Reading Motion as a Connected Story
A motion problem may provide a position graph, a velocity graph, an acceleration graph, a table of measurements, or a combination of these. The goal is not simply to read off numbers. It is to connect the information carefully: position describes location, velocity describes the rate and direction of position change, and acceleration describes the rate of velocity change.
These quantities answer different questions. A positive position does not necessarily mean an object is moving in the positive direction. Positive acceleration does not necessarily mean the object is speeding up. And a table of values at a few times does not, by itself, reveal exactly what happens between those times.
In the examples below, position is measured in meters and time in seconds. Thus velocity is measured in meters per second (\(\text{m/s}\)), and acceleration in meters per second squared (\(\text{m/s}^2\)). The same reasoning applies with other consistent units.
What Each Kind of Data Can Tell You
| Information provided | Direct interpretation | Important limitation |
|---|---|---|
| Position \(x(t)\) | Where the object is at time \(t\) | Position values alone do not specify motion between measured times. |
| Velocity \(v(t)\) | Direction and instantaneous rate of position change | A velocity value is not a position value. |
| Acceleration \(a(t)\) | Rate of velocity change | Acceleration alone does not specify velocity without an initial velocity. |
| A finite table | Values at listed times; differences can give average rates | It generally does not determine the exact graph or instantaneous rates between entries. |
On a position-time graph, the slope gives velocity: an upward-sloping segment indicates positive velocity, a downward-sloping segment indicates negative velocity, and a horizontal segment indicates zero velocity. On a velocity-time graph, the slope gives acceleration. The sign of the graph’s height and the sign of its slope answer different questions.
Worked Example: Position Measurements and Average Velocity
Worked Example: Position Measurements and Average Velocity
An object’s recorded positions are shown below. What can be concluded about its average velocity on each time interval? Can the table establish the object’s exact direction at every instant?
| Time \(t\) (s) | Position \(x(t)\) (m) |
|---|---|
| 0 | 0 |
| 2 | 12 |
| 4 | 20 |
| 6 | 20 |
For a time interval from \(t_1\) to \(t_2\), average velocity is the change in position divided by the change in time:
From \(0\) to \(2\) seconds, position changes by \(12-0=12\) meters and time changes by \(2-0=2\) seconds. The average velocity is \(12/2=6\ \text{m/s}\).
From \(2\) to \(4\) seconds, position changes by \(20-12=8\) meters and time changes by \(4-2=2\) seconds. The average velocity is \(8/2=4\ \text{m/s}\).
From \(4\) to \(6\) seconds, position changes by \(20-20=0\) meters and time changes by \(6-4=2\) seconds. The average velocity is \(0/2=0\ \text{m/s}\).
Conclusion: The average velocities on the three intervals are \(6\ \text{m/s}\), \(4\ \text{m/s}\), and \(0\ \text{m/s}\), respectively. These are interval averages, not necessarily instantaneous velocities. The table does not establish whether the object moved forward and then backward between recorded times, or remained still throughout the final interval. Many different motions can pass through the same listed positions.
Check: For each interval, multiplying the calculated average velocity by the interval duration gives the position change: \(6(2)=12\) m, \(4(2)=8\) m, and \(0(2)=0\) m. Each matches the table.
Direction and Speed Are Different Questions
Velocity’s sign gives direction along the chosen coordinate axis: \(v(t)>0\) means motion in the positive direction, and \(v(t)<0\) means motion in the negative direction. Speed is the magnitude of velocity, \(|v(t)|\), and is never negative. An object can have negative velocity and positive speed.
Acceleration describes how velocity changes, not directly how fast the object moves. To determine whether speed is increasing or decreasing, consider both the signs of velocity and acceleration.
Why the Sign Rule Works
Speed is \(|v(t)|\). Whenever \(v(t)\ne 0\), its sign remains fixed in a sufficiently small neighborhood of that time. If \(v(t)>0\), then \(|v(t)|=v(t)\), so the rate of change of speed is \(a(t)\). If \(v(t)<0\), then \(|v(t)|=-v(t)\), so the rate of change of speed is \(-a(t)\).
Consequently, when \(v\) and \(a\) have the same sign, the applicable expression is positive and speed is increasing. When their signs differ, it is negative and speed is decreasing. At \(v=0\), \(|v|\) may not be differentiable, so the behavior must be determined from nearby values or from the motion description.
Worked Example: Reading Velocity and Acceleration Together
Worked Example: Reading Velocity and Acceleration Together
Suppose an object’s velocity, in \(\text{m/s}\), is \(v(t)=t^2-4t+3\) for \(0\le t\le4\). Determine its direction and whether its speed is increasing or decreasing across the interval.
First find acceleration by differentiating velocity: \(a(t)=v'(t)=2t-4\ \text{m/s}^2\). The velocity factors as \(v(t)=(t-1)(t-3)\), so it is zero at \(t=1\) and \(t=3\). Acceleration is zero at \(t=2\). These times divide the interval into regions on which the relevant signs are constant.
For \(0\le t<1\), velocity is positive: for instance, \(v(0)=3\ \text{m/s}\). Acceleration is negative because \(2t-4<0\). The object moves in the positive direction and slows down.
For \(1<t<2\), velocity is negative: \(v(1.5)=2.25-6+3=-0.75\ \text{m/s}\). Acceleration is also negative, since \(a(1.5)=3-4=-1\ \text{m/s}^2\). The object moves in the negative direction and speeds up.
For \(2<t<3\), velocity remains negative: \(v(2.5)=6.25-10+3=-0.75\ \text{m/s}\). Acceleration is positive: \(a(2.5)=5-4=1\ \text{m/s}^2\). The object moves in the negative direction and slows down.
For \(3<t\le4\), velocity is positive: \(v(4)=16-16+3=3\ \text{m/s}\). Acceleration is positive because \(2t-4>0\). The object moves in the positive direction and speeds up.
At \(t=1\) and \(t=3\), velocity is zero and changes sign, so the object reverses direction at each time. At \(t=2\), acceleration is zero, but velocity is \(v(2)=4-8+3=-1\ \text{m/s}\); the object is still moving in the negative direction at that instant.
Check: Differentiating \(t^2-4t+3\) gives \(2t-4\), the acceleration used above. Also, the velocity signs on either side of \(t=1\) are positive then negative, while those on either side of \(t=3\) are negative then positive, confirming both reversals.
What Acceleration Alone Does Not Determine
Acceleration tells how velocity changes, but an acceleration graph without an initial velocity does not tell which velocity values the object has. If a constant acceleration \(a_0\) is known throughout an interval and the initial velocity is known, then velocity is determined there: \(v(t)=v(t_0)+a_0(t-t_0)\). Without \(v(t_0)\), the velocity could differ by an arbitrary constant. Position also requires an initial position.
Care is also needed with sampled acceleration. Values such as \(a(0)\), \(a(1)\), and \(a(2)\) specify acceleration at those times, not necessarily throughout the intervals between them. Unless the data say more, they do not establish the exact velocity change over an interval.
Worked Example: Constant Acceleration and a Known Initial Velocity
Worked Example: Constant Acceleration and a Known Initial Velocity
An object has constant acceleration \(a(t)=2\ \text{m/s}^2\) for \(0\le t\le3\), and its initial velocity is \(v(0)=-3\ \text{m/s}\). Find its velocity and interpret its direction and speed changes. Is its position determined?
Because \(v'(t)=2\), velocity has the form \(v(t)=2t+C\): differentiating this expression gives \(2\). Apply the initial condition: \(v(0)=2(0)+C=-3\), so \(C=-3\). Therefore \(v(t)=2t-3\ \text{m/s}\).
Velocity is zero when \(2t-3=0\), which gives \(t=1.5\) seconds. For \(0\le t<1.5\), velocity is negative, so the object moves in the negative direction. Since acceleration is positive while velocity is negative, speed decreases. For \(1.5<t\le3\), velocity is positive and acceleration is positive, so the object moves in the positive direction and speeds up. At the endpoint, \(v(3)=6-3=3\ \text{m/s}\).
The velocity changes from \(-3\ \text{m/s}\) to \(3\ \text{m/s}\) over three seconds. The change is \(3-(-3)=6\ \text{m/s}\), consistent with constant acceleration \(2\ \text{m/s}^2\) over \(3\) seconds.
Conclusion and check: Differentiating \(v(t)=2t-3\) gives \(a(t)=2\ \text{m/s}^2\), and substituting \(t=0\) gives the stated initial velocity. The object reverses direction at \(1.5\) seconds. Its position is not determined by the given data: an initial position is also needed to specify where it is.
A Common Mistake: Positive Acceleration Means Speeding Up
Consider the concrete case \(v=-3\ \text{m/s}\) and \(a=2\ \text{m/s}^2\). A mistaken interpretation says, “Acceleration is positive, so the object is speeding up.” But velocity is negative and acceleration is positive, so their signs differ. The object is moving in the negative direction while its velocity is becoming less negative; its speed is decreasing.
For example, with constant acceleration, one second later the velocity is \(-3+2(1)=-1\ \text{m/s}\). Speed has changed from \(3\ \text{m/s}\) to \(1\ \text{m/s}\), so the object has slowed down. The corrected rule is to compare the signs of velocity and acceleration, not to infer speed change from acceleration alone.
A Reliable Interpretation Checklist
When given motion data, proceed in a consistent order:
Distinguish a position, velocity, and acceleration value before interpreting its sign.
Separate values at particular times from information given throughout an interval.
Positive velocity indicates positive-direction motion; negative velocity indicates negative-direction motion.
At nonzero velocity, matching signs mean speed increases and opposite signs mean it decreases.
Do not claim an exact instantaneous rate or between-measurement behavior unless the supplied data justify it.
This disciplined reading prevents several common confusions: position is not direction, acceleration is not speed, and a finite set of measurements is not a complete motion record. A strong interpretation reports the conclusion and its limits together.
Check Your Understanding
Use signs, units, and the stated information carefully when answering.
- A position table gives \(x(1)=4\) m and \(x(3)=10\) m. What is the average velocity from \(t=1\) s to \(t=3\) s, and what does this value not establish?
- An object has \(v=-5\ \text{m/s}\) and \(a=-2\ \text{m/s}^2\). What is its direction of motion, and is its speed increasing or decreasing?
- If acceleration is known exactly throughout an interval but the initial velocity is unknown, what motion information remains undetermined?
- Why does zero velocity at one time not, by itself, establish that an object remains at rest?
- What additional information is needed to determine position from a specified velocity function?
