From Acceleration to Motion
A particle’s acceleration describes how its velocity changes over time. If the acceleration vector is known, integrating it gives the velocity vector. Integrating velocity then gives the position vector. Initial velocity and position determine the constants that appear at the two stages.
The key is to work with every component and every initial condition. A vector integration constant has one constant for each component, and the velocity and position integrations produce separate sets of constants.
Acceleration, Velocity, and Position
For a particle moving in two or three dimensions, let \(\mathbf{r}(t)\) denote its position vector. Its velocity and acceleration are the successive derivatives
For instance, in two dimensions we can write \(\mathbf{a}(t)=\langle a_x(t),a_y(t)\rangle\). Integration is performed component by component: an antiderivative of \(\mathbf{a}\) is a vector whose derivative is \(\mathbf{a}\).
Each component needs an integration constant. In three dimensions there are three constants in each vector antiderivative. When the initial time is \(t_0\), a condition such as \(\mathbf{v}(t_0)=\mathbf{v}_0\) specifies the whole velocity vector at that time, component by component.
Why Two Integrations Recover the Motion
The Fundamental Theorem for vector integrals applies component by component. For each coordinate, the derivative of the accumulated integral of acceleration is acceleration itself. Therefore adding that integral to the initial velocity produces a function whose derivative is \(\mathbf{a}(t)\), and whose value at \(t_0\) is \(\mathbf{v}(t_0)\).
To see why this determines velocity uniquely, suppose two velocity functions both have derivative \(\mathbf{a}(t)\) and the same value at \(t_0\). Their difference has derivative \(\mathbf{0}\) in every component, so each component of that difference is constant. Since the difference is zero at \(t_0\), those constants are zero. The two velocity functions must agree. The same argument applied to position, whose derivative is the now-determined velocity, proves that the initial position fixes position uniquely.
For indefinite integration, the same result appears as two sets of constants. First, the velocity constants are set by the initial velocity. The position constants are then set by the initial position. The conditions are used in that order because position is found by integrating the velocity function.
Worked Examples
Worked Example: Polynomial Acceleration with Initial Data at \(t=0\)
Suppose \(\mathbf{a}(t)=\langle 6t,-4\rangle\), \(\mathbf{v}(0)=\langle 2,3\rangle\), and \(\mathbf{r}(0)=\langle 1,-2\rangle\). Find \(\mathbf{v}(t)\) and \(\mathbf{r}(t)\).
Integrate the acceleration components separately:
Use the velocity condition. Substituting \(t=0\) gives \(\langle C_1,C_2\rangle=\langle 2,3\rangle\), so \(C_1=2\) and \(C_2=3\). Thus \(\mathbf{v}(t)=\langle 3t^2+2,-4t+3\rangle\).
Now integrate velocity:
Since \(\mathbf{r}(0)=\langle 1,-2\rangle\), substitution gives \(D_1=1\) and \(D_2=-2\). Therefore \(\mathbf{r}(t)=\langle t^3+2t+1,-2t^2+3t-2\rangle\).
Check by differentiating: \(\mathbf{r}'(t)=\langle 3t^2+2,-4t+3\rangle=\mathbf{v}(t)\), and \(\mathbf{v}'(t)=\langle 6t,-4\rangle=\mathbf{a}(t)\). Also, \(\mathbf{v}(0)=\langle 2,3\rangle\) and \(\mathbf{r}(0)=\langle 1,-2\rangle\), as required.
Worked Example: Trigonometric Acceleration and a Nonzero Initial Time
Let \(\mathbf{a}(t)=\langle 2\cos t,-6\sin t\rangle\), with \(\mathbf{v}(\pi/2)=\langle 1,-2\rangle\) and \(\mathbf{r}(\pi/2)=\langle 0,4\rangle\). Find the velocity and position.
Integrate each component of acceleration:
At \(t=\pi/2\), \(\sin(\pi/2)=1\) and \(\cos(\pi/2)=0\). Thus \(\langle 2+C_1,C_2\rangle=\langle 1,-2\rangle\), which gives \(C_1=-1\) and \(C_2=-2\). Hence \(\mathbf{v}(t)=\langle 2\sin t-1,6\cos t-2\rangle\).
Integrate this velocity:
Substitute \(t=\pi/2\) and use \(\mathbf{r}(\pi/2)=\langle 0,4\rangle\). The first coordinate gives \(-\pi/2+D_1=0\), so \(D_1=\pi/2\). The second gives \(6-\pi+D_2=4\), so \(D_2=\pi-2\). Therefore
Check: differentiating this position gives \(\langle 2\sin t-1,6\cos t-2\rangle\), and differentiating again gives \(\langle 2\cos t,-6\sin t\rangle\). At \(t=\pi/2\), the velocity is \(\langle 1,-2\rangle\), and the position is \(\langle 0,4\rangle\). Both initial conditions hold.
Worked Example: Linear Acceleration and Initial Data at \(t=1\)
Suppose \(\mathbf{a}(t)=\langle 4,2t\rangle\), \(\mathbf{v}(1)=\langle 0,5\rangle\), and \(\mathbf{r}(1)=\langle 3,-1\rangle\). Reconstruct the motion.
First integrate acceleration:
At \(t=1\), this is \(\langle 4+C_1,1+C_2\rangle=\langle 0,5\rangle\). Thus \(C_1=-4\) and \(C_2=4\), so \(\mathbf{v}(t)=\langle 4t-4,t^2+4\rangle\).
Next, integrate velocity:
Use the position condition at \(t=1\). The first coordinate gives \(2-4+D_1=3\), so \(D_1=5\). The second gives \(1/3+4+D_2=-1\), so \(D_2=-16/3\). The position is therefore \(\mathbf{r}(t)=\langle 2t^2-4t+5,t^3/3+4t-16/3\rangle\).
Check: \(\mathbf{r}'(t)=\langle 4t-4,t^2+4\rangle\), and \(\mathbf{r}''(t)=\langle 4,2t\rangle\). At \(t=1\), the velocity is \(\langle 0,5\rangle\), while the position is \(\langle 3,-1\rangle\). The reconstructed functions satisfy the acceleration and both initial conditions.
Constants, Conditions, and Checks
There is one integration constant for each coordinate at each integration. For example, in two dimensions the first integration gives two velocity constants, and the second gives two position constants. The initial velocity determines the first pair; the initial position determines the second pair. The constants do not have to be the same at the two stages.
Initial data can be given at any time in the interval, not only at \(t=0\). Substitute the stated time into the antiderivative before solving for its constants. If acceleration is not continuous throughout the interval, the integral formulas require more care; the theorem above assumes continuity on an interval containing the initial time and the time being considered.
Units can help keep the quantities distinct. If time is measured in seconds and position in metres, then velocity has units of metres per second and acceleration has units of metres per second squared. Integrating acceleration with respect to time gives velocity units; integrating velocity with respect to time gives position units.
Common Mistake: Losing an Integration Constant
Consider constant acceleration \(\mathbf{a}(t)=\langle 2,0\rangle\), with \(\mathbf{v}(0)=\langle 1,3\rangle\) and \(\mathbf{r}(0)=\langle 0,0\rangle\). A common incorrect step is to integrate acceleration and write \(\mathbf{v}(t)=\langle 2t,0\rangle\), as if the initial velocity were unnecessary. This gives \(\mathbf{v}(0)=\langle 0,0\rangle\), which contradicts the given \(\langle 1,3\rangle\).
The correct antiderivative is \(\mathbf{v}(t)=\langle 2t+C_1,C_2\rangle\). Applying the initial velocity gives \(C_1=1\) and \(C_2=3\), so \(\mathbf{v}(t)=\langle 2t+1,3\rangle\). Integrating again gives \(\mathbf{r}(t)=\langle t^2+t+D_1,3t+D_2\rangle\). The initial position implies \(D_1=D_2=0\). Differentiation returns the given velocity and acceleration, and the initial position and velocity now both check.
Omitting constants can make a formula look simpler, but it generally selects only one possible motion among many. The initial conditions are what identify the correct one.
Check Your Understanding
For each problem, find velocity and position, showing how the initial conditions determine the constants. Check by differentiation and substitution.
- If \(\mathbf{a}(t)=\langle 2t,6\rangle\), \(\mathbf{v}(0)=\langle 0,1\rangle\), and \(\mathbf{r}(0)=\langle 2,0\rangle\), what are \(\mathbf{v}(t)\) and \(\mathbf{r}(t)\)?
- Why does a two-dimensional acceleration vector produce two constants when integrated to find velocity?
- For data specified at \(t=2\), at what time should you substitute when determining the velocity constants?
- What two derivative checks should be performed on a proposed position function?
- Explain why the velocity initial condition and position initial condition determine different sets of constants.
