Draw and interpret motion graphs
Build the graphs as well as calculating the rates. Choose the correct quantity, retain its physical interval, and use your own exact tables, plotted points and strokes to show how the motion changes.
Section 4 · Functions (all14rows linked;self-check practice)
Exam scope and assessment notes
Original joint 4N teaching and practice across three lessons,25 cases and all twelve lesson/stage groups. Students enter term-by-term first and second derivatives,notation,units,all mathematical velocity candidates,original-domain/piece/context filtering,rest intervals and isolated/endpoint rest,reversal signs,every monotone leg,total distance,displacement and both averages. The two construction lessons require32 original graph constructions across their cases,with independently entered exact tables and axis quantities/units/scales,actual unsnapped points and stroke vertices,exact open-boundary decisions,separate branches at idealised speed jumps,and retained physical branches and ends beyond a cropped frame. Positive acceleration need not increase speed;speed cusps and mathematical two-sided versus physical one-sided gradients remain distinct. Practical rates retain their actual dependent/input units,positive magnitude,signed instantaneous value,original finite differences/averages,open physical ends and cancelled original holes. Correct independent answers cannot replace incorrect actual working. All three complete original teaching/worked/practice records are required for one structural 4N link;All14Functions source rows have reviewed structural teaching/practice links;other Mathematics B source gaps remain explicit in the current course map. The wider Pearson curriculum remains incomplete. This finite reference supports at most six continuous position pieces,eight terms per expression,powers -2 through6,complete cleared velocity degree at most five and exact rational/quadratic-field roots. These are software bounds,not syllabus exclusions;unresolved cases require review. Actual graph comparisons use finite position tolerances and65 probes per physical branch,including off-window portions;they do not prove a globally correct smooth curve. Automated calculation and drawing controls are checked;fresh browser interaction remains unverified. Original invented models and metadata-only bookISBN9780435044107 do not establish real physical/device/price validity,a full book body/chapter/Hub map,practical competence or specialist approval. No marks,mastery or formal assessment readiness. Single Higher-tier linear4MB1,exam entry context2026/27only;no MathematicsA/Foundation/modular inheritance.
Before you begin
- Signed velocity, speed and monotone-leg distance calculations
- Graph coordinates, units and scales
- Original domain and open or closed physical endpoints
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What belongs on the vertical axis of a total-distance/time graph?
What you will learn
- Construct original distance/time, speed/time and related signed motion graphs from independently entered exact tables.
- Choose quantities, units and scales; place actual own points and stroke vertices without snapping or clamping.
- Distinguish included reversals, isolated rest, rest intervals, speed cusps and undefined idealised corners.
- Retain open ends and off-window physical branches, and explain the limits of finite graph comparisons.
Draw and interpret motion graphs: Worked
Build the graphs as well as calculating the rates. Choose the correct quantity, retain its physical interval, and use your own exact tables, plotted points and strokes to show how the motion changes.
Study the method, examples and limits
Choose a quantity before choosing a shape
A position/time graph has signed position on its vertical axis. Its gradient is signed velocity. A distance/time graph records total distance since the stated start; its gradient is speed on the differentiable motion pieces. It cannot decrease. A horizontal interval represents rest, while a single horizontal tangent can be only an instant of rest.
A velocity/time graph can extend below zero and has signed acceleration as its gradient. A speed/time graph is nonnegative. Where velocity is negative, its speed-graph gradient is the negative of acceleration; where velocity is positive, it agrees with acceleration. At a smooth reversal with nonzero acceleration, speed has a cusp although velocity is differentiable.
Construct from your own table and geometry
Label time and the selected vertical quantity with units. Use the stated scale and axis window, then enter your own exact table values and separate domain/context decisions. Place actual points at those values and draw actual strokes with enough vertices to express curvature. A correct derivative or table does not draw the curve for you.
Retain actual coordinates without snapping a wrong point onto a solution. Use coordinate entry for points beyond the current frame; a cropped view does not shorten the physical trip. Closed physical endpoints remain included even when off screen.
Show continuity, corners and excluded ends
A continuous position corner remains a point on the position or distance graph, but its gradient can fail to exist. For the idealised stopping model, draw separate speed branches with open markers and open stroke ends at each jump. A vertical or diagonal connector would assign speeds the original instantaneous model does not supply.
Distinguish that jump from a continuous zero-speed cusp at a smooth reversal. The zero of a cusp is included. At a physical starting or ending time, a one-sided graph gradient describes the permitted side; do not confuse it with a two-sided gradient on the mathematical extension.
Explain what the graph demonstrates
Use the drawn slope, horizontal intervals, signs and corners to explain the corresponding motion. Compare position and travelled distance after a reversal, and signed velocity and speed before it. Preserve original reciprocal exclusions in the mathematical account even when they lie outside the visible physical graph.
Finite checks compare actual points, stroke segments and selected visible locations. They support self-checking but do not prove that every point of a drawn smooth curve is correct. Written graph interpretation and physical modelling still require teacher or specialist review.
In this stage: 0 of 2 current case attempts recorded; 0 fixed comparisons agree. These counts award no marks or mastery.
Original givens
- Piece 1: s=2+4*t, mathematical piece bounds -infinity to infinity. At joins, compare one-sided derivatives.
Physical time interval: [0, 3] s. Position unit: m.
Named inputs: 0, 1, 2, 3. Retain the original expression’s mathematical exclusions separately from physical permission.
Draw both graphs from your own exact tables and actual points/strokes. Distance travelled starts at 0 even though the original position starts at 2 m.
- For each graph choose quantity, units, bounds and scale explicitly; complete its own exact table with physical-admission decisions.
- Enter and edit your own points and stroke vertices. For an excluded finite end, choose an open marker, its exact boundary input and an open stroke end.
- Retain valid off-window points and keep discontinuous branches separate. Give an independent explanation of gradients, rest, reversals and corners.
- Check each construction separately. Correct numerical answers cannot rescue an absent or incorrectly joined drawing.
Use explicit * and ^ in formulas, such as 3*t^2-2/t. Exact values may use fractions and sqrt(2). Enter undefined for an undefined mathematical reading. Formula pieces are numbered from one. For a constant derivative, enter zero coefficient and leave its surviving power blank. For a rest interval use “not applicable” for the two individual side-sign fields; for endpoint rest use “no physical side” on the missing side.
My actual working
Original domain and derivative expressions
Term-by-term first and second differentiation
Named-input quantity, rate and context readings
Complete mathematical velocity equations
Every cleared candidate before physical filtering
Physical rest sets and reversal decisions
Every monotone leg and its length
Exact tables for each separate graph
Quantity, units, bounds and scales for each graph
Construct my two original graphs
Complete your own axis and exact-value tables above as well as these actual drawings. Record each current construction before recording the full case. Included points at the physical ends and excluded open ends must be retained even when a frame crops their view.
My construction: Distance travelled from the start
Given frame: input 0 to 3 in steps of 1; distance travelled from the start 0 to 14 m in steps of 2. This is an empty construction surface until you add your own points and strokes.
Add a point or stroke, choose an item to place, then click the graph. Use arrow keys to move that item; hold Shift for smaller steps. Enter coordinates directly for points outside the frame. A stroke joins your consecutive vertices; use enough vertices to represent a curve. Choose open or closed ends yourself.
No point or vertex selected.
My current geometry is not recorded, or has changed.
Drawing and review limits
- These controls record only actual drawing geometry. The full learning task also requires the learner’s own axes, exact table, calculation, interpretation and separate answers. Geometry agreement alone awards no task pass, marks, mastery or syllabus coverage.
- Points and stroke vertices remain the learner’s actual coordinates. A click or arrow key changes the chosen item; it does not snap it to a reference curve. Exact open-boundary decisions are entered separately.
- No reference curve, solution points or completed stroke is supplied at the start. Finite shape comparisons support a self-check; specialist graph judgment and formal assessment remain pending.
My construction: Speed
Given frame: input 0 to 3 in steps of 1; speed 0 to 6 m/s in steps of 1. This is an empty construction surface until you add your own points and strokes.
Add a point or stroke, choose an item to place, then click the graph. Use arrow keys to move that item; hold Shift for smaller steps. Enter coordinates directly for points outside the frame. A stroke joins your consecutive vertices; use enough vertices to represent a curve. Choose open or closed ends yourself.
No point or vertex selected.
My current geometry is not recorded, or has changed.
Drawing and review limits
- These controls record only actual drawing geometry. The full learning task also requires the learner’s own axes, exact table, calculation, interpretation and separate answers. Geometry agreement alone awards no task pass, marks, mastery or syllabus coverage.
- Points and stroke vertices remain the learner’s actual coordinates. A click or arrow key changes the chosen item; it does not snap it to a reference curve. Exact open-boundary decisions are entered separately.
- No reference curve, solution points or completed stroke is supplied at the start. Finite shape comparisons support a self-check; specialist graph judgment and formal assessment remain pending.
Current working not recorded, or has changed.
My separate final answers
Enter these independently. They cannot supply or repair missing term steps, candidate roots, legs, graph tables or actual drawing geometry.
Separate answers: Original domain and derivative expressions
Separate answers: Named-input quantity, rate and context readings
Separate answers: Every cleared candidate before physical filtering
Separate answers: Physical rest sets and reversal decisions
Separate answers: Exact tables for each separate graph
Current separate answers not recorded, or have changed.
Record your current calculation, table and actual graph working for this case.
What the self-checks cover
- All calculation and answer fields start blank. Original formulas, piece intervals, sample inputs and drawing frames are givens, not recorded learner work. Term rows refer to original formula pieces numbered from one.
- Record your own rewritten terms, coefficient/power differentiation, first and second derivatives, complete mathematical candidates, original-domain/piece/context decisions, rest sets and every physical monotone leg. Correct totals cannot replace incorrect working.
- Exact rational and quadratic-surd values are required in these authored cases. A derivative identity has a whole rest interval. Candidate rows retain roots outside the physical trip; original holes and idealised corners stay explicit.
- Each graph needs your own exact table, quantity/unit/scale decisions and actual construction. Geometry agreement alone does not pass the case; a separate correct numerical answer cannot supply missing points, strokes or open-end decisions.
- A complete current attempt can be studied against an explained reference even when it is wrong. Fixed comparisons do not award marks or mastery. Free written reasoning, completeness explanations, physical validity, specialist review and formal assessment remain for human review.
- Original continuous position formulas and stated physical time intervals are separate. Mathematical extensions do not validate negative times or extrapolated physical predictions.
- The general integer-power rule is taught;this finite reference uses at most six position pieces,eight terms per piece,powers -2 through6 and a complete cleared velocity equation of degree at most five. These are tool bounds,not syllabus exclusions.
- Every finite velocity-root certificate accounts for the whole real cleared equation before original-domain,piece and physical-context filtering. Exact quadratic-field roots and complete sign partitions support these authored distance calculations;unresolved algebraic cases require review.
- Total distance is the sum of absolute position changes over every certified monotone leg and rest piece. It is not a signed displacement and is not obtained by pretending that finite sampled points prove no reversal.
- Speed is the magnitude of signed velocity on the supported differentiable motion pieces. Increasing speed depends on both velocity and acceleration signs. A speed-graph cusp at a smooth reversal does not make the signed acceleration undefined.
- Two-sided speed slopes on the mathematical extension are separate from one-sided readings at the closed physical start/end. At an admitted endpoint with zero velocity and nonzero acceleration,the initial one-sided speed slope is positive magnitude of acceleration and the final one-sided slope is its negative magnitude.
- At idealised position or velocity corners,the two one-sided derivatives are checked explicitly. No finite acceleration is assigned across a velocity jump. Equal-speed instantaneous reversals at a corner require an explicit convention and human review.
- Actual distance/time,speed/time and other practical graph drawing remains a separate required learner strand. No source completion or drawing credit comes from this reference engine alone. Written reasoning,physical model validity,specialist review and formal assessment remain pending.
- Every graph needs the learner’s own exact table,actual point markers and actual stroke vertices. No ideal reference curve or correct separate answer substitutes for construction.
- Open boundary labels are exact input decisions,separate from finite drawing-position tolerances. Every finite open end needs an actual open marker and an explicitly open stroke end. Idealised velocity/speed jumps must use separate branches,not a diagonal or vertical connecting segment.
- Actual points and vertices remain unsnapped and uncropped in storage. A valid physical point outside the drawing window must be retained;the window does not change the original domain or model context.
- All retained vertices and three interior samples per segment are compared. Each physical branch has65fixed probe positions,including off-window portions,and both original included or excluded stroke ends must be retained. Visible and whole-physical finite comparisons are reported separately;they do not prove a globally correct smooth curve or approve human drawing judgment.
- Motion and practical models preserve their original exclusions. Quantity height,signed rate,speed magnitude,physical context and plotted position are separate. Written reasoning,model validity,specialist review and formal assessment remain pending.
Specialist review, practical competence and formal assessment readiness remain unverified. No proprietary book text or past-paper questions are reproduced.
Choose a quantity before choosing a shape
A position/time graph has signed position on its vertical axis. Its gradient is signed velocity. A distance/time graph records total distance since the stated start; its gradient is speed on the differentiable motion pieces. It cannot decrease. A horizontal interval represents rest, while a single horizontal tangent can be only an instant of rest.
A velocity/time graph can extend below zero and has signed acceleration as its gradient. A speed/time graph is nonnegative. Where velocity is negative, its speed-graph gradient is the negative of acceleration; where velocity is positive, it agrees with acceleration. At a smooth reversal with nonzero acceleration, speed has a cusp although velocity is differentiable.
Construct from your own table and geometry
Label time and the selected vertical quantity with units. Use the stated scale and axis window, then enter your own exact table values and separate domain/context decisions. Place actual points at those values and draw actual strokes with enough vertices to express curvature. A correct derivative or table does not draw the curve for you.
Retain actual coordinates without snapping a wrong point onto a solution. Use coordinate entry for points beyond the current frame; a cropped view does not shorten the physical trip. Closed physical endpoints remain included even when off screen.
Show continuity, corners and excluded ends
A continuous position corner remains a point on the position or distance graph, but its gradient can fail to exist. For the idealised stopping model, draw separate speed branches with open markers and open stroke ends at each jump. A vertical or diagonal connector would assign speeds the original instantaneous model does not supply.
Distinguish that jump from a continuous zero-speed cusp at a smooth reversal. The zero of a cusp is included. At a physical starting or ending time, a one-sided graph gradient describes the permitted side; do not confuse it with a two-sided gradient on the mathematical extension.
Explain what the graph demonstrates
Use the drawn slope, horizontal intervals, signs and corners to explain the corresponding motion. Compare position and travelled distance after a reversal, and signed velocity and speed before it. Preserve original reciprocal exclusions in the mathematical account even when they lie outside the visible physical graph.
Finite checks compare actual points, stroke segments and selected visible locations. They support self-checking but do not prove that every point of a drawn smooth curve is correct. Written graph interpretation and physical modelling still require teacher or specialist review.
Sources for this lesson
Pause and explain
Does a correct numerical table alone count as constructing the required graph?
Worked example
Original position piece 1: s=2+4*t on mathematical piece bounds -infinity to infinity. Physical trip [0,3] s; position unit m. Given inputs 0, 1, 2, 3. Draw both graphs from your own exact tables and actual points/strokes. Distance travelled starts at 0 even though the original position starts at 2 m. Construct Distance travelled from the start (m), horizontal frame 0, 3, 1, vertical frame 0, 14, 2. Construct Speed (m/s), horizontal frame 0, 3, 1, vertical frame 0, 6, 1. For each graph choose quantity, units, bounds and scale explicitly; complete its own exact table with physical-admission decisions. Enter and edit your own points and stroke vertices. For an excluded finite end, choose an open marker, its exact boundary input and an open stroke end. Retain valid off-window points and keep discontinuous branches separate. Give an independent explanation of gradients, rest, reversals and corners. Check each construction separately. Correct numerical answers cannot rescue an absent or incorrectly joined drawing. Original position piece 1: s=t^2 on mathematical piece bounds -infinity to infinity. Physical trip [0,3] s; position unit m. Given inputs 0, 1, 2, 3. Construct the curved distance/time graph and the straight speed/time graph. Relate their different gradients to speed and acceleration without replacing the drawings by a correct derivative. Construct Distance travelled (m), horizontal frame 0, 3, 1, vertical frame 0, 10, 2. Construct Speed (m/s), horizontal frame 0, 3, 1, vertical frame 0, 8, 2. For each graph choose quantity, units, bounds and scale explicitly; complete its own exact table with physical-admission decisions. Enter and edit your own points and stroke vertices. For an excluded finite end, choose an open marker, its exact boundary input and an open stroke end. Retain valid off-window points and keep discontinuous branches separate. Give an independent explanation of gradients, rest, reversals and corners. Check each construction separately. Correct numerical answers cannot rescue an absent or incorrectly joined drawing.
Show the worked solution
- Case 1: From s=2+4t and v=4, the distance travelled from t=0 is D=4t on [0,3]. Its table at 0,1,2,3 is 0,4,8,12 m.
- Case 1: Draw the straight distance/time graph through your own points, starting at (0,0). Its gradient is 4 m/s.
- Case 1: Draw a separate horizontal speed/time graph at 4 m/s with both physical endpoints included. Its gradient is zero.
- Case 2: On [0,3], s=t^2, v=2t and a=2. Since v is nonnegative, D=t^2 and speed is 2t.
- Case 2: At 0,1,2,3 the distance values are 0,1,4,9 m and speed values are 0,2,4,6 m/s. Draw a curved distance graph and a straight speed graph.
- Case 2: The distance graph becomes steeper; the speed graph has constant gradient 2 m/s^2. At its physical start, use the right-hand speed gradient.
Answer Case 1: From s=2+4t and v=4, the distance travelled from t=0 is D=4t on [0,3]. Its table at 0,1,2,3 is 0,4,8,12 m. Draw the straight distance/time graph through your own points, starting at (0,0). Its gradient is 4 m/s. Draw a separate horizontal speed/time graph at 4 m/s with both physical endpoints included. Its gradient is zero. The original initial position 2 m belongs on a position graph, not as the initial distance travelled. Correct tables and derivative answers do not substitute for the learner’s actual points, axes and strokes. Case 2: On [0,3], s=t^2, v=2t and a=2. Since v is nonnegative, D=t^2 and speed is 2t. At 0,1,2,3 the distance values are 0,1,4,9 m and speed values are 0,2,4,6 m/s. Draw a curved distance graph and a straight speed graph. The distance graph becomes steeper; the speed graph has constant gradient 2 m/s^2. At its physical start, use the right-hand speed gradient. Constant acceleration produces a changing distance gradient, not a straight distance graph. Connect sufficiently many own stroke vertices to express curvature. Sparse straight chords can disagree between otherwise correct table points.
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Original position piece 1: s=2*t on mathematical piece bounds -infinity to 1. Original position piece 2: s=2 on mathematical piece bounds 1 to 2. Original position piece 3: s=3*t-4 on mathematical piece bounds 2 to infinity. Physical trip [0,4] s; position unit m. Given inputs 0, 1/2, 1, 3/2, 2, 3, 4. Show the horizontal rest interval on distance/time and the zero-speed interval on speed/time. At the idealised corners, record undefined instantaneous velocity/acceleration and draw open step ends without a false diagonal joining them. Construct Distance travelled (m), horizontal frame 0, 4, 1, vertical frame 0, 10, 2. Construct Speed (m/s), horizontal frame 0, 4, 1, vertical frame 0, 4, 1. For each graph choose quantity, units, bounds and scale explicitly; complete its own exact table with physical-admission decisions. Enter and edit your own points and stroke vertices. For an excluded finite end, choose an open marker, its exact boundary input and an open stroke end. Retain valid off-window points and keep discontinuous branches separate. Give an independent explanation of gradients, rest, reversals and corners. Check each construction separately. Correct numerical answers cannot rescue an absent or incorrectly joined drawing. Original position piece 1: s=t^2-4*t+3 on mathematical piece bounds -infinity to infinity. Physical trip [0,4] s; position unit m. Given inputs 0, 1, 2, 3, 4. Draw signed velocity separately from nonnegative speed. At 2 s, the velocity is smooth but speed has a corner; its left/right gradients are not a single signed acceleration. Construct Signed velocity (m/s), horizontal frame 0, 4, 1, vertical frame -6, 6, 2. Construct Speed magnitude (m/s), horizontal frame 0, 4, 1, vertical frame 0, 6, 1. For each graph choose quantity, units, bounds and scale explicitly; complete its own exact table with physical-admission decisions. Enter and edit your own points and stroke vertices. For an excluded finite end, choose an open marker, its exact boundary input and an open stroke end. Retain valid off-window points and keep discontinuous branches separate. Give an independent explanation of gradients, rest, reversals and corners. Check each construction separately. Correct numerical answers cannot rescue an absent or incorrectly joined drawing.
Give me a hint
For each graph choose quantity, units, bounds and scale explicitly; complete its own exact table with physical-admission decisions. Enter and edit your own points and stroke vertices. For an excluded finite end, choose an open marker, its exact boundary input and an open stroke end. Retain valid off-window points and keep discontinuous branches separate. Give an independent explanation of gradients, rest, reversals and corners. Check each construction separately. Correct numerical answers cannot rescue an absent or incorrectly joined drawing.
Compare my reasoning
- Case 1: The position pieces are 2t before 1, 2 between 1 and 2, and 3t-4 after 2. Position is continuous and always nondecreasing, so D equals this position on [0,4].
- Case 1: Draw distance from (0,0) to (1,2), horizontally to (2,2), then to (4,8). The two corner positions are included.
- Case 1: Draw speed 2 on [0,1), zero on (1,2), and 3 on (2,4]. The rest interval is (1,2); neither idealised corner has a two-sided velocity or acceleration.
- Case 1: Use open markers and open stroke ends at (1,2), (1,0), (2,0) and (2,3) on the speed graph. Keep the three branches separate; do not join a jump vertically or diagonally.
- Case 2: For s=t^2-4t+3, draw signed velocity v=2t-4 through (0,-4), (2,0) and (4,4). This is one straight line of gradient 2.
- Case 2: Draw speed |2t-4| through (0,4), (2,0) and (4,4), with two straight arms meeting at an included point.
- Case 2: Before 2 the speed gradient is -2; after 2 it is 2. At 2 there is no single two-sided speed gradient, although acceleration is 2.
Case 1: The position pieces are 2t before 1, 2 between 1 and 2, and 3t-4 after 2. Position is continuous and always nondecreasing, so D equals this position on [0,4]. Draw distance from (0,0) to (1,2), horizontally to (2,2), then to (4,8). The two corner positions are included. Draw speed 2 on [0,1), zero on (1,2), and 3 on (2,4]. The rest interval is (1,2); neither idealised corner has a two-sided velocity or acceleration. Use open markers and open stroke ends at (1,2), (1,0), (2,0) and (2,3) on the speed graph. Keep the three branches separate; do not join a jump vertically or diagonally. A position corner can have an admitted position and accumulated distance while instantaneous velocity is undefined. A speed jump is an idealisation, not a finite constant acceleration. This convention derives speed from the differentiable position pieces. Real transitions require physical modelling and specialist review. Case 2: For s=t^2-4t+3, draw signed velocity v=2t-4 through (0,-4), (2,0) and (4,4). This is one straight line of gradient 2. Draw speed |2t-4| through (0,4), (2,0) and (4,4), with two straight arms meeting at an included point. Before 2 the speed gradient is -2; after 2 it is 2. At 2 there is no single two-sided speed gradient, although acceleration is 2. Speed reflects the negative part of the velocity graph above the time axis. Signed acceleration is the velocity gradient, not always the speed gradient. The reversal is a continuous zero-speed point, not an open jump.
Look for these in your work
- My original formulas, term-by-term derivatives and domain/context decisions are explicit.
- I retained every relevant candidate, rest decision and monotone leg, or each practical finite comparison.
- My graph quantities, units, scales and exact tables are independently entered.
- My actual points, strokes and included/excluded ends retain the physical trip, including off-window portions.
- My separate final answers cannot repair incorrect own working. Written reasoning and model validity still need human review.
Original position piece 1: s=6*t-t^2 on mathematical piece bounds -infinity to infinity. Physical trip [0,5] s; position unit m. Given inputs 0, 1, 2, 3, 4, 5. Construct both graphs through and after the reversal. The position may decrease; the accumulated distance travelled cannot decrease. Construct Signed position relative to the origin (m), horizontal frame 0, 5, 1, vertical frame 0, 10, 2. Construct Total distance travelled (m), horizontal frame 0, 5, 1, vertical frame 0, 14, 2. For each graph choose quantity, units, bounds and scale explicitly; complete its own exact table with physical-admission decisions. Enter and edit your own points and stroke vertices. For an excluded finite end, choose an open marker, its exact boundary input and an open stroke end. Retain valid off-window points and keep discontinuous branches separate. Give an independent explanation of gradients, rest, reversals and corners. Check each construction separately. Correct numerical answers cannot rescue an absent or incorrectly joined drawing. Original position piece 1: s=(t-1)^3 on mathematical piece bounds -infinity to infinity. Physical trip [0,2] s; position unit m. Given inputs 0, 1/2, 1, 3/2, 2. Draw the flat instant on the increasing distance graph and the zero of the speed graph. Explain the difference between a single instant of rest and an interval of rest. Construct Distance travelled (m), horizontal frame 0, 2, 1/2, vertical frame 0, 3, 1/2. Construct Speed (m/s), horizontal frame 0, 2, 1/2, vertical frame 0, 4, 1. For each graph choose quantity, units, bounds and scale explicitly; complete its own exact table with physical-admission decisions. Enter and edit your own points and stroke vertices. For an excluded finite end, choose an open marker, its exact boundary input and an open stroke end. Retain valid off-window points and keep discontinuous branches separate. Give an independent explanation of gradients, rest, reversals and corners. Check each construction separately. Correct numerical answers cannot rescue an absent or incorrectly joined drawing.
Give me a hint
For each graph choose quantity, units, bounds and scale explicitly; complete its own exact table with physical-admission decisions. Enter and edit your own points and stroke vertices. For an excluded finite end, choose an open marker, its exact boundary input and an open stroke end. Retain valid off-window points and keep discontinuous branches separate. Give an independent explanation of gradients, rest, reversals and corners. Check each construction separately. Correct numerical answers cannot rescue an absent or incorrectly joined drawing.
Compare my reasoning
- Case 1: For s=6t-t^2 on [0,5], v=6-2t is zero at 3. Position values at 0,1,2,3,4,5 are 0,5,8,9,8,5 m.
- Case 1: Before the reversal, D=6t-t^2. After 3, D=9+(9-s)=t^2-6t+18. Distance values at the same inputs are 0,5,8,9,10,13 m.
- Case 1: Construct the position curve turning down after (3,9), and the accumulated-distance curve continuing upwards. Both have zero gradient at the rest instant.
- Case 2: For s=(t-1)^3 on [0,2], v=3(t-1)^2 is never negative. Since the starting position is -1, D=(t-1)^3+1.
- Case 2: At 0,1/2,1,3/2,2, distances are 0,7/8,1,9/8,2 m and speeds are 3,3/4,0,3/4,3 m/s.
- Case 2: Draw the increasing distance graph with a horizontal tangent at (1,1). Draw the curved speed graph touching zero at 1. Neither graph has a horizontal interval of positive duration.
Case 1: For s=6t-t^2 on [0,5], v=6-2t is zero at 3. Position values at 0,1,2,3,4,5 are 0,5,8,9,8,5 m. Before the reversal, D=6t-t^2. After 3, D=9+(9-s)=t^2-6t+18. Distance values at the same inputs are 0,5,8,9,10,13 m. Construct the position curve turning down after (3,9), and the accumulated-distance curve continuing upwards. Both have zero gradient at the rest instant. Decreasing position after a reversal contributes additional positive distance. Accumulated distance cannot decrease. A physical position graph and a total-distance graph need independent constructions even when some points coincide. Case 2: For s=(t-1)^3 on [0,2], v=3(t-1)^2 is never negative. Since the starting position is -1, D=(t-1)^3+1. At 0,1/2,1,3/2,2, distances are 0,7/8,1,9/8,2 m and speeds are 3,3/4,0,3/4,3 m/s. Draw the increasing distance graph with a horizontal tangent at (1,1). Draw the curved speed graph touching zero at 1. Neither graph has a horizontal interval of positive duration. A single flat instant can occur within a continually increasing distance graph. It does not establish a stop lasting an interval. The distance curve must not be replaced by a flat segment near 1 just because the exact gradient there is zero.
Look for these in your work
- My original formulas, term-by-term derivatives and domain/context decisions are explicit.
- I retained every relevant candidate, rest decision and monotone leg, or each practical finite comparison.
- My graph quantities, units, scales and exact tables are independently entered.
- My actual points, strokes and included/excluded ends retain the physical trip, including off-window portions.
- My separate final answers cannot repair incorrect own working. Written reasoning and model validity still need human review.
Original position piece 1: s=t^4-8*t^2 on mathematical piece bounds -infinity to infinity. Physical trip [0,3] s; position unit m. Given inputs 0, 1/2, 1, 3/2, 2, 5/2, 3. Retain all physical table points and strokes, including the valid point at 3 s outside the stated drawing window. Do not clamp it to a frame edge or redefine the physical trip. Construct Signed position (m), horizontal frame 0, 5/2, 1/2, vertical frame -20, 12, 4. Construct Total distance travelled (m), horizontal frame 0, 5/2, 1/2, vertical frame 0, 32, 4. For each graph choose quantity, units, bounds and scale explicitly; complete its own exact table with physical-admission decisions. Enter and edit your own points and stroke vertices. For an excluded finite end, choose an open marker, its exact boundary input and an open stroke end. Retain valid off-window points and keep discontinuous branches separate. Give an independent explanation of gradients, rest, reversals and corners. Check each construction separately. Correct numerical answers cannot rescue an absent or incorrectly joined drawing. Original position piece 1: s=t+4/t on mathematical piece bounds -infinity to infinity. Physical trip [1,4] s; position unit m. Given inputs 1, 3/2, 2, 5/2, 3, 4. Draw the original reciprocal position and the speed magnitude on the stated physical interval. Keep its original mathematical zero exclusion even though zero is outside this drawing. Construct Signed position (m), horizontal frame 1, 4, 1/2, vertical frame 3, 6, 1/2. Construct Speed (m/s), horizontal frame 1, 4, 1/2, vertical frame 0, 4, 1. For each graph choose quantity, units, bounds and scale explicitly; complete its own exact table with physical-admission decisions. Enter and edit your own points and stroke vertices. For an excluded finite end, choose an open marker, its exact boundary input and an open stroke end. Retain valid off-window points and keep discontinuous branches separate. Give an independent explanation of gradients, rest, reversals and corners. Check each construction separately. Correct numerical answers cannot rescue an absent or incorrectly joined drawing.
Give me a hint
For each graph choose quantity, units, bounds and scale explicitly; complete its own exact table with physical-admission decisions. Enter and edit your own points and stroke vertices. For an excluded finite end, choose an open marker, its exact boundary input and an open stroke end. Retain valid off-window points and keep discontinuous branches separate. Give an independent explanation of gradients, rest, reversals and corners. Check each construction separately. Correct numerical answers cannot rescue an absent or incorrectly joined drawing.
Compare my reasoning
- Case 1: For s=t^4-8t^2 on [0,3], position decreases to -16 at 2 and increases afterwards. Before 2, D=-s; after 2, D=16+(s+16)=s+32.
- Case 1: At t=5/2, s=-175/16 m and D=337/16 m. At t=3, s=9 m and D=41 m.
- Case 1: The displayed frame stops at t=5/2, but the physical trip ends at 3. Retain both true points at 3 and the actual coordinates of your strokes outside the frame.
- Case 1: Show the visible portions within the original axes. A point at the edge t=5/2 cannot stand in for the physical point at t=3.
- Case 2: For s=t+4/t on [1,4], the original mathematical domain excludes 0. Velocity is 1-4/t^2, so speed is |1-4/t^2|.
- Case 2: At 1,2,4, position is 5,4,5 m and speed is 3,0,3/4 m/s. Position has a smooth minimum at 2; speed has a cusp at its included zero.
- Case 2: Draw the decreasing then increasing position curve and the two curved speed arms. Keep 0 excluded in the mathematical account even though it lies outside this physical graph.
Case 1: For s=t^4-8t^2 on [0,3], position decreases to -16 at 2 and increases afterwards. Before 2, D=-s; after 2, D=16+(s+16)=s+32. At t=5/2, s=-175/16 m and D=337/16 m. At t=3, s=9 m and D=41 m. The displayed frame stops at t=5/2, but the physical trip ends at 3. Retain both true points at 3 and the actual coordinates of your strokes outside the frame. Show the visible portions within the original axes. A point at the edge t=5/2 cannot stand in for the physical point at t=3. A drawing window crops the view. It does not crop the mathematical domain, the physical context or the stored construction. Finite drawing checks assess sampled agreement; off-window retention does not certify a globally correct curve. Case 2: For s=t+4/t on [1,4], the original mathematical domain excludes 0. Velocity is 1-4/t^2, so speed is |1-4/t^2|. At 1,2,4, position is 5,4,5 m and speed is 3,0,3/4 m/s. Position has a smooth minimum at 2; speed has a cusp at its included zero. Draw the decreasing then increasing position curve and the two curved speed arms. Keep 0 excluded in the mathematical account even though it lies outside this physical graph. A reciprocal law can produce equal endpoint positions and unequal endpoint speeds. The two graphs carry different quantities. Draw curved arms from enough actual vertices; a correct root or table alone does not verify their shape.
Look for these in your work
- My original formulas, term-by-term derivatives and domain/context decisions are explicit.
- I retained every relevant candidate, rest decision and monotone leg, or each practical finite comparison.
- My graph quantities, units, scales and exact tables are independently entered.
- My actual points, strokes and included/excluded ends retain the physical trip, including off-window portions.
- My separate final answers cannot repair incorrect own working. Written reasoning and model validity still need human review.
Common mistakes
- Equating signed displacement with the total distance travelled.
- Reading the acceleration sign alone as increasing or decreasing speed.
- Deleting mathematical velocity candidates before the original-domain, piece and context decisions.
- Treating every zero velocity as a reversal, or an idealised corner as finite acceleration.
- Using a correct final answer or table in place of actual derivative working and graph construction.
- Clamping physical points to a drawing frame, joining speed jumps, or filling original holes after simplification.
- Equating a marginal instantaneous rate with a finite change, or omitting the actual input unit.
What can you explain now?
Does a correct numerical table alone count as constructing the required graph?
Compare with the explanation
No.
Retain your own axes, actual points, stroke vertices and inclusion decisions; correct numbers do not replace drawing.
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Make it stick
Does a correct numerical table alone count as constructing the required graph? No. Retain your own axes, actual points, stroke vertices and inclusion decisions; correct numbers do not replace drawing. Do fixed self-checks establish specialist approval, practical competence, exam marks or formal assessment readiness? No; written reasoning and model validity still need human review.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.