Position, velocity, speed and acceleration
Describe motion with a signed position and a stated time interval. Differentiate to obtain velocity and acceleration, then connect those signed rates to direction, speed and the actual distance travelled.
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
- Integer-power differentiation and exact substitution
- Signed coordinates, real roots and original domains
- Rates, units and physical time intervals
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
An object returns to its initial position. Must its travelled distance be zero?
What you will learn
- Differentiate the original position twice with notation, units and retained mathematical restrictions.
- Distinguish signed position/displacement and velocity from accumulated distance and speed.
- Account for all velocity candidates before piece and physical filtering; classify rest intervals, endpoint rest and reversals.
- Sum every monotone leg and compare signed displacement, total distance, average velocity and average speed.
Position, velocity, speed and acceleration: Worked
Describe motion with a signed position and a stated time interval. Differentiate to obtain velocity and acceleration, then connect those signed rates to direction, speed and the actual distance travelled.
Study the method, examples and limits
Start with position and its original domain
Write s(t) with its position unit, time unit and physical interval. Position uses a coordinate origin. Displacement from one time to another is final position minus initial position. Distance travelled accumulates the lengths of the movements; it is nonnegative and starts at zero at the stated beginning of the trip.
Retain exclusions in the original formula even after rewriting integer powers or cancelling a factor. A formula can have a mathematical value outside the physical interval without justifying a physical prediction there. A mathematical hole is a different issue from a context restriction.
Differentiate twice and interpret the signs
Apply the integer-power rule term by term: the derivative of c*t^n is c*n*t^(n-1). Constants contribute zero. Velocity v=ds/dt is signed; speed is |v|. Acceleration a=dv/dt=d^2s/dt^2 measures change in signed velocity. If position is in metres and time in seconds, their units are m/s and m/s^2.
Positive velocity gives motion in the positive coordinate direction, and negative velocity gives the opposite direction. At a nonzero velocity, acceleration with the same sign increases speed and acceleration with the opposite sign decreases it. Zero acceleration gives an instantaneously unchanged speed. At zero velocity, examine neighbouring times instead of inferring a whole stopping interval from a single value.
Account for every rest time and every leg
Solve the complete velocity-zero equation before filtering its roots by the original domain, the formula piece and the physical interval. Clearing a reciprocal denominator retains its original exclusions. Count a repeated root once as a distinct time. If velocity is identically zero on an interval, state that entire rest interval rather than inventing a finite root list.
For each admitted isolated rest instant, examine velocity on both sides within the trip. Opposite signs give a reversal; matching signs give rest without reversal. A rest time at the physical start or end has only one neighbouring physical side, so identify it as endpoint rest. Include formula boundaries as well as every interior reversal in a monotone-leg partition. For these fixed self-checks, split at every admitted interior velocity zero, including rest without reversal. This keeps every candidate visible and can split one monotone movement into smaller monotone legs.
On each leg with no missed reversal, subtract original endpoint positions and take the absolute value for its length. Add every length, including any zero-length rest piece. Total distance divided by elapsed time gives average speed. Net displacement divided by elapsed time gives average velocity. These averages require positive elapsed time and need not equal instantaneous values.
Keep idealisation and checking limits explicit
At an idealised position corner, the one-sided velocities can differ. The position is then admitted but there is no two-sided instantaneous velocity or finite acceleration at the corner. Where velocity is continuous but its gradient jumps, velocity can be admitted while acceleration is undefined.
The power rule is general for integer powers on the original domain. The current reference tool supports only its declared finite expression and exact-root bounds; those software bounds are not syllabus exclusions. Physical realism, written explanations and assessment readiness need human 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, -1. Retain the original expression’s mathematical exclusions separately from physical permission.
Calculate position, velocity, speed and acceleration. The initial position is 2 m; measure total distance from the stated start rather than from the coordinate origin. Keep mathematical extrapolation separate from the physical interval.
- Rewrite each original position piece while preserving its domain. Enter your term-by-term first and second derivative working, notation and units before a separate final answer.
- For every named input, record original-domain and physical-context decisions, position, signed velocity, speed, signed acceleration, direction and speed trend.
- Give the complete mathematical velocity-root account, filter candidates explicitly, state isolated or interval rest and construct every monotone leg with original endpoint positions and absolute length.
- Calculate net displacement, total distance and both averages independently. A correct final distance cannot replace missing or incorrect leg working.
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
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
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.
Specialist review, practical competence and formal assessment readiness remain unverified. No proprietary book text or past-paper questions are reproduced.
Start with position and its original domain
Write s(t) with its position unit, time unit and physical interval. Position uses a coordinate origin. Displacement from one time to another is final position minus initial position. Distance travelled accumulates the lengths of the movements; it is nonnegative and starts at zero at the stated beginning of the trip.
Retain exclusions in the original formula even after rewriting integer powers or cancelling a factor. A formula can have a mathematical value outside the physical interval without justifying a physical prediction there. A mathematical hole is a different issue from a context restriction.
Differentiate twice and interpret the signs
Apply the integer-power rule term by term: the derivative of c*t^n is c*n*t^(n-1). Constants contribute zero. Velocity v=ds/dt is signed; speed is |v|. Acceleration a=dv/dt=d^2s/dt^2 measures change in signed velocity. If position is in metres and time in seconds, their units are m/s and m/s^2.
Positive velocity gives motion in the positive coordinate direction, and negative velocity gives the opposite direction. At a nonzero velocity, acceleration with the same sign increases speed and acceleration with the opposite sign decreases it. Zero acceleration gives an instantaneously unchanged speed. At zero velocity, examine neighbouring times instead of inferring a whole stopping interval from a single value.
Account for every rest time and every leg
Solve the complete velocity-zero equation before filtering its roots by the original domain, the formula piece and the physical interval. Clearing a reciprocal denominator retains its original exclusions. Count a repeated root once as a distinct time. If velocity is identically zero on an interval, state that entire rest interval rather than inventing a finite root list.
For each admitted isolated rest instant, examine velocity on both sides within the trip. Opposite signs give a reversal; matching signs give rest without reversal. A rest time at the physical start or end has only one neighbouring physical side, so identify it as endpoint rest. Include formula boundaries as well as every interior reversal in a monotone-leg partition. For these fixed self-checks, split at every admitted interior velocity zero, including rest without reversal. This keeps every candidate visible and can split one monotone movement into smaller monotone legs.
On each leg with no missed reversal, subtract original endpoint positions and take the absolute value for its length. Add every length, including any zero-length rest piece. Total distance divided by elapsed time gives average speed. Net displacement divided by elapsed time gives average velocity. These averages require positive elapsed time and need not equal instantaneous values.
Keep idealisation and checking limits explicit
At an idealised position corner, the one-sided velocities can differ. The position is then admitted but there is no two-sided instantaneous velocity or finite acceleration at the corner. Where velocity is continuous but its gradient jumps, velocity can be admitted while acceleration is undefined.
The power rule is general for integer powers on the original domain. The current reference tool supports only its declared finite expression and exact-root bounds; those software bounds are not syllabus exclusions. Physical realism, written explanations and assessment readiness need human review.
Sources for this lesson
Pause and explain
Can positive acceleration accompany decreasing speed?
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, -1. Calculate position, velocity, speed and acceleration. The initial position is 2 m; measure total distance from the stated start rather than from the coordinate origin. Keep mathematical extrapolation separate from the physical interval. Rewrite each original position piece while preserving its domain. Enter your term-by-term first and second derivative working, notation and units before a separate final answer. For every named input, record original-domain and physical-context decisions, position, signed velocity, speed, signed acceleration, direction and speed trend. Give the complete mathematical velocity-root account, filter candidates explicitly, state isolated or interval rest and construct every monotone leg with original endpoint positions and absolute length. Calculate net displacement, total distance and both averages independently. A correct final distance cannot replace missing or incorrect leg working. 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 1, 2, 3. Find every admitted rest instant and decide whether direction reverses. Compare signed net displacement with total distance and explain why positive acceleration can accompany decreasing speed. Rewrite each original position piece while preserving its domain. Enter your term-by-term first and second derivative working, notation and units before a separate final answer. For every named input, record original-domain and physical-context decisions, position, signed velocity, speed, signed acceleration, direction and speed trend. Give the complete mathematical velocity-root account, filter candidates explicitly, state isolated or interval rest and construct every monotone leg with original endpoint positions and absolute length. Calculate net displacement, total distance and both averages independently. A correct final distance cannot replace missing or incorrect leg working.
Show the worked solution
- Case 1: The original position is s=2+4t metres. Differentiating gives v=4 m/s and a=0 m/s^2; speed is |v|=4 m/s.
- Case 1: At t=0, position is 2 m but distance travelled since the start is 0 m. At t=1, position is 6 m and accumulated distance is 4 m.
- Case 1: On the physical interval [0,3], displacement and distance are both 12 m. Average velocity and average speed are both 4 m/s.
- Case 1: The formula gives position -2 m at t=-1, with the same mathematical derivative. That input lies outside the physical trip; no accumulated trip distance is assigned there.
- Case 2: For s=t^2-4t+3, v=2t-4 and a=2. The complete velocity equation gives t=2, inside [0,4].
- Case 2: At t=1,2,3, positions are 0,-1,0 m; velocities are -2,0,2 m/s and speeds are 2,0,2 m/s.
- Case 2: Velocity is negative before 2 and positive after 2, so direction reverses. Position changes from 3 to -1 to 3 m: distance is 4+4=8 m, while net displacement is 0 m.
- Case 2: Average speed over 4 s is 2 m/s and average velocity is 0 m/s. Before the reversal, positive acceleration opposes negative velocity, so speed decreases. After it, speed increases.
Answer Case 1: The original position is s=2+4t metres. Differentiating gives v=4 m/s and a=0 m/s^2; speed is |v|=4 m/s. At t=0, position is 2 m but distance travelled since the start is 0 m. At t=1, position is 6 m and accumulated distance is 4 m. On the physical interval [0,3], displacement and distance are both 12 m. Average velocity and average speed are both 4 m/s. The formula gives position -2 m at t=-1, with the same mathematical derivative. That input lies outside the physical trip; no accumulated trip distance is assigned there. Position uses an origin. Travelled distance measures movement from the stated starting time. A nonzero initial position is not an initial travelled distance. The straight-line model is invented; its mathematical extension does not validate physical predictions before the trip. Case 2: For s=t^2-4t+3, v=2t-4 and a=2. The complete velocity equation gives t=2, inside [0,4]. At t=1,2,3, positions are 0,-1,0 m; velocities are -2,0,2 m/s and speeds are 2,0,2 m/s. Velocity is negative before 2 and positive after 2, so direction reverses. Position changes from 3 to -1 to 3 m: distance is 4+4=8 m, while net displacement is 0 m. Average speed over 4 s is 2 m/s and average velocity is 0 m/s. Before the reversal, positive acceleration opposes negative velocity, so speed decreases. After it, speed increases. Acceleration describes change in signed velocity. Its positive sign alone does not tell you that speed is increasing. At the interior reversal the speed graph has a corner. Signed acceleration remains 2 m/s^2; a two-sided speed-graph gradient is undefined there.
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Original position piece 1: s=3 on mathematical piece bounds -infinity to infinity. Physical trip [0,4] s; position unit m. Given inputs 0, 2, 4. Differentiate the constant position. State the entire physical rest set rather than inventing a finite list of isolated stops. Rewrite each original position piece while preserving its domain. Enter your term-by-term first and second derivative working, notation and units before a separate final answer. For every named input, record original-domain and physical-context decisions, position, signed velocity, speed, signed acceleration, direction and speed trend. Give the complete mathematical velocity-root account, filter candidates explicitly, state isolated or interval rest and construct every monotone leg with original endpoint positions and absolute length. Calculate net displacement, total distance and both averages independently. A correct final distance cannot replace missing or incorrect leg working. 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. Solve the complete velocity equation and retain its repeated root once. Explain why an isolated rest instant does not necessarily reverse the motion. Rewrite each original position piece while preserving its domain. Enter your term-by-term first and second derivative working, notation and units before a separate final answer. For every named input, record original-domain and physical-context decisions, position, signed velocity, speed, signed acceleration, direction and speed trend. Give the complete mathematical velocity-root account, filter candidates explicitly, state isolated or interval rest and construct every monotone leg with original endpoint positions and absolute length. Calculate net displacement, total distance and both averages independently. A correct final distance cannot replace missing or incorrect leg working.
Give me a hint
Rewrite each original position piece while preserving its domain. Enter your term-by-term first and second derivative working, notation and units before a separate final answer. For every named input, record original-domain and physical-context decisions, position, signed velocity, speed, signed acceleration, direction and speed trend. Give the complete mathematical velocity-root account, filter candidates explicitly, state isolated or interval rest and construct every monotone leg with original endpoint positions and absolute length. Calculate net displacement, total distance and both averages independently. A correct final distance cannot replace missing or incorrect leg working.
Compare my reasoning
- Case 1: For the constant position s=3, both derivatives are zero: v=0 and a=0. Speed is zero at every physical time.
- Case 1: The equation v=0 is an identity, so the rest set is every t in [0,4], including its endpoints. A selected finite list would miss almost all rest times.
- Case 1: Every position change is zero. Distance, displacement, average speed and average velocity are all zero.
- Case 2: Expand s=(t-1)^3 or differentiate the cube: v=3(t-1)^2 and a=6(t-1). The sole distinct velocity root is t=1.
- Case 2: At t=0,1,2, position is -1,0,1 m, velocity is 3,0,3 m/s, and acceleration is -6,0,6 m/s^2.
- Case 2: Velocity is positive on both sides of 1. There is a single instant of rest without reversal. The increasing position has a horizontal tangent there.
- Case 2: On [0,2], displacement and distance are both 2 m, and both averages are 1 m/s.
Case 1: For the constant position s=3, both derivatives are zero: v=0 and a=0. Speed is zero at every physical time. The equation v=0 is an identity, so the rest set is every t in [0,4], including its endpoints. A selected finite list would miss almost all rest times. Every position change is zero. Distance, displacement, average speed and average velocity are all zero. A constant nonzero position can represent complete rest. Being 3 m from the origin does not mean travelling 3 m. The rest set is an interval, not a set of isolated reversal events. Case 2: Expand s=(t-1)^3 or differentiate the cube: v=3(t-1)^2 and a=6(t-1). The sole distinct velocity root is t=1. At t=0,1,2, position is -1,0,1 m, velocity is 3,0,3 m/s, and acceleration is -6,0,6 m/s^2. Velocity is positive on both sides of 1. There is a single instant of rest without reversal. The increasing position has a horizontal tangent there. On [0,2], displacement and distance are both 2 m, and both averages are 1 m/s. A zero of velocity needs neighbouring signs to decide whether direction changes. A repeated root is counted once as a distinct instant. A zero instantaneous speed does not by itself establish an interval of rest.
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 [3,5] s; position unit m. Given inputs 3, 4, 5. The physical trip starts at 3 s. Keep the mathematical velocity root and the physical endpoint decision separate. Determine speeding up or slowing down from both signed velocity and acceleration. Rewrite each original position piece while preserving its domain. Enter your term-by-term first and second derivative working, notation and units before a separate final answer. For every named input, record original-domain and physical-context decisions, position, signed velocity, speed, signed acceleration, direction and speed trend. Give the complete mathematical velocity-root account, filter candidates explicitly, state isolated or interval rest and construct every monotone leg with original endpoint positions and absolute length. Calculate net displacement, total distance and both averages independently. A correct final distance cannot replace missing or incorrect leg working. Original position piece 1: s=t^3-6*t on mathematical piece bounds -infinity to infinity. Physical trip [0,2] s; position unit m. Given inputs 0, 1, 2. Retain the exact irrational reversal time and original position. Partition every monotone leg before comparing net displacement, total distance, average velocity and average speed. Rewrite each original position piece while preserving its domain. Enter your term-by-term first and second derivative working, notation and units before a separate final answer. For every named input, record original-domain and physical-context decisions, position, signed velocity, speed, signed acceleration, direction and speed trend. Give the complete mathematical velocity-root account, filter candidates explicitly, state isolated or interval rest and construct every monotone leg with original endpoint positions and absolute length. Calculate net displacement, total distance and both averages independently. A correct final distance cannot replace missing or incorrect leg working. Original position piece 1: s=t^6-6*t on mathematical piece bounds -infinity to infinity. Physical trip [0,2] s; position unit m. Given inputs 0, 1, 2. Apply the integer-power rule beyond cubics. Account for every real root of the velocity equation, then use all admitted monotone legs for the physical trip. Rewrite each original position piece while preserving its domain. Enter your term-by-term first and second derivative working, notation and units before a separate final answer. For every named input, record original-domain and physical-context decisions, position, signed velocity, speed, signed acceleration, direction and speed trend. Give the complete mathematical velocity-root account, filter candidates explicitly, state isolated or interval rest and construct every monotone leg with original endpoint positions and absolute length. Calculate net displacement, total distance and both averages independently. A correct final distance cannot replace missing or incorrect leg working.
Give me a hint
Rewrite each original position piece while preserving its domain. Enter your term-by-term first and second derivative working, notation and units before a separate final answer. For every named input, record original-domain and physical-context decisions, position, signed velocity, speed, signed acceleration, direction and speed trend. Give the complete mathematical velocity-root account, filter candidates explicitly, state isolated or interval rest and construct every monotone leg with original endpoint positions and absolute length. Calculate net displacement, total distance and both averages independently. A correct final distance cannot replace missing or incorrect leg working.
Compare my reasoning
- Case 1: For s=6t-t^2, v=6-2t and a=-2. Its mathematical rest input is 3, the physical start of this trip [3,5].
- Case 1: At t=3,4,5, positions are 9,8,5 m; velocities are 0,-2,-4 m/s and speeds are 0,2,4 m/s.
- Case 1: After the start, negative acceleration and negative velocity have the same sign, so speed increases. The physical trip supplies no interval before its starting endpoint on which to claim a reversal.
- Case 1: Net displacement is -4 m, distance is 4 m, average velocity is -2 m/s and average speed is 2 m/s.
- Case 2: For s=t^3-6t, v=3t^2-6 and a=6t. The complete mathematical root list is -sqrt(2), sqrt(2); only the positive root belongs to [0,2].
- Case 2: At that reversal, s=t(t^2-6)=-4sqrt(2). The endpoint positions are s(0)=0 and s(2)=-4.
- Case 2: The first leg has length 4sqrt(2); the return leg has length 4sqrt(2)-4. Total distance is 8sqrt(2)-4 m and net displacement is -4 m.
- Case 2: Average speed is 4sqrt(2)-2 m/s and average velocity is -2 m/s. Keep the exact reversal and leg lengths before giving any decimal approximation.
- Case 3: For s=t^6-6t, the integer-power rule gives v=6t^5-6 and a=30t^4. Since t^5 is strictly increasing, the sole real velocity root is 1.
- Case 3: Position is 0 at 0, -5 at 1 and 52 at 2. Velocity is negative before 1 and positive afterwards.
- Case 3: Travelled distance is |(-5)-0|+|52-(-5)|=5+57=62 m. Displacement is 52 m over 2 s.
- Case 3: Average speed is 31 m/s and average velocity is 26 m/s. The acceleration is nonnegative throughout, yet speed decreases on the negative-velocity leg.
Case 1: For s=6t-t^2, v=6-2t and a=-2. Its mathematical rest input is 3, the physical start of this trip [3,5]. At t=3,4,5, positions are 9,8,5 m; velocities are 0,-2,-4 m/s and speeds are 0,2,4 m/s. After the start, negative acceleration and negative velocity have the same sign, so speed increases. The physical trip supplies no interval before its starting endpoint on which to claim a reversal. Net displacement is -4 m, distance is 4 m, average velocity is -2 m/s and average speed is 2 m/s. Negative velocity means motion in the negative coordinate direction. It does not mean a negative speed or necessarily slowing down. The mathematical speed extension has a cusp at 3. The physical right-hand speed gradient at its start is 2 m/s^2; this is a one-sided reading. Case 2: For s=t^3-6t, v=3t^2-6 and a=6t. The complete mathematical root list is -sqrt(2), sqrt(2); only the positive root belongs to [0,2]. At that reversal, s=t(t^2-6)=-4sqrt(2). The endpoint positions are s(0)=0 and s(2)=-4. The first leg has length 4sqrt(2); the return leg has length 4sqrt(2)-4. Total distance is 8sqrt(2)-4 m and net displacement is -4 m. Average speed is 4sqrt(2)-2 m/s and average velocity is -2 m/s. Keep the exact reversal and leg lengths before giving any decimal approximation. The final position being negative does not mean every part of the trip has negative velocity. The complete reversal partition controls the distance. A rounded reversal marker is suitable for drawing, but does not replace the exact value used in calculation. Case 3: For s=t^6-6t, the integer-power rule gives v=6t^5-6 and a=30t^4. Since t^5 is strictly increasing, the sole real velocity root is 1. Position is 0 at 0, -5 at 1 and 52 at 2. Velocity is negative before 1 and positive afterwards. Travelled distance is |(-5)-0|+|52-(-5)|=5+57=62 m. Displacement is 52 m over 2 s. Average speed is 31 m/s and average velocity is 26 m/s. The acceleration is nonnegative throughout, yet speed decreases on the negative-velocity leg. The same integer-power differentiation rule applies beyond cubics. Completeness of the velocity-root list still matters. This sixth-power teaching example is an invented model, not evidence of a realistic physical trajectory.
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/t on mathematical piece bounds -infinity to infinity. Physical trip [1,4] s; position unit m. Given inputs 0, 1, 2, 4. Keep the original zero exclusion and reject mathematical roots outside the physical interval. Equal final and initial positions do not imply zero travelled distance. Rewrite each original position piece while preserving its domain. Enter your term-by-term first and second derivative working, notation and units before a separate final answer. For every named input, record original-domain and physical-context decisions, position, signed velocity, speed, signed acceleration, direction and speed trend. Give the complete mathematical velocity-root account, filter candidates explicitly, state isolated or interval rest and construct every monotone leg with original endpoint positions and absolute length. Calculate net displacement, total distance and both averages independently. A correct final distance cannot replace missing or incorrect leg working. 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, 5/2, 3. Keep the complete mathematical velocity-root list. Distinguish a rest instant at the physical starting endpoint from a later reversal, and calculate the entire trip rather than only its net change. Rewrite each original position piece while preserving its domain. Enter your term-by-term first and second derivative working, notation and units before a separate final answer. For every named input, record original-domain and physical-context decisions, position, signed velocity, speed, signed acceleration, direction and speed trend. Give the complete mathematical velocity-root account, filter candidates explicitly, state isolated or interval rest and construct every monotone leg with original endpoint positions and absolute length. Calculate net displacement, total distance and both averages independently. A correct final distance cannot replace missing or incorrect leg working.
Give me a hint
Rewrite each original position piece while preserving its domain. Enter your term-by-term first and second derivative working, notation and units before a separate final answer. For every named input, record original-domain and physical-context decisions, position, signed velocity, speed, signed acceleration, direction and speed trend. Give the complete mathematical velocity-root account, filter candidates explicitly, state isolated or interval rest and construct every monotone leg with original endpoint positions and absolute length. Calculate net displacement, total distance and both averages independently. A correct final distance cannot replace missing or incorrect leg working.
Compare my reasoning
- Case 1: For s=t+4/t, the original domain excludes 0. On that domain v=1-4/t^2 and a=8/t^3.
- Case 1: Clearing t^2 in the velocity-zero equation is valid only for t!=0. The mathematical candidates are -2 and 2; only 2 is in [1,4].
- Case 1: Positions at 1,2,4 are 5,4,5 m. The two monotone leg lengths are 1 and 1 m, giving total distance 2 m and displacement 0 m.
- Case 1: Average speed is 2/3 m/s and average velocity is 0 m/s. At t=0 the original position and rates are undefined, not merely physically outside the trip.
- Case 2: For s=t^4-8t^2, v=4t(t^2-4) and a=12t^2-16. The complete mathematical velocity roots are -2,0,2.
- Case 2: The physical interval [0,3] admits 0 and 2. At the start 0 is endpoint rest; at 2 the velocity changes from negative to positive, giving an interior reversal.
- Case 2: Position is 0 at 0, -16 at 2 and 9 at 3. Total distance is 16+25=41 m, while net displacement is 9 m.
- Case 2: Average speed is 41/3 m/s and average velocity is 3 m/s. At t=5/2, position is -175/16 m, velocity 45/2 m/s and acceleration 59 m/s^2.
Case 1: For s=t+4/t, the original domain excludes 0. On that domain v=1-4/t^2 and a=8/t^3. Clearing t^2 in the velocity-zero equation is valid only for t!=0. The mathematical candidates are -2 and 2; only 2 is in [1,4]. Positions at 1,2,4 are 5,4,5 m. The two monotone leg lengths are 1 and 1 m, giving total distance 2 m and displacement 0 m. Average speed is 2/3 m/s and average velocity is 0 m/s. At t=0 the original position and rates are undefined, not merely physically outside the trip. Returning to the initial position gives zero displacement and can still require positive travelled distance. An algebraic rewrite cannot fill the reciprocal hole; the physical interval happens to avoid it. Case 2: For s=t^4-8t^2, v=4t(t^2-4) and a=12t^2-16. The complete mathematical velocity roots are -2,0,2. The physical interval [0,3] admits 0 and 2. At the start 0 is endpoint rest; at 2 the velocity changes from negative to positive, giving an interior reversal. Position is 0 at 0, -16 at 2 and 9 at 3. Total distance is 16+25=41 m, while net displacement is 9 m. Average speed is 41/3 m/s and average velocity is 3 m/s. At t=5/2, position is -175/16 m, velocity 45/2 m/s and acceleration 59 m/s^2. A physical endpoint and an interior rest instant need different descriptions. Include the entire final leg when calculating the trip. Negative-time mathematical roots stay in the complete root account but are rejected from this physical trip.
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?
Can positive acceleration accompany decreasing speed?
Compare with the explanation
Yes.
When velocity is negative and acceleration positive, speed decreases until a possible rest instant; inspect both signs.
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Make it stick
Can positive acceleration accompany decreasing speed? Yes. When velocity is negative and acceleration positive, speed decreases until a possible rest instant; inspect both signs. 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.