One Position Record, Several Motion Questions
A position record can be given as a formula, a graph, or a table. Each form can answer some questions about a particle’s motion, but the amount of detail available differs. A formula may allow exact differentiation; a smooth graph can show slope and concavity; a table gives positions at selected times but does not reveal everything that happens between them.
For a particle moving along a line, let \(s(t)\) denote its position at time \(t\). Its velocity is \(v(t)=s'(t)\), and its acceleration is \(a(t)=v'(t)=s''(t)\) wherever the derivatives exist. Positive velocity means motion in the positive direction, negative velocity means motion in the negative direction, and zero velocity means the particle is instantaneously at rest.
A useful analysis separates these quantities. Displacement depends only on the initial and final positions. Total distance can depend on every reversal of direction along the way.
A Reliable Order of Analysis
When a position equation is provided, differentiate it to find velocity and acceleration. Use velocity’s sign to determine direction and solve \(v(t)=0\) to identify candidate rest times. Check the sign on either side of each zero: a zero is a direction change only if the sign changes. Then evaluate position at the interval endpoints and at any direction changes. Those positions give displacement and, when all relevant changes are accounted for, total distance.
Record the start and end times, including units. Restrict all sign conclusions to the stated interval.
For an equation, differentiate \(s(t)\). For a graph, use slope for velocity and concavity for acceleration.
Find where velocity is zero, then test its sign between those times.
Use endpoint positions for displacement; split the path at direction changes to calculate total distance.
A table of isolated positions usually does not determine instantaneous motion between its sample times.
Worked Example: Full Analysis from a Position Equation
Worked Example: Full Analysis from a Position Equation
A particle’s position, in metres, is \(s(t)=t^3-6t^2+9t\), where \(0\le t\le5\) seconds. Find its direction of motion, rest times, acceleration behavior, displacement, and total distance.
1. Differentiate to find velocity and acceleration.
The velocity is zero at \(t=1\) and \(t=3\). These are the rest times to check. The acceleration is zero at \(t=2\).
2. Determine direction from the velocity sign. The factors \(t-1\) and \(t-3\) give the signs on the three intervals:
The particle moves in the positive direction on \((0,1)\), in the negative direction on \((1,3)\), and in the positive direction on \((3,5)\). Since velocity changes sign at both zeros, the particle changes direction at \(t=1\) and \(t=3\).
3. Describe acceleration and speed behavior. Since \(a(t)=6(t-2)\), acceleration is negative for \(t<2\), zero at \(t=2\), and positive for \(t>2\). To decide whether speed increases or decreases, compare the signs of \(v\) and \(a\): matching signs mean speeding up, and opposite signs mean slowing down. On \((0,1)\), \(v>0\) and \(a<0\), so the particle slows down. On \((1,2)\), both are negative, so it speeds up. On \((2,3)\), \(v<0\) and \(a>0\), so it slows down. On \((3,5)\), both are positive, so it speeds up.
4. Evaluate positions and calculate displacement and distance.
The displacement over the full interval is final position minus initial position. The particle’s path changes direction at \(t=1\) and \(t=3\), so total distance is the sum of the magnitudes of the three position changes:
Check: The particle ends \(20\) m in the positive direction from where it started, while it travels \(4+4+20=28\) m along its path. Distance exceeds the magnitude of displacement because the particle reverses direction twice.
Why Position Changes Give Distance
The calculation above works because, between consecutive direction changes, velocity keeps one sign. Position is therefore monotonic on each such interval: it moves steadily in one direction, even if its speed varies. On a monotonic interval, the distance traveled is the magnitude of the endpoint position change.
In calculus terms, the Fundamental Theorem of Calculus gives displacement as the integral of velocity. Total distance instead accumulates the magnitude of velocity. Splitting at velocity zeros allows the sign of velocity to remain constant on each piece, so the magnitude of the position change on that piece gives its distance.
Thus, if \(t_0,t_1,\ldots,t_n\) are endpoints and direction-change times in order, then the distance is the sum of the absolute position changes across those intervals. This method requires that the selected times capture every direction change; missing one can make the distance calculation too small.
Worked Example: Reading a Smooth Position Graph
Worked Example: Reading a Smooth Position Graph
Suppose a smooth position graph on \(0\le t\le6\) seconds passes through \((0,2)\), \((1,6)\), \((4,-2)\), and \((6,3)\), where the second coordinate is position in metres. The graph rises to a local maximum at \(t=1\), falls until a local minimum at \(t=4\), and rises after that. It is concave down on \((0,2)\) and concave up on \((2,6)\). Analyze its direction, rest times, acceleration behavior, displacement, and total distance.
The slope of a position graph represents velocity. It is positive while the graph rises and negative while it falls. At a smooth local maximum or minimum, the tangent is horizontal, so velocity is zero. Therefore, the particle moves in the positive direction on \((0,1)\), the negative direction on \((1,4)\), and the positive direction on \((4,6)\). It is at rest at \(t=1\) and \(t=4\), and changes direction at both times.
Concavity gives the sign of acceleration: concave down means \(a<0\), and concave up means \(a>0\). Thus \(a<0\) on \((0,2)\), \(a>0\) on \((2,6)\), and the graph changes concavity at \(t=2\). Combining these signs with velocity, the particle slows down on \((0,1)\), speeds up on \((1,2)\), slows down on \((2,4)\), and speeds up on \((4,6)\).
The initial and final positions are \(s(0)=2\) m and \(s(6)=3\) m, so displacement is \(3-2=1\) m. The graph’s direction changes at the maximum and minimum, whose positions are \(6\) m and \(-2\) m. Total distance is
Check: The particle finishes \(1\) m above its starting position, and the three path segments have lengths \(4\) m, \(8\) m, and \(5\) m, totaling \(17\) m. The graph supplies signs and positions, but unless its scale or an equation provides more information, it does not give exact numerical values for velocity or acceleration.
Worked Example: What a Position Table Can Establish
Worked Example: What a Position Table Can Establish
A table reports the following positions, in metres, at times measured in seconds:
| \(t\) (s) | 0 | 1 | 2 |
|---|---|---|---|
| \(s(t)\) (m) | 0 | 1 | 0 |
The displacement from \(t=0\) to \(t=2\) is exactly \(s(2)-s(0)=0-0=0\) m. The position changes recorded from one sample to the next have magnitudes \(|1-0|=1\) m and \(|0-1|=1\) m. Their sum, \(2\) m, is the distance indicated by the sampled positions if the motion is monotonic between consecutive samples. Without that assumption, it is only a lower bound on the actual distance: unobserved movement between samples could add more travel.
The average velocities over the two sample intervals can be calculated exactly:
These are interval averages, not instantaneous velocities at every time in those intervals. In particular, the table alone does not tell us the exact rest times or acceleration behavior.
To see why, two different smooth position functions can fit all three entries. One is \(s_A(t)=2t-t^2\). Another is \(s_B(t)=2t-t^2+\frac{\sin(2\pi t)}{2\pi}\). The sine term is zero at \(t=0,1,2\), so both functions give the listed positions. Yet their velocities at \(t=\frac12\) differ:
Check: Substitution into either position formula gives positions \(0,1,0\) at the three listed times, but at \(t=\frac12\) one particle is moving and the other is instantaneously at rest. The table does not determine the in-between motion.
A Common Mistake: Treating Displacement as Distance
A common error is to calculate total distance by taking the absolute value of the net displacement. That only measures how far apart the starting and ending positions are; it ignores any backtracking. In the first example, the incorrect calculation would be
This is not the total distance. The particle first travels from \(0\) m to \(4\) m, then back to \(0\) m, and finally to \(20\) m. The correct distance is \(|4-0|+|0-4|+|20-0|=28\) m. To avoid this error, find all direction changes, evaluate position at each one, and add the magnitudes of the successive position changes.
What to Report—and What Not to Claim
A complete motion analysis should connect each conclusion to the information that supports it:
- Use the sign of velocity for direction; a zero of velocity identifies a rest time, while a sign change confirms a direction change.
- Use the sign of acceleration for acceleration behavior; combine it with velocity’s sign to decide whether speed increases or decreases.
- Use final minus initial position for displacement, with its sign and units.
- For total distance, add absolute position changes across all direction changes. If the data do not establish all such changes, state that exact distance is not determined.
- For tables, distinguish average velocity across a sample interval from instantaneous velocity, and do not infer unobserved behavior without an explicit model or assumption.
Check Your Understanding
For each question, identify what the given position information supports before calculating.
- If a particle’s position graph rises, what is the sign of its velocity? What does a horizontal tangent imply?
- A particle moves from position \(2\) m to \(7\) m and then to \(1\) m. Find its displacement and the distance traveled along those two segments.
- What additional information is needed to use positions at selected times to determine exact total distance?
- Why can a table of position values fail to determine the instantaneous velocity between sample times?
- If velocity is negative and acceleration is positive throughout an interval, is the particle speeding up or slowing down? Explain using the signs.
