Why Check Motion Data for Consistency?
A motion description may provide position, velocity, and acceleration in different forms: a position table, a velocity graph, or a list of acceleration values. These quantities are related, so they cannot vary independently. Checking their consistency means asking whether the reported rates agree with the slopes of the reported functions.
The central relationships are \(v(t)=x'(t)\) and \(a(t)=v'(t)=x''(t)\). Here \(x(t)\) is position, \(v(t)\) is velocity, and \(a(t)\) is acceleration. The derivatives describe instantaneous rates at a time. A slope computed between two distinct data points is instead an average rate over an interval. That difference matters: a table can provide useful evidence about consistency without determining every instantaneous derivative exactly.
In this tutorial, position is measured in metres, time in seconds, velocity in metres per second, and acceleration in metres per second squared. The units are part of the check: a position slope must have velocity units, and a velocity slope must have acceleration units.
What Slopes Should Be Compared?
For two times \(t_1\) and \(t_2\), the position secant slope is
This is average velocity over the interval from \(t_1\) to \(t_2\). It is not, in general, the instantaneous velocity at either endpoint. Likewise,
is average acceleration over that interval; it need not equal the acceleration at either endpoint.
A graph gives a related check. At a time where the tangent exists, the tangent slope of the position graph should equal the velocity there. The tangent slope of the velocity graph should equal the acceleration there. A secant line estimates these tangent slopes when its endpoints are close, but it is not automatically the tangent line.
Why the Derivative Gives the Instantaneous Slope
The derivative is defined by the limit of nearby secant slopes. For position, if the limit exists,
Each quotient inside the limit is an average velocity over the short interval from \(t\) to \(t+h\). As the interval shrinks toward \(t\), those average velocities approach the instantaneous velocity \(x'(t)\). Applying the same definition to the velocity function gives
when the limit exists. This is why the correct comparisons are position slope against velocity and velocity slope against acceleration. It also explains why one finite interval slope cannot generally be substituted for a derivative value.
A Practical Consistency Check
Use the same time interval when comparing changes. First check that the position data and velocity data use the same time units and that position changes have the right direction relative to velocity. Then compare position secant slopes with velocity information over those intervals. Finally, compare changes in velocity with acceleration information. If graphs are available, check tangent slopes and whether their signs and relative steepness agree with the values.
Compute position changes divided by elapsed time. Treat the results as average velocities, not automatically as endpoint velocities.
Compute velocity changes divided by elapsed time. Compare these average accelerations with the reported acceleration or its trend.
Confirm that slopes have the appropriate units and that each comparison uses the interval or time at which the data are claimed to apply.
Worked Examples
Worked Example: Checking a Motion Table
A proposed motion table gives the following values. Decide whether they agree with the model \(x(t)=t^2+t\).
| \(t\) (s) | \(x(t)\) (m) | Reported \(v(t)\) (m/s) | Reported \(a(t)\) (m/s²) |
|---|---|---|---|
| 0 | 0 | 1 | 2 |
| 1 | 2 | 3 | 2 |
| 2 | 6 | 5 | 2 |
Differentiate the position model:
Thus the model's velocity values are \(x'(0)=1\), \(x'(1)=3\), and \(x'(2)=5\) metres per second. Differentiate again:
The model's acceleration is \(2\) metres per second squared at each listed time. These derivative values match the reported velocity and acceleration entries.
Now check the position secant slopes. On \([0,1]\), the slope is \((2-0)/(1-0)=2\) metres per second. On \([1,2]\), it is \((6-2)/(2-1)=4\) metres per second. Neither slope equals both endpoint velocities. Instead, the first is \((1+3)/2=2\), and the second is \((3+5)/2=4\). This agrees with the model's linearly changing velocity on each interval. The position entries also match the model: \(x(0)=0\), \(x(1)=1+1=2\), and \(x(2)=4+2=6\). All entries and interval slopes are consistent.
Worked Example: Finding a Contradiction in a Proposed Model
Suppose a report claims that the position model is still \(x(t)=t^2+t\), but gives \(x(2)=8\), \(v(2)=7\) metres per second, and \(a(2)=2\) metres per second squared. Check the entries at \(t=2\).
The position model gives
not \(8\) metres. Its velocity function is \(x'(t)=2t+1\), so
not \(7\) metres per second. Its acceleration is \(x''(t)=2\), which does agree with the reported acceleration. Therefore the three reported quantities are not all consistent with the stated position model: position and velocity contradict it, while acceleration agrees.
As a further check, the model's position slope from \(t=1\) to \(t=2\) is \((6-2)/(2-1)=4\) metres per second. The model's endpoint velocities are \(3\) and \(5\) metres per second, whose average is \((3+5)/2=4\). The reported \(x(2)=8\) would instead make that secant slope \((8-2)/(2-1)=6\). This is another warning that the altered position entry does not belong to the stated model.
Worked Example: Reading Consistency from Graph Slopes
Suppose the position graph is the graph of \(x(t)=10+3t-t^2\), and the velocity graph is claimed to be \(v(t)=3-2t\). Check their relationship, then check a claimed constant acceleration of \(-2\) metres per second squared.
Differentiate position:
This is exactly the claimed velocity function. Its graph should therefore give the tangent slopes of the position graph. For example, at \(t=1\), the position is \(10+3-1=12\) metres and the velocity is \(3-2=1\) metre per second. The position graph has tangent slope \(1\) there.
Differentiate velocity:
Thus the velocity graph has slope \(-2\) metres per second squared at every time, matching the claimed acceleration. Direct substitution at \(t=1\) gives \(v(1)=1\) metre per second, and the velocity function's slope remains \(-2\). The position, velocity, and acceleration descriptions agree. In a qualitative graph check, the downward-sloping velocity graph signals negative acceleration, while the position graph's tangent slopes decrease with time.
A Common Mistake: Treating a Secant Slope as an Endpoint Velocity
Consider \(x(t)=t^2\). Between \(t=0\) and \(t=1\), the position secant slope is
A mistaken conclusion would be: “Therefore the velocity at \(t=0\) is \(1\) metre per second.” But differentiating gives \(v(t)=x'(t)=2t\), so \(v(0)=0\) metres per second. The secant slope \(1\) is the average velocity over the whole interval, not the instantaneous velocity at its start. The correction is to compare that interval average with velocity behavior across the interval, or use the derivative or tangent slope to identify a velocity at a particular time.
What a Consistency Check Can Establish
A contradiction can decisively show that a set of claims cannot all describe the same differentiable motion. For example, a stated position formula fixes its derivative, so a different reported velocity value at the same time conflicts with that formula. Agreement among a few sampled values, on the other hand, does not prove that the underlying functions are consistent between the samples. More information about the functions, or sufficiently detailed local slope information, may be needed.
Keep the levels of comparison clear: position changes give average velocity; velocity changes give average acceleration; derivatives and graph tangent slopes give instantaneous rates. Matching units and using the same time intervals make the comparisons meaningful. These checks help distinguish a genuine inconsistency from a difference between an average rate and an instantaneous one.
Check Your Understanding
Use the derivative relationships and interval definitions to answer the following questions.
- What quantity does the position secant slope between two times represent, and why is it not necessarily the velocity at either endpoint?
- If \(x(t)=3t^2\), find \(v(t)\) and \(a(t)\), then check both at \(t=2\).
- A velocity graph has a constant negative slope. What does that say about acceleration?
- For \(v(1)=4\) m/s and \(v(3)=10\) m/s, calculate the average acceleration over \([1,3]\) and give its units.
- Why can a few matching position, velocity, and acceleration entries fail to prove consistency for all times?
