 m/s at 0 s to 0 m/s at 2 s and −2 m/s at 4 s. What is true at exactly 2 s?", "options": [ "It has zero displacement so far and is moving backward.", "It has 4 m displacement so far and is moving forward.", "It is momentarily at rest with 2 m displacement so far.", "It has been at rest for the entire first two seconds." ], "answer": 2, "explanation": "At that instant v(2) = 0, but the earlier positive triangle has area (1/2) × 2 × 2 = 2 m. Zero instantaneous velocity does not erase earlier travel or imply an interval of rest. Immediately before and after 2 s, the velocity signs differ, so this particular curve reverses direction.", "transfer": "Compare with vertices (0,2), (2,0), (4,2). Does reaching zero in that curve cause a reversal?" }, { "id": "ra-11", "concept": "Reading a continuous curve", "prerequisites": [ "Signed accumulation", "Local and interval rates" ], "prompt": "A continuous velocity graph joins (0 s,0 m/s), (2 s,4 m/s), and (4 s,0 m/s). At 2 s, which derivative statement is correct?", "options": [ "Position has no derivative; acceleration is +2 m/s².", "Position has derivative 4 m/s; acceleration has unequal one-sided limits.", "Position has derivative 0 m/s; acceleration is −2 m/s².", "Position has no derivative; acceleration is 0 m/s²." ], "answer": 1, "explanation": "Velocity is continuous at 2 s and equals 4 m/s, so the accumulated position has derivative 4 m/s there. The velocity graph's left slope is +2 m/s² and its right slope is −2 m/s². Those unequal slopes mean the two-sided derivative of velocity, acceleration, does not exist at the corner. The corner is in velocity, not in position.", "transfer": "Add a vertex in the middle of a straight velocity segment. Does that new table row alone create a derivative failure?" }, { "id": "ra-12", "concept": "Reading a continuous curve", "prerequisites": [ "Signed accumulation", "Local and interval rates" ], "prompt": "A curve begins at time 0 s with velocity +3 m/s. At that initial moment, what are displacement since the start, instantaneous velocity, and average velocity over [0,0]?", "options": [ "0 m; 0 m/s; 0 m/s", "3 m; 3 m/s; 3 m/s", "0 m; 3 m/s; 0 m/s", "0 m; 3 m/s; undefined over zero duration" ], "answer": 3, "explanation": "No time has elapsed, so accumulated displacement is zero. The supplied rate at the initial moment is still +3 m/s. Average velocity over [0,0] would divide zero displacement by zero duration, so that interval average is undefined. A right-sided limiting average is a different question and equals +3 m/s for this continuous curve.", "transfer": "Explain why an interface should not show 0 m/s as the average merely because the elapsed duration is zero." } ] }