Reading Motion Information in a Table
A table of motion data can place position, velocity, and acceleration side by side. The values describe related features of motion, but they are not interchangeable: position tells where the particle is, velocity tells how position is changing, and acceleration tells how velocity is changing.
A table may also invite conclusions about an interval between listed times. Some interval quantities, such as displacement, can be calculated from endpoint positions. Other claims, such as whether a particle moves in one direction for the entire interval, generally need information about what happens between the entries.
What the Entries Tell You at a Listed Time
The sign and size of each entry answer different questions. If position is measured in meters, velocity in meters per second, and acceleration in meters per second squared, then the units help identify the role of each value.
- Position \(s(t)\): gives the particle’s coordinate at time \(t\). A negative position means it is on the negative side of the chosen origin; it does not mean the particle is moving left.
- Velocity \(v(t)\): gives the instantaneous rate of change of position. Positive velocity means motion in the positive direction, negative velocity means motion in the negative direction, and zero velocity means the particle is momentarily at rest.
- Acceleration \(a(t)\): gives the instantaneous rate of change of velocity. Its sign tells whether velocity is increasing or decreasing, not by itself whether speed is increasing or decreasing.
To decide whether speed is increasing or decreasing at a time when \(v(t)\ne 0\), compare the signs of velocity and acceleration. Speed is the nonnegative quantity \(|v(t)|\). If velocity and acceleration have the same sign, speed is increasing at that instant; if their signs are opposite, speed is decreasing.
Why the Speed Rule Works
The rule follows by differentiating speed locally. At any time when \(v(t)>0\), velocity remains positive for times sufficiently close to \(t\), so speed equals \(v\) there. Its rate of change is therefore \(a(t)\). When \(v(t)<0\), velocity remains negative nearby, so speed equals \(-v\), and its rate of change is \(-a(t)\).
Thus, if \(v(t)>0\), speed increases when \(a(t)>0\) and decreases when \(a(t)<0\). If \(v(t)<0\), speed increases when \(a(t)<0\) and decreases when \(a(t)>0\). In both cases the test is whether \(v(t)a(t)\) is positive or negative. At \(v(t)=0\), the absolute-value function has a corner as a function of velocity, so this derivative argument does not apply.
Worked Example: Describe Motion at Listed Times
Worked Example: Describe Motion at Listed Times
A particle moves on a line. The table gives its position, velocity, and acceleration, with position in meters and time in seconds. Describe its motion at \(t=0\), \(t=2\), and \(t=4\).
| \(t\) (s) | \(s(t)\) (m) | \(v(t)\) (m/s) | \(a(t)\) (m/s²) |
|---|---|---|---|
| 0 | 3 | 2 | -1 |
| 2 | 7 | 0 | -2 |
| 4 | 11 | -3 | 1 |
At \(t=0\), the position is \(3\) m, so the particle is at coordinate \(3\). Its velocity is \(2\) m/s, so it is moving in the positive direction. Since \(v(0)=2>0\) and \(a(0)=-1<0\), the signs are opposite: its speed is decreasing at that instant.
At \(t=2\), the position is \(7\) m and velocity is \(0\) m/s. Thus the particle is momentarily at rest. Its acceleration is \(-2\) m/s², meaning its velocity is decreasing at that instant. The speed test using \(v(t)a(t)\) cannot be used at \(t=2\), because \(v(2)=0\).
At \(t=4\), the position is \(11\) m and velocity is \(-3\) m/s, so the particle is moving in the negative direction. Here \(v(4)a(4)=(-3)(1)=-3<0\), so its speed is decreasing at that instant.
The entries also distinguish position from motion: the particle’s position increases from \(3\) m to \(11\) m across the listed times, even though at \(t=4\) its velocity is negative. That endpoint comparison does not establish its direction throughout the interval.
What Endpoint Values Can Tell You About an Interval
If positions are known at two times \(t_1\) and \(t_2\), displacement over the interval is the final position minus the initial position. Average velocity is displacement divided by elapsed time. Similarly, average acceleration is the change in velocity divided by elapsed time.
These are interval averages, not instantaneous values at either endpoint. For example, an average velocity of \(-2\) m/s does not say that the particle’s velocity was \(-2\) m/s at every moment in the interval.
Worked Example: Calculate Displacement and Average Rates
Worked Example: Calculate Displacement and Average Rates
At \(t=1\) s and \(t=3\) s, a particle has the following data. Find its displacement, average velocity, and average acceleration over \([1,3]\).
| \(t\) (s) | \(s(t)\) (m) | \(v(t)\) (m/s) | \(a(t)\) (m/s²) |
|---|---|---|---|
| 1 | 6 | 3 | -1 |
| 3 | 2 | -2 | -1 |
First, the elapsed time is \(3-1=2\) s. The displacement is final position minus initial position:
The negative displacement means the net change in position is in the negative direction. Average velocity is:
Average acceleration is the change in velocity divided by elapsed time:
Check the calculations: \((-2\text{ m/s})(2\text{ s})=-4\text{ m}\), which matches the displacement. Also, \((-2.5\text{ m/s}^2)(2\text{ s})=-5\text{ m/s}\), which matches the change from \(3\) m/s to \(-2\) m/s. The listed endpoint accelerations are both \(-1\) m/s², but that does not make the average acceleration \(-1\) m/s²; the average is determined by the endpoint velocities.
Because the endpoint velocities have opposite signs, if velocity is continuous between \(t=1\) and \(t=3\), it must equal zero at least once in between. The table alone does not identify the time of that zero or show how many times velocity is zero.
Sampled Values Are Not a Record of Every Instant
A table reports values at particular times. Unless it also states a condition holding throughout an interval, those entries do not give the signs of velocity or acceleration at every time between the listed rows. In particular, endpoint velocity signs can sometimes guarantee a zero in between, but they do not generally determine the exact path of the motion.
This distinction matters when describing direction or speed over an interval. A negative average velocity establishes a negative net displacement per unit time; it does not establish that the particle moved in the negative direction at every instant. Likewise, acceleration values at two endpoints do not tell you that acceleration kept the same sign between them.
Worked Example: Separate Endpoint Facts from an Interval Conclusion
Worked Example: Separate Endpoint Facts from an Interval Conclusion
A particle has the following velocity and acceleration values. Determine its direction and whether its speed is increasing or decreasing at the listed times. Then state what can be concluded about a possible rest time between \(t=0\) and \(t=2\).
| \(t\) (s) | \(v(t)\) (m/s) | \(a(t)\) (m/s²) |
|---|---|---|
| 0 | -4 | 3 |
| 1 | -1 | 3 |
| 2 | 2 | 3 |
At \(t=0\), \(v(0)=-4<0\), so the particle moves in the negative direction. The product is \(v(0)a(0)=(-4)(3)=-12<0\), so its speed is decreasing at that instant.
At \(t=1\), \(v(1)=-1<0\), so it still moves in the negative direction. The product \(v(1)a(1)=(-1)(3)=-3<0\), so its speed is also decreasing at that instant.
At \(t=2\), \(v(2)=2>0\), so the particle moves in the positive direction. The product \(v(2)a(2)=(2)(3)=6>0\), so its speed is increasing at that instant.
If velocity is continuous on \([0,2]\), then the change from \(v(0)=-4\) to \(v(2)=2\) guarantees at least one time between \(0\) and \(2\) when velocity is zero. Therefore the particle is momentarily at rest at least once in that interval. The table does not establish the exact time of rest. The endpoint acceleration entries do not, on their own, prove that acceleration equals \(3\) m/s² throughout the interval.
A Common Mistake: Confusing Acceleration with Speed
A positive acceleration does not always mean that a particle is speeding up. For example, at \(t=0\) in the last table, acceleration is \(3\) m/s², but velocity is \(-4\) m/s. The signs differ, so speed is decreasing, not increasing.
The mistaken calculation is: “\(a(0)=3>0\), therefore speed is increasing.” The correction is to check both signs:
A related error is to treat a listed acceleration as the average acceleration over an interval. Average acceleration comes from the endpoint velocity change divided by elapsed time. A listed acceleration is an instantaneous rate at its particular time.
Check Your Understanding
Use the distinctions between position, instantaneous rates, and interval averages to answer these questions.
- A table gives \(s(2)=5\) m and \(v(2)=-3\) m/s. What does each value describe, and in which direction is the particle moving at \(t=2\)?
- At a listed time, \(v=4\) m/s and \(a=-2\) m/s². Is speed increasing or decreasing at that instant? Show the sign test.
- A particle has positions \(s(1)=7\) m and \(s(5)=-1\) m. Find its displacement and average velocity over \([1,5]\).
- Endpoint velocities are \(-3\) m/s and \(5\) m/s. What can be concluded about a rest time between the endpoints if velocity is continuous? What cannot be determined from those entries alone?
- Why does a negative average velocity not prove that velocity was negative at every instant in the interval?
