Why Check a Motion Formula?
A position formula can look plausible and still contain an error: a coefficient may have the wrong units, a derivative may have a sign mistake, or an integration constant may fail to match the stated initial position or velocity. Checking a proposed motion means testing these features together rather than trusting the appearance of the formula.
For motion in three-dimensional space, write the position as \(\mathbf r(t)=\langle x(t),y(t),z(t)\rangle\). Its velocity and acceleration are \(\mathbf v(t)=\mathbf r'(t)\) and \(\mathbf a(t)=\mathbf v'(t)=\mathbf r''(t)\), wherever the derivatives exist. These equations apply component by component.
Units as a First Test
If position is measured in metres and time in seconds, then each position coordinate has units of metres, each velocity coordinate has units of metres per second, and each acceleration coordinate has units of metres per second squared. In symbols, these units are \(\mathrm m\), \(\mathrm{m/s}\), and \(\mathrm{m/s^2}\). The same principle holds with other length and time units, provided the units are used consistently.
Every term added to another term must have the same units. For example, a coordinate expression may contain a constant position, a velocity multiplied by time, and an acceleration multiplied by time squared. Each of those terms has units of length. A bare numerical coefficient does not automatically have the right units: its units must make the whole term a position.
The derivative relationships also explain the expected units. A difference quotient for position divides a change in length by a change in time, so velocity has units of length per time. Differentiating velocity divides length per time by time again, giving acceleration units of length per time squared. Thus units provide a quick check before any detailed substitution.
A Reliable Checking Procedure
Check the formulas in a fixed order. First inspect the units of every term. Next, differentiate position and compare the result with the proposed velocity. Then differentiate velocity and compare it with the proposed acceleration. Finally, substitute the initial time into the position and velocity components and compare those values with the given data.
Verify that position, velocity, and acceleration terms have units of length, length per time, and length per time squared, respectively.
Differentiate each position component and compare it with the corresponding velocity component; then differentiate velocity and compare with acceleration.
Substitute the specified initial time into each position and velocity component and compare with the given vectors.
All three checks matter. A formula can have consistent units but the wrong initial velocity. A position formula can meet the initial conditions but have a derivative that does not match the claimed velocity. Passing only one test does not validate the complete set of motion formulas.
Worked Example: Checking a Position Formula and Its Derivatives
Worked Example: Checking a Position Formula and Its Derivatives
Suppose \(t\) is measured in seconds and the proposed position is
Each term in each coordinate has units of metres. For example, \((2\,\mathrm{m/s})t\) has units \((\mathrm{m/s})(\mathrm s)=\mathrm m\), and \((0.5\,\mathrm{m/s^2})t^2\) has units \((\mathrm{m/s^2})(\mathrm{s^2})=\mathrm m\).
Differentiate the first coordinate:
Differentiate the second and third coordinates:
Therefore the velocity is
Differentiate again to obtain the acceleration:
At \(t=0\,\mathrm s\), the position is \(\mathbf r(0)=\langle1,-3,5\rangle\,\mathrm m\), and the velocity is \(\mathbf v(0)=\langle2,4,0\rangle\,\mathrm{m/s}\). These are the initial values implied by the formulas. The derivatives reproduce the velocity and acceleration shown, so the formulas are mutually consistent.
Checking Initial Data Component by Component
Initial conditions specify values at a particular time, often \(t=0\), but not necessarily. If the given data are \(\mathbf r(t_0)=\mathbf r_0\) and \(\mathbf v(t_0)=\mathbf v_0\), evaluate the position and velocity at \(t_0\). Compare vectors coordinate by coordinate: the \(x\)-coordinate must match the \(x\)-coordinate, and likewise for \(y\) and \(z\). Matching one coordinate does not establish that the entire vector condition holds.
Worked Example: Verifying Motion from Constant Acceleration
Worked Example: Verifying Motion from Constant Acceleration
Let \(t\) be in seconds. Suppose the proposed formulas are
Each term of \(\mathbf r(t)\) has units of metres; each term of \(\mathbf v(t)\) has units of metres per second. Differentiate the position coordinates:
These are exactly the three proposed velocity coordinates. Differentiating them gives \(x''(t)=2\,\mathrm{m/s^2}\), \(y''(t)=-6\,\mathrm{m/s^2}\), and \(z''(t)=4\,\mathrm{m/s^2}\), which agrees with the proposed acceleration.
At \(t=0\,\mathrm s\), substitute into position: \(\mathbf r(0)=\langle4,1,-1\rangle\,\mathrm m\). Substitution into velocity gives \(\mathbf v(0)=\langle1,-2,3\rangle\,\mathrm{m/s}\). Thus the formulas pass the derivative checks and have the initial position and velocity \(\langle4,1,-1\rangle\,\mathrm m\) and \(\langle1,-2,3\rangle\,\mathrm{m/s}\), respectively.
Why the Checks Determine the Motion
There is a useful reason to check both acceleration and initial data. On an interval, suppose two twice-differentiable position functions have the same acceleration and the same position and velocity at time \(t_0\). Subtract them, and call the difference \(\mathbf q(t)\). Then \(\mathbf q''(t)=\mathbf 0\), \(\mathbf q(t_0)=\mathbf 0\), and \(\mathbf q'(t_0)=\mathbf 0\).
Each coordinate of \(\mathbf q'\) has derivative zero, so it is constant on the interval by the mean value theorem. Since \(\mathbf q'(t_0)=\mathbf 0\), that constant is zero. Therefore \(\mathbf q'(t)=\mathbf 0\) throughout the interval. Each coordinate of \(\mathbf q\) is now constant, and \(\mathbf q(t_0)=\mathbf 0\) forces that constant to be zero. Hence \(\mathbf q(t)=\mathbf 0\) throughout: the two position functions agree.
This conclusion assumes the functions are defined and twice differentiable on the interval in question. A formula check confirms consistency with the stated kinematics and initial data; it does not, by itself, establish that a proposed acceleration is the correct physical model for a particular situation.
Worked Example: Finding an Initial-Data Error
Worked Example: Finding an Initial-Data Error
Suppose \(t\) is in seconds and a proposed position is
All terms have the correct position units. Differentiating gives
At \(t=0\,\mathrm s\), the position is \(\langle2,-1,4\rangle\,\mathrm m\), while the velocity is \(\langle3,2,-2\rangle\,\mathrm{m/s}\). Suppose the required initial velocity is instead \(\langle3,0,-2\rangle\,\mathrm{m/s}\). The \(x\)- and \(z\)-components match, but the \(y\)-component is \(2\,\mathrm{m/s}\), not \(0\,\mathrm{m/s}\). The initial-data check fails.
To correct the \(y\)-coordinate while retaining its initial position and its \(t^2\) term, remove the linear term. The corrected coordinate is \(y(t)=-1\,\mathrm m+(1\,\mathrm{m/s^2})t^2\). Its derivative is \(y'(t)=(2\,\mathrm{m/s^2})t\), so \(y'(0)=0\,\mathrm{m/s}\), as required. Also \(y(0)=-1\,\mathrm m\), so the initial position is preserved. The corrected \(y\)-acceleration is \(2\,\mathrm{m/s^2}\), obtained by differentiating \(y'\). Thus the revised formula meets the specified initial data, and its acceleration must use this corrected \(y\)-component.
A Common Units Mistake
A frequent error is to check only the numerical coefficients, ignoring their units. For instance, writing \(x(t)=5\,\mathrm m+(2\,\mathrm{m/s})t+(3\,\mathrm m)t^2\) is not dimensionally consistent: the final term has units \(\mathrm{m\,s^2}\), not metres. Differentiating it formally produces \(6\,\mathrm m\,t\), whose units are \(\mathrm{m\,s}\), not velocity. The coefficient of \(t^2\) must have units of acceleration. For example, replacing it with \(3\,\mathrm{m/s^2}\) gives a position term with units of metres and a derivative term with units of metres per second.
Why This Validation Matters
Motion formulas often combine several components and integration constants. A systematic check catches errors that are difficult to notice by inspection: a wrong sign in one coordinate, a missing time factor, an inconsistent coefficient unit, or an initial value copied incorrectly. The method is especially useful when formulas have been obtained by integrating acceleration, since differentiation checks the recovered functions while substitution checks the constants.
Keep the tests distinct. Units test dimensional consistency. Differentiation tests the relationships between position, velocity, and acceleration. Initial substitution tests whether the formulas describe the specified starting state. A reliable validation uses all three.
Check Your Understanding
Use units, componentwise differentiation, and substitution to answer these questions.
- If \(t\) is measured in seconds and \(x(t)\) is a position in metres, what units must the coefficients of \(t\) and \(t^2\) have in a quadratic position formula?
- For \(x(t)=6\,\mathrm m-(2\,\mathrm{m/s})t+(3\,\mathrm{m/s^2})t^2\), find \(x'(t)\), \(x''(t)\), \(x(0)\), and \(x'(0)\), including units.
- A proposed position formula has the correct initial position but its derivative at the initial time does not equal the stated initial velocity. Does it pass the initial-data check? Explain.
- Why must position and velocity initial conditions both be checked when validating a motion with a specified acceleration?
- Give a term involving \(t^2\) that has units of position, and explain the units required for its coefficient.