From Velocity Data to Displacement
A velocity graph or table describes how an object's position changes over time. The accumulated change depends not only on how large the velocity is, but also on whether the object moves in the positive or negative direction. That is why displacement is found using signed accumulation: velocity above the time axis contributes positively, and velocity below it contributes negatively.
Sometimes the graph gives a clear shape, so areas can be calculated directly. A table gives velocity at selected times, but not necessarily what happens between those times. In that case, numerical integration can provide an estimate, provided the method and its assumptions are stated.
Reading Signed Area from a Graph
For a velocity graph, split the time interval into sections whose areas can be calculated. Regions above the horizontal axis have positive signed area; regions below it have negative signed area. Add these signed areas to find displacement. The units also follow from the graph: velocity in meters per second multiplied by time in seconds gives displacement in meters.
If the graph consists of straight segments, each section is a triangle, rectangle, or trapezoid. If the graph is curved, a geometric area may not be exact unless its shape is specified sufficiently. A graph that is only a sketch may support a visual estimate, not a precise numerical answer.
Worked Example: Adding Signed Areas on a Piecewise-Linear Graph
A particle's velocity is shown by straight-line segments through the points \((0,3)\), \((2,3)\), \((5,-1)\), and \((7,2)\), where time is in seconds and velocity is in meters per second. Find the displacement from \(t=0\) to \(t=7\).
From \(t=0\) to \(t=2\), the graph is a rectangle of height \(3\) and width \(2\):
From \(t=2\) to \(t=5\), the endpoint velocities are \(3\) and \(-1\). The signed trapezoid area is:
This segment crosses the time axis, but its positive and negative parts are already combined with their signs in that calculation. From \(t=5\) to \(t=7\):
Therefore, the total displacement is:
Check: the graph has a net positive signed area, so a positive displacement is consistent. Because the graph is specified to consist of straight segments, the trapezoid areas are exact; the displacement is \(10\text{ m}\).
Estimating from a Table: The Trapezoidal Rule
A table usually gives velocities at times \(t_0,t_1,\ldots,t_n\). Over each interval \([t_i,t_{i+1}]\), the trapezoidal rule estimates the signed area by treating the velocity as if it changed linearly between the two recorded values. The width of that trapezoid is the time difference \(t_{i+1}-t_i\), and its signed height is the average of the endpoint velocities.
Worked Example: Unequal Time Intervals in a Velocity Table
A table gives the following data. Estimate the particle's displacement from \(0\) to \(6\) seconds with the trapezoidal rule.
| Time \(t\) (s) | Velocity \(v\) (m/s) |
|---|---|
| 0 | 2 |
| 1 | 4 |
| 3 | 1 |
| 6 | -2 |
The time widths are \(1-0=1\), \(3-1=2\), and \(6-3=3\) seconds. Calculate each signed trapezoid:
Add the results:
Check: the final interval has a negative average endpoint velocity, so it contributes \(-1.5\text{ m}\), not \(+1.5\text{ m}\). The computed estimate is positive overall and has units of meters. It is an estimate unless the velocity is known to be linear on each interval.
Why the Trapezoidal Rule Works
The rule follows by integrating the straight-line model between two consecutive data points. Suppose the recorded velocities at times \(t_i\) and \(t_{i+1}\) are \(v_i\) and \(v_{i+1}\). If velocity changes linearly between them, its equation on that interval is:
Let \(h=t_{i+1}-t_i\). Integrate this model across the interval. With \(u=t-t_i\), the integration limits become \(0\) and \(h\):
This is the signed trapezoid formula. It also explains its limitation: with only the two endpoint values, the calculation uses a linear model for what happens between them. If the actual velocity curve bends, the model's area may differ from the actual accumulated displacement.
Other Numerical Estimates
A table can also be used to form left- or right-endpoint sums. For each time interval, a left sum uses the velocity at the interval's starting time; a right sum uses the velocity at its ending time. For interval widths \(\Delta t_i=t_{i+1}-t_i\), these are:
These are approximations based on a different assumption: each interval's velocity is treated as constant at the selected endpoint value. The direction of error cannot be decided from the table alone. For instance, a left sum is an underestimate on an interval where velocity is increasing throughout, but that conclusion requires knowing that it is increasing throughout—not just seeing that the two recorded endpoint values differ.
Worked Example: Comparing Endpoint and Trapezoidal Estimates
Suppose the table records \(v=1,3,2,4\) meters per second at \(t=0,2,4,6\) seconds, respectively. Each interval has width \(2\) seconds. The left sum is:
The right sum is:
The trapezoidal estimate averages the left and right endpoint estimates on each interval:
Check: here \(T=(L+R)/2=(12+18)/2=15\text{ m}\), as expected when all intervals have equal width. The three numbers are methods of estimation from the table, not necessarily three bounds on the true displacement.
What the Data Can—and Cannot—Tell You
A graph or table must be interpreted according to how much information it provides. A stated piecewise-linear graph specifies velocity between its listed points, so its trapezoidal areas are exact. A table of samples alone gives the velocity only at selected times. Unless additional information is supplied—for example, that velocity is linear between samples—the exact integral is not determined by the sample values.
Worked Example: Identical Table Values, Different Displacements
Suppose a table records \(v(0)=2\), \(v(2)=4\), and \(v(4)=2\), with time in seconds and velocity in meters per second. The trapezoidal estimate across the two intervals is:
That estimate is the exact displacement if velocity is linear on each interval. In that case, the linear models are \(v(t)=2+t\) for \(0\leq t\leq2\) and \(v(t)=6-t\) for \(2\leq t\leq4\); their integrals are \(6\text{ m}\) and \(6\text{ m}\).
But the same table values are also consistent with a different continuous velocity: use \(v(t)=2+t+t(2-t)\) on \(0\leq t\leq2\) and \(v(t)=6-t\) on \(2\leq t\leq4\). At \(t=0\) and \(t=2\), the added term \(t(2-t)\) is zero, so the listed values remain unchanged. The added displacement on the first interval is:
The second velocity has not changed, so this particle's total displacement is \(12+\frac{4}{3}=\frac{40}{3}\text{ m}\), not \(12\text{ m}\). Check: both velocity functions match all three table entries, yet their integrals differ. Thus \(12\text{ m}\) is a trapezoidal estimate based on linear interpolation; the table alone does not establish the exact displacement.
A Common Mistake: Ignoring the Sign
A frequent error is to count every region's area as positive when the question asks for displacement. For the piecewise-linear graph in the first example, the interval from \(t=2\) to \(t=5\) crosses the axis. Its positive triangle has width \(2.25\) seconds and height \(3\) meters per second, so its area is \(\frac12(2.25)(3)=3.375\text{ m}\). Its negative triangle has width \(0.75\) seconds and height \(1\) meter per second, so its signed area is \(-\frac12(0.75)(1)=-0.375\text{ m}\). The interval's displacement is \(3.375-0.375=3\text{ m}\), not \(3.75\text{ m}\).
Likewise, from \(t=5\) to \(t=7\), the velocity rises from \(-1\) to \(2\), crossing zero after \(\frac{2}{3}\) second. Its positive triangle has area \(\frac12(\frac43)(2)=\frac43\text{ m}\); its negative triangle has signed area \(-\frac12(\frac23)(1)=-\frac13\text{ m}\). The signed area is \(1\text{ m}\), not \(\frac53\text{ m}\). Adding all areas as positive would give \(6+\frac{15}{4}+\frac53=\frac{137}{12}\text{ m}\), rather than the displacement \(10\text{ m}\). Keep the negative sign for regions below the axis.
A Reliable Workflow
Record the initial and final times, and combine velocity and time units to determine the displacement units.
Determine whether the graph gives an exact shape, states that segments are linear, or only supplies selected values.
Use geometric areas for a specified graph, or a named numerical rule for tabulated data. Keep each interval's time width and sign.
Say whether the displacement is exact under the given graph description or an estimate based on an interpolation or summation method.
Check Your Understanding
For each question, pay attention to whether the data specify an exact graph or only sampled values.
- A straight-line velocity segment goes from \(5\text{ m/s}\) at \(t=1\text{ s}\) to \(1\text{ m/s}\) at \(t=4\text{ s}\). What signed displacement does the trapezoidal area give?
- A velocity table lists values at \(t=0,2,5\). What time widths should be used in the trapezoidal rule?
- When is the trapezoidal rule exact on an interval?
- Why can two continuous velocity functions with the same table values have different displacements?
- What is the difference between displacement and the sum of the unsigned areas on a velocity graph?
