Reconstructing Motion from Acceleration Workshop

From Acceleration to a Complete Motion

A prescribed acceleration describes how velocity changes, but it does not by itself determine a unique trajectory. To reconstruct a particular motion, we also need the velocity and position at a specified time. The task is to use those initial data to determine every integration constant in all three coordinates.

For a particle moving in space, write its position as \(\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle\). Its velocity and acceleration are the first and second derivatives of position:

\mathbf{v}(t)=\mathbf{r}'(t) \qquad\text{and}\qquad \mathbf{a}(t)=\mathbf{r}''(t)=\mathbf{v}'(t).

These vector equations mean that the corresponding equations hold in each coordinate. If acceleration and initial data are given, we integrate to find velocity, then integrate again to find position. At both stages, initial conditions determine constants separately in each coordinate.

Definition: A motion reconstructed from acceleration \(\mathbf{a}(t)\), initial velocity \(\mathbf{v}(t_0)=\mathbf{v}_0\), and initial position \(\mathbf{r}(t_0)=\mathbf{r}_0\) is a function \(\mathbf{r}(t)\) satisfying \(\mathbf{r}''(t)=\mathbf{a}(t)\), \(\mathbf{r}'(t_0)=\mathbf{v}_0\), and \(\mathbf{r}(t_0)=\mathbf{r}_0\).

Here \(t_0\) is the time at which the data are specified. The acceleration is assumed continuous on the time interval under consideration, so it can be integrated there. The initial position and velocity are vectors, with three components each. Neither one can be replaced by the other: acceleration determines changes in velocity, while the two initial vectors select the particular motion.

The Reconstruction Procedure

Suppose \(\mathbf{a}(t)=\langle a_x(t),a_y(t),a_z(t)\rangle\). First integrate each component to obtain velocity, including an integration constant for each component. Apply \(\mathbf{v}(t_0)=\mathbf{v}_0\) to determine those constants. Then integrate the resulting velocity components and use \(\mathbf{r}(t_0)=\mathbf{r}_0\) to determine the position constants.

Integrate acceleration.
Find a general velocity vector and include one constant for each coordinate.

Apply the initial velocity.
Substitute \(t=t_0\) and set the result equal to \(\mathbf{v}_0\).

Integrate velocity.
Find a general position vector, again including one constant for each coordinate.

Apply the initial position and check.
Use \(\mathbf{r}(t_0)=\mathbf{r}_0\), then differentiate and test both initial conditions.

Writing constants as vectors is also possible, but the coordinate form makes it harder to overlook one. Constants in the velocity and position integrations are distinct; they are determined by different initial data.

Worked Example: Polynomial Acceleration from Time Zero

Let \(\mathbf{a}(t)=\langle 6t,4,-2t\rangle\), with \(\mathbf{v}(0)=\langle 1,-3,5\rangle\) and \(\mathbf{r}(0)=\langle 2,0,-1\rangle\). Find the motion.

Integrating each acceleration component gives a general velocity:

\mathbf{v}(t) = \left\langle 3t^2+C_x,\;4t+C_y,\;-t^2+C_z \right\rangle.

At \(t=0\), this becomes \(\langle C_x,C_y,C_z\rangle\). Matching the given initial velocity gives \(C_x=1\), \(C_y=-3\), and \(C_z=5\). Thus \(\mathbf{v}(t)=\langle 3t^2+1,4t-3,-t^2+5\rangle\).

Integrating that velocity gives

\mathbf{r}(t) = \left\langle t^3+t+D_x,\;2t^2-3t+D_y,\;-\frac{t^3}{3}+5t+D_z \right\rangle.

At \(t=0\), the position is \(\langle D_x,D_y,D_z\rangle\). Therefore \(D_x=2\), \(D_y=0\), and \(D_z=-1\), so

\mathbf{r}(t) = \left\langle t^3+t+2,\;2t^2-3t,\;-\frac{t^3}{3}+5t-1 \right\rangle.

Check by differentiating: \(\mathbf{r}'(t)=\langle 3t^2+1,4t-3,-t^2+5\rangle\), and \(\mathbf{r}''(t)=\langle 6t,4,-2t\rangle\), the prescribed acceleration. At \(t=0\), velocity is \(\langle 1,-3,5\rangle\) and position is \(\langle 2,0,-1\rangle\), as required.

Initial Data at a Nonzero Time

The starting time need not be zero. When \(t_0\ne 0\), evaluate the general antiderivatives at \(t_0\), not at zero. A convenient alternative is to write integrals with lower limit \(t_0\); these automatically give zero change at the initial time.

Worked Example: Applying Data at Time One

Suppose \(\mathbf{a}(t)=\langle 2,6t,-4\rangle\), \(\mathbf{v}(1)=\langle 0,2,1\rangle\), and \(\mathbf{r}(1)=\langle 3,-1,4\rangle\). Integrate acceleration to obtain

\mathbf{v}(t) = \left\langle 2t+C_x,\;3t^2+C_y,\;-4t+C_z \right\rangle.

At \(t=1\), the components are \(2+C_x\), \(3+C_y\), and \(-4+C_z\). Equating them to \(0,2,1\), respectively, gives \(C_x=-2\), \(C_y=-1\), and \(C_z=5\). Hence \(\mathbf{v}(t)=\langle 2t-2,3t^2-1,-4t+5\rangle\).

Integrating this velocity gives a general position:

\mathbf{r}(t) = \left\langle t^2-2t+D_x,\;t^3-t+D_y,\;-2t^2+5t+D_z \right\rangle.

At \(t=1\), these components are \(D_x-1\), \(D_y\), and \(D_z+3\). Matching \(\langle 3,-1,4\rangle\) gives \(D_x=4\), \(D_y=-1\), and \(D_z=1\). Therefore

\mathbf{r}(t) = \left\langle t^2-2t+4,\;t^3-t-1,\;-2t^2+5t+1 \right\rangle.

Differentiate to check: \(\mathbf{r}'(t)=\langle 2t-2,3t^2-1,-4t+5\rangle\), and \(\mathbf{r}''(t)=\langle 2,6t,-4\rangle\). At \(t=1\), the velocity is \(\langle 0,2,1\rangle\), and the position is \(\langle 3,-1,4\rangle\). All three prescribed conditions hold.

A Direct Formula and Why It Works

The two integrations can be summarized in one expression. For continuous acceleration, the reconstructed position is

\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}_0(t-t_0) + \int_{t_0}^{t}(t-s)\mathbf{a}(s)\,ds.

The integral is evaluated component by component. Its variable \(s\) is the integration variable, while \(t\) is the time at which position is being evaluated. The formula applies when \(t\) and \(t_0\) lie in the interval on which the acceleration is continuous.

To verify the formula, call its right-hand side \(\mathbf{R}(t)\). The fundamental theorem of calculus and differentiation of an integral with a variable upper limit give

\mathbf{R}'(t) = \mathbf{v}_0+\int_{t_0}^{t}\mathbf{a}(s)\,ds, \qquad \mathbf{R}''(t)=\mathbf{a}(t).

In the first derivative, the upper-limit term from the integral is zero because its factor \(t-s\) is zero when \(s=t\). Differentiating the remaining integral gives \(\mathbf{a}(t)\). At \(t=t_0\), the integral vanishes, so \(\mathbf{R}'(t_0)=\mathbf{v}_0\) and \(\mathbf{R}(t_0)=\mathbf{r}_0\). Thus the expression has the required acceleration and both initial values.

It is also the only such position function. If two functions satisfy the same acceleration and initial data, their difference \(\mathbf{q}(t)\) has \(\mathbf{q}''(t)=\mathbf{0}\). Therefore \(\mathbf{q}'\) is constant; since \(\mathbf{q}'(t_0)=\mathbf{0}\), it is zero. Then \(\mathbf{q}\) is constant, and \(\mathbf{q}(t_0)=\mathbf{0}\) forces that constant to be zero. So the two motions agree throughout the interval.

Worked Example: Trigonometric Acceleration

Let \(\mathbf{a}(t)=\langle \cos t,-\sin t,3t^2\rangle\), \(\mathbf{v}(0)=\langle 0,1,-2\rangle\), and \(\mathbf{r}(0)=\langle 1,2,0\rangle\). Integrating acceleration gives

\mathbf{v}(t) = \left\langle \sin t+C_x,\;\cos t+C_y,\;t^3+C_z \right\rangle.

At zero, this vector is \(\langle C_x,1+C_y,C_z\rangle\). The initial velocity gives \(C_x=0\), \(C_y=0\), and \(C_z=-2\), so \(\mathbf{v}(t)=\langle\sin t,\cos t,t^3-2\rangle\).

Integrating once more gives \(\mathbf{r}(t)=\langle-\cos t+D_x,\sin t+D_y,\frac{t^4}{4}-2t+D_z\rangle\). At zero, the position is \(\langle -1+D_x,D_y,D_z\rangle\). Equating this to \(\langle 1,2,0\rangle\) gives \(D_x=2\), \(D_y=2\), and \(D_z=0\). Thus

\mathbf{r}(t) = \left\langle 2-\cos t,\;2+\sin t,\;\frac{t^4}{4}-2t \right\rangle.

Differentiation gives \(\mathbf{r}'(t)=\langle\sin t,\cos t,t^3-2\rangle\) and \(\mathbf{r}''(t)=\langle\cos t,-\sin t,3t^2\rangle\). At zero, these give \(\mathbf{r}'(0)=\langle 0,1,-2\rangle\) and \(\mathbf{r}(0)=\langle 1,2,0\rangle\), verifying the acceleration and initial data.

A Common Integration Error

A frequent mistake is to find an antiderivative of acceleration and treat it as the velocity, without applying the initial velocity. For example, if \(a_x(t)=6t\), then \(3t^2\) is an antiderivative. But if \(v_x(0)=1\), setting \(v_x(t)=3t^2\) gives \(v_x(0)=0\), which contradicts the data. Integrating that incorrect velocity and using \(x(0)=2\) gives \(x(t)=t^3+2\); its derivative still has the wrong initial value.

The correction is to include the constant: \(v_x(t)=3t^2+C\). The condition \(v_x(0)=1\) forces \(C=1\), so \(v_x(t)=3t^2+1\). A second integration gives \(x(t)=t^3+t+D\), and \(x(0)=2\) forces \(D=2\). Differentiating confirms \(x'(0)=1\) and \(x''(t)=6t\).

Key check: After reconstructing a motion, verify both derivatives and both initial values: \(\mathbf{r}''(t)=\mathbf{a}(t)\), \(\mathbf{r}'(t_0)=\mathbf{v}_0\), and \(\mathbf{r}(t_0)=\mathbf{r}_0\). Checking acceleration alone does not confirm that the constants match the specified motion.

Check Your Understanding

For each problem, keep the three coordinates separate and use the initial conditions at the stated time.

If \(\mathbf{a}(t)=\langle 0,2,0\rangle\), \(\mathbf{v}(0)=\langle 1,0,-1\rangle\), and \(\mathbf{r}(0)=\langle 0,2,3\rangle\), what is \(\mathbf{r}(t)\)?
Why does integrating a known acceleration introduce three independent constants when finding velocity?
For the direct position formula, what does the integral term contribute at \(t=t_0\)?
How can differentiation detect that an integration constant was omitted?
What conditions must two reconstructed motions share for the uniqueness argument to show that they are identical?