AP Calculus AB/BC · Unit 4 · Topic 4.2
Straight-Line Motion: Connecting Position, Velocity, and Acceleration
Relate position, velocity, speed, and acceleration along a line.
1. Topic Focus
Interpret derivatives as rates in context, connect motion quantities, solve related-rate models, linearize, and evaluate indeterminate limits.
This topic: Relate position, velocity, speed, and acceleration along a line.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
A particle changes direction where velocity changes sign, not merely where acceleration is zero.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
1. Position Locates the Particle on an Oriented Line
The position function \(s(t)\) gives a signed coordinate at time \(t\). Before interpreting motion, identify which direction is positive. A negative position means the particle lies on the negative side of the origin; it does not mean the particle is moving in the negative direction.
Position, direction of motion, and distance from the origin are different ideas:
2. Velocity Is the Rate of Change of Position
Velocity is the slope of the position graph. Its sign determines direction, not location:
- \(v(t)>0\): motion in the positive direction, such as right or up.
- \(v(t)<0\): motion in the negative direction, such as left or down.
- \(v(t)=0\): the particle is instantaneously at rest.
3. Speed Is the Magnitude of Velocity
Velocity includes direction and may be negative; speed is nonnegative. If \(v(4)=-7\) meters per second, the particle moves in the negative direction with speed 7 meters per second.
4. Acceleration Is the Rate of Change of Velocity
Acceleration is the slope of the velocity graph and records how velocity changes. Positive acceleration means velocity is increasing numerically; negative acceleration means velocity is decreasing numerically. Acceleration does not directly state the direction of motion.
5. Keep the Quantities and Units Distinct
| Quantity | Relationship | If position is meters and time is seconds |
|---|---|---|
| Position | \(s(t)\) | meters |
| Velocity | \(v(t)=s'(t)\) | meters per second |
| Speed | \(|v(t)|\) | meters per second |
| Acceleration | \(a(t)=v'(t)=s''(t)\) | meters per second squared |
6. Rest Is Not Automatically a Direction Change
A particle is at rest wherever \(v(t)=0\). It changes direction at a rest time only if velocity changes sign across that time. A velocity graph may touch the time axis and remain on the same side, producing a momentary stop without a reversal.
The reverse implication is false: \(v(c)=0\) alone is insufficient.
7. Speeding Up and Slowing Down
Speed increases when velocity and acceleration have the same sign because acceleration pushes velocity farther from zero. Speed decreases when their signs differ because acceleration pushes velocity toward zero.
| Velocity | Acceleration | Motion direction | Speed behavior |
|---|---|---|---|
| positive | positive | positive direction | speeding up |
| positive | negative | positive direction | slowing down |
| negative | negative | negative direction | speeding up |
| negative | positive | negative direction | slowing down |
8. Build a Motion Sign Chart
- Find \(v=s'\) and \(a=v'\).
- Solve \(v(t)=0\) for rest-time candidates.
- Solve \(a(t)=0\) for possible changes in speed behavior.
- Place all relevant times on one timeline within the physical domain.
- Test the signs of \(v\) and \(a\) on every interval.
- Translate each sign pair into direction and speed behavior.
Zeros of velocity divide direction intervals; zeros of velocity or acceleration can divide speeding-up and slowing-down intervals.
9. Connect the Three Graphs
- The slope of the position graph is the value of velocity.
- A local maximum or minimum of position can occur where velocity changes sign through zero.
- The slope of the velocity graph is the value of acceleration.
- A local maximum or minimum of velocity can occur where acceleration changes sign through zero.
- Concavity of the position graph has the same sign as acceleration.
A horizontal position graph means the particle remains at one position over an interval. A single horizontal tangent means only instantaneous rest.
10. Average and Instantaneous Velocity
Average velocity over \([a,b]\) is displacement divided by elapsed time, while instantaneous velocity at \(a\) is the derivative:
A particle can return to its starting position, giving zero displacement and zero average velocity, even though it moved and had nonzero speed during the interval.
11. Interpret a Given Velocity or Acceleration Model
If velocity is given directly, it determines direction, rest times, and acceleration after differentiation, but it does not determine absolute position without one position value. Likewise, knowing acceleration alone does not reveal velocity or direction without additional velocity information.
12. Common AP Motion Errors
- Using the sign of position to determine direction.
- Reporting negative speed instead of taking \(|v|\).
- Calling every zero of velocity a direction change without a sign test.
- Using \(a>0\) to conclude that the particle moves right.
- Claiming that \(a=0\) means the particle is at rest.
- Testing only velocity when asked whether speed increases.
- Ignoring the stated time domain, especially \(t\ge0\).
6. Detailed Worked Example and Error Check
Example 1: Complete motion analysis from position. Let
Differentiate and factor:
The critical times are 1 and 3 from velocity and 2 from acceleration.
| Interval | Sign of \(v\) | Sign of \(a\) | Conclusion |
|---|---|---|---|
| \((0,1)\) | positive | negative | moves right and slows down |
| \((1,2)\) | negative | negative | moves left and speeds up |
| \((2,3)\) | negative | positive | moves left and slows down |
| \((3,\infty)\) | positive | positive | moves right and speeds up |
Velocity changes sign at both \(t=1\) and \(t=3\), so the particle changes direction at both times. It moves from \(s(0)=0\) to \(s(1)=4\), reverses and returns to \(s(3)=0\), then reverses again.
Example 2: Rest without reversal. Suppose \(s(t)=(t-2)^3\). Then
The particle is at rest at \(t=2\), but \(v(t)>0\) on both sides. It pauses instantaneously and continues in the positive direction; it does not change direction.
Example 3: Start from velocity. Let \(v(t)=t^2-4t+3=(t-1)(t-3)\), \(t\ge0\). Then \(a(t)=2t-4\). The particle moves right on \([0,1)\) and \((3,\infty)\), moves left on \((1,3)\), and is at rest at 1 and 3. It speeds up on \((1,2)\) and \((3,\infty)\), where \(v\) and \(a\) have the same sign.
Without a value such as \(s(0)\), this velocity model cannot identify the particle's coordinate.
Example 4: Compare average and instantaneous velocity. A ball's height is \(s(t)=64-16t^2\) feet until it reaches the ground. Solving \(s(t)=0\) gives impact at \(t=2\). Since
its impact velocity is \(-64\) feet per second and impact speed is 64 feet per second. Its average velocity over the full fall is
The average velocity and final instantaneous velocity differ because the velocity changes throughout the fall.
Example 5: Read motion from a position graph. Suppose a differentiable position graph rises on \((0,2)\), has a horizontal tangent at \(t=2\), falls on \((2,5)\), and is constant on \([5,7]\). Then velocity is positive on \((0,2)\), zero at 2, negative on \((2,5)\), and zero throughout \([5,7]\). The particle reverses at \(t=2\) and remains at one fixed position from time 5 through time 7.
7. AP Reasoning Routine
Name variables and units, write the relationship before differentiating, substitute values at the correct time, and interpret the sign in context.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Let \(s(t)=t^3-3t^2-9t\), for \(t\ge0\), unless a part states otherwise.
(a) Find velocity and acceleration, all rest times, and every interval of positive- and negative-direction motion.
(b) Find the particle's position, velocity, speed, and acceleration at \(t=2\), with appropriate distinctions among them.
(c) Determine every interval on which the particle is speeding up or slowing down.
(d) Does the particle change direction at each rest time? Justify using velocity signs.
(e) For a different particle with \(v(t)=(t-2)^2\), explain what happens at \(t=2\).
(f) A position graph is decreasing and concave up at \(t=5\). State the signs of velocity and acceleration and decide whether speed is increasing or decreasing.
(g) If position is measured in kilometers and time in hours, state the units of \(s\), \(v\), \(|v|\), and \(a\).
(h) Explain why \(a(4)=0\) does not establish that a particle is at rest or changes direction at \(t=4\).
Check the solution
In part (a), \(v(t)=3t^2-6t-9=3(t-3)(t+1)\) and \(a(t)=6t-6\). In the domain \(t\ge0\), the only rest time is \(t=3\). Velocity is negative on \([0,3)\) and positive on \((3,\infty)\), so the particle moves in the negative direction before 3 and the positive direction after 3. In part (b), \(s(2)=-22\) is the position, \(v(2)=-9\) is the velocity, \(|v(2)|=9\) is the speed, and \(a(2)=6\) is the acceleration. In part (c), acceleration is negative on \((0,1)\) and positive on \((1,\infty)\). Thus \(v\) and \(a\) have the same sign on \((0,1)\) and \((3,\infty)\), where the particle speeds up; their signs differ on \((1,3)\), where it slows down. In part (d), velocity changes from negative to positive at \(t=3\), so the particle does change direction there. In part (e), velocity is zero at 2 but positive on both sides, so the particle rests instantaneously without reversing direction. In part (f), decreasing position gives \(v(5)<0\), and concave up gives \(a(5)>0\). Opposite signs mean the particle is slowing down. In part (g), position is kilometers, velocity and speed are kilometers per hour, and acceleration is kilometers per hour squared. In part (h), \(a(4)=0\) says only that velocity has zero instantaneous rate of change at that time. Rest and direction change depend on the value and sign behavior of velocity.