Reading a Motion Table as Evidence
A motion table gives values at particular times. Its entries can support clear statements about a particle’s position, direction of motion, and changes in speed. The key is to match each claim to the quantity that supports it: position describes location, velocity describes direction and instantaneous rate of position change, and acceleration describes instantaneous rate of velocity change.
A table does not automatically describe every moment between its listed times. A conclusion about a specified time can use the entry at that time. A conclusion about an entire interval needs enough information to justify what happens throughout that interval. This distinction helps make a response both accurate and appropriately limited.
What Each Column Can Tell You
Position identifies where the particle is relative to the chosen origin. Comparing two position entries gives displacement: final position minus initial position. Displacement is signed and is not necessarily the distance traveled.
Velocity gives direction at the listed time through its sign. Positive velocity means motion in the positive direction; negative velocity means motion in the negative direction. A zero velocity means the particle is instantaneously at rest. The sign does not tell you where the particle is; use position for that.
Acceleration describes how velocity is changing at the listed time. Positive acceleration means velocity is increasing; negative acceleration means velocity is decreasing. Acceleration’s sign alone does not identify direction of motion or whether speed is increasing. For those claims, compare velocity and acceleration together.
| Table entry | Directly supports | Does not establish by itself |
|---|---|---|
| Position \(x(t)\) | Location at the listed time | Direction at that time or the path between entries |
| Velocity \(v(t)\) | Direction and instantaneous rest at the listed time | Position or motion direction throughout an interval |
| Acceleration \(a(t)\) | Whether velocity is increasing or decreasing at the listed time | Direction or whether speed is increasing without velocity data |
| Two position values | Displacement and average velocity between those times | Instantaneous velocity or total distance traveled |
Speeding Up and Slowing Down
Speed is the magnitude of velocity, \(|v(t)|\). When \(v\) is positive, increasing velocity increases speed; when \(v\) is negative, decreasing velocity makes its magnitude larger. Thus the signs of velocity and acceleration must be considered together. This rule applies at a time when \(v(t)\ne 0\); if velocity is zero, the rule alone does not classify the particle as speeding up or slowing down at that instant.
Here is why the sign test works. If velocity is positive near the time in question, speed equals velocity, so its rate of change is \(a\). If velocity is negative, speed equals \(-v\), so its rate of change is \(-a\). In either case, the rate of change of speed has the sign of \(va\): positive means speed is increasing and negative means it is decreasing.
Worked Example: Interpreting Velocity and Acceleration at Listed Times
Worked Example: Interpreting Velocity and Acceleration at Listed Times
A particle moves along a line. A table gives its velocity and acceleration at two times:
| Time \(t\) (s) | Velocity \(v(t)\) (m/s) | Acceleration \(a(t)\) (m/s²) |
|---|---|---|
| 5 | 2.5 | -0.8 |
| 7 | -1.2 | -0.4 |
At \(t=5\), \(v(5)=2.5>0\), so the particle moves in the positive direction. Also, \(a(5)=-0.8\) m/s², so velocity is decreasing at that time. To classify the speed, compute the product of the signs: \(v(5)a(5)=(2.5)(-0.8)=-2.0\) m²/s³, which is negative. Therefore, the particle is slowing down at \(t=5\).
At \(t=7\), \(v(7)=-1.2<0\), so the particle moves in the negative direction. The acceleration is also negative, \(a(7)=-0.4\) m/s². Their product is \(v(7)a(7)=(-1.2)(-0.4)=0.48\) m²/s³, which is positive. Therefore, the particle is speeding up at \(t=7\).
Check: At \(t=5\), negative acceleration opposes positive velocity, matching a decrease in speed. At \(t=7\), negative acceleration has the same direction as negative velocity, matching an increase in speed. These conclusions describe the listed instants; the two rows alone do not determine the motion at every time between 5 and 7 seconds.
Using Position Entries to Find Average Velocity
If a table gives position at two times \(t_1\) and \(t_2\), the displacement over the interval is \(x(t_2)-x(t_1)\). The average velocity is displacement divided by elapsed time. It describes the net change in position per unit time over that interval; it is not necessarily the particle’s velocity at either endpoint or at any particular instant inside the interval.
Worked Example: Finding Average Velocity from Position Data
A particle’s position is \(3\) m at \(t=1\) s and \(-6\) m at \(t=4\) s. Find its displacement and average velocity from \(t=1\) to \(t=4\).
The displacement is final position minus initial position:
The elapsed time is \(4\text{ s}-1\text{ s}=3\text{ s}\). Therefore, the average velocity is
The negative average velocity means that the net displacement is in the negative direction. It does not establish that the particle moved in the negative direction at every moment during the interval; the particle could have changed direction and still ended at \(-6\) m.
Check: Multiplying average velocity by elapsed time gives \((-3\text{ m/s})(3\text{ s})=-9\text{ m}\), the displacement calculated from the position entries.
What Can Be Claimed About an Interval?
A table may show values at selected times, while a question asks about motion over a time interval. Read the wording carefully. A claim “at \(t=3\)” concerns an instantaneous value; a claim “from \(t=1\) to \(t=4\)” concerns the behavior across an interval. A few sampled entries do not generally prove that a sign or behavior persists between samples.
For example, positive velocity at both endpoints does not, from those endpoint values alone, establish positive velocity throughout the interval. Likewise, a negative acceleration entry at one listed time does not show that acceleration is negative throughout a longer interval. If a problem explicitly states that a value or sign holds on an interval, use that information. Otherwise, state only what the table supports.
Determine whether the question asks about one listed instant, an interval average, or behavior throughout an interval.
Use position for location and displacement, velocity for direction, and velocity together with acceleration for speeding up or slowing down.
Quote the relevant table values, including their signs and units, and show any calculation used.
Do not extend a conclusion at one time to an entire interval unless the given information justifies it.
Worked Example: Combining Position, Velocity, and Acceleration Data
Worked Example: Combining Position, Velocity, and Acceleration Data
The table gives a particle’s position, velocity, and acceleration at three times. Position is measured in meters, velocity in meters per second, and acceleration in meters per second squared.
| Time \(t\) (s) | Position \(x(t)\) (m) | Velocity \(v(t)\) (m/s) | Acceleration \(a(t)\) (m/s²) |
|---|---|---|---|
| 1 | 2 | 4 | -2 |
| 3 | 8 | 0 | -1 |
| 5 | 5 | -3 | 1 |
At \(t=1\), the particle is at position \(2\) m. Since \(v(1)=4\) m/s is positive, it moves in the positive direction. Since \(a(1)=-2\) m/s² is negative, velocity and acceleration have opposite signs, so the particle is slowing down at this instant.
At \(t=3\), the position is \(8\) m and the velocity is \(0\) m/s. Thus the particle is instantaneously at rest at that time. The acceleration is \(-1\) m/s², so its velocity is decreasing at that instant. The table does not, by itself, tell us whether the particle is moving in the positive or negative direction at every nearby time.
At \(t=5\), the particle is at \(5\) m and \(v(5)=-3\) m/s, so it moves in the negative direction. Because \(a(5)=1\) m/s² is positive, velocity and acceleration have opposite signs. The particle is therefore slowing down at \(t=5\).
We can also calculate average velocity from \(t=1\) to \(t=5\). The displacement is \(5\text{ m}-2\text{ m}=3\text{ m}\), and the elapsed time is \(5\text{ s}-1\text{ s}=4\text{ s}\). Hence,
Check: The average velocity is positive because the final position, \(5\) m, is \(3\) m greater than the initial position, \(2\) m. The table also lists negative velocity at \(t=5\); this does not contradict the positive average velocity, which summarizes net displacement over the entire four-second interval rather than direction at the endpoint.
A Common Mistake: Treating a Sampled Value as an Interval Claim
Suppose a table lists \(v(2)=3\) m/s and \(v(6)=3\) m/s. A mistaken conclusion would be: “The particle moves in the positive direction throughout \(2\le t\le6\), because velocity is positive at both endpoints.” The calculation behind that claim uses only \(v(2)>0\) and \(v(6)>0\), which say nothing by themselves about the values between those times. Velocity could be negative at some intermediate time and return to a positive value at \(t=6\).
The corrected statement is: “The particle moves in the positive direction at \(t=2\) and \(t=6\), since velocity is \(3\) m/s at both listed times. These entries alone do not establish its direction throughout the interval.” If the problem separately states that velocity remains positive on the interval, then the stronger interval conclusion is justified by that additional information.
Another frequent error is to use the sign of acceleration alone to classify speed. For instance, negative acceleration does not always mean slowing down: when velocity is negative too, the particle is speeding up. Check both signs, and include the time at which the claim applies.
Writing a Supported Motion Conclusion
A strong response makes the evidence and conclusion explicit. For a statement about direction or speed change, name the time and cite velocity and acceleration with their units. For a statement about average velocity, show the position difference and elapsed time. For an interval claim, make clear whether the table gives endpoint values, interval-wide information, or only an average.
Check Your Understanding
Use the definitions and evidence limits in this tutorial to answer each question.
- A table gives \(v(4)=-2\) m/s and \(a(4)=-3\) m/s². What is the particle’s direction at \(t=4\), and is it speeding up or slowing down then?
- A particle’s position changes from \(7\) m at \(t=2\) s to \(1\) m at \(t=5\) s. Find its displacement and average velocity over that interval.
- At a listed time, \(v(t)=0\) and \(a(t)>0\). What can you conclude at that instant, and what does the speed-up sign test not determine there?
- A table gives positive velocity at the first and last times of an interval. Does that alone prove positive direction throughout the interval? Explain.
- Why is acceleration’s sign alone insufficient to decide whether a particle is speeding up?
