One Motion, Three Coordinate Equations
A particle moving through space has one position at each time, but that position is described by three coordinate functions. Its motion can therefore be studied through three related one-dimensional equations: one for each coordinate. The equations can have different accelerations, velocities, and initial data, yet they all use the same time parameter.
This separation is useful when the forces or prescribed motion affect different directions in different ways. It lets us reconstruct a spatial trajectory without treating the coordinates as if they had to follow identical rules. The coordinates are solved separately, then assembled into a single vector-valued function.
Thus, the acceleration information amounts to three scalar equations: \(x''(t)=a_x(t)\), \(y''(t)=a_y(t)\), and \(z''(t)=a_z(t)\). To determine a position coordinate from its acceleration, we need two initial conditions for that coordinate: its position and velocity at a specified time. Acceleration alone does not determine a unique trajectory.
Why Each Coordinate Can Be Solved Separately
The vector derivative is defined component by component. Consequently, the vector equation \(\mathbf{r}''(t)=\mathbf{a}(t)\) holds exactly when each of its coordinate equations holds. If the initial position is \(\mathbf{r}(t_0)=\langle x_0,y_0,z_0\rangle\) and the initial velocity is \(\mathbf{v}(t_0)=\langle u_0,v_0,w_0\rangle\), these data specify the initial position and velocity for each scalar equation.
Here is why the procedure works. For the \(x\)-coordinate, the Fundamental Theorem of Calculus gives \(x'(t)-x'(t_0)=\int_{t_0}^{t}a_x(s)\,ds\), because \(x''(s)=a_x(s)\). A second application gives \(x(t)-x(t_0)=\int_{t_0}^{t}x'(s)\,ds\). The same reasoning applies to \(y\) and \(z\). Since each equation uses its own acceleration and initial data, solving one coordinate does not require changing the equation for another. Once the three solutions are combined, differentiating the resulting vector recovers the specified acceleration and initial velocity.
These equations also show why all coordinates must use the same time variable. At a particular time \(t\), the position vector is assembled from \(x(t)\), \(y(t)\), and \(z(t)\), not from values evaluated at three different times.
Worked Example: Different Coordinate Accelerations
Worked Example: Different Coordinate Accelerations
Suppose \(t\) is measured in seconds and position in metres. The accelerations are \(a_x(t)=2\), \(a_y(t)=6t\), and \(a_z(t)=0\), in units of metres per second squared. At \(t=0\), the position is \(\langle 1,-2,3\rangle\) metres and the velocity is \(\langle 3,1,-4\rangle\) metres per second. Find the position.
For the \(x\)-coordinate, \(x''(t)=2\). Integrating once gives \(x'(t)=2t+C_1\). From \(x'(0)=3\), \(C_1=3\), so \(x'(t)=2t+3\). Integrating again gives \(x(t)=t^2+3t+C_2\). From \(x(0)=1\), \(C_2=1\). Thus \(x(t)=t^2+3t+1\).
For the \(y\)-coordinate, \(y''(t)=6t\). Therefore \(y'(t)=3t^2+C_3\). Since \(y'(0)=1\), \(C_3=1\). Integrating again gives \(y(t)=t^3+t+C_4\). Since \(y(0)=-2\), \(C_4=-2\), so \(y(t)=t^3+t-2\).
For the \(z\)-coordinate, \(z''(t)=0\), so \(z'(t)=C_5\). Since \(z'(0)=-4\), \(z'(t)=-4\). Then \(z(t)=-4t+C_6\), and \(z(0)=3\) gives \(C_6=3\).
The reconstructed position is therefore \(\mathbf{r}(t)=\langle t^2+3t+1,\ t^3+t-2,\ -4t+3\rangle\) metres. To check it, differentiate: \(\mathbf{v}(t)=\langle 2t+3,\ 3t^2+1,\ -4\rangle\), and \(\mathbf{a}(t)=\langle 2,\ 6t,\ 0\rangle\). At \(t=0\), these give the specified initial position \(\langle1,-2,3\rangle\) and velocity \(\langle3,1,-4\rangle\), and the acceleration matches all three specifications.
Recovering Position from Coordinate Velocities
If the velocities are given directly, each coordinate requires one integration rather than two. For instance, if \(x'(t)=u(t)\), then \(x(t)=x(t_0)+\int_{t_0}^{t}u(s)\,ds\). The initial coordinate value fixes the integration constant. This is the coordinate-by-coordinate form of recovering position from velocity.
Worked Example: Integrating Three Velocity Components
Let \(t\) be in seconds, and suppose the velocity components, in metres per second, are \(x'(t)=6t\), \(y'(t)=4-2t\), and \(z'(t)=3t^2\). The initial position at \(t=0\) is \(\langle-1,5,2\rangle\) metres. Find the position and its value at \(t=2\).
For \(x\), integration gives \(x(t)=3t^2+C_1\). The condition \(x(0)=-1\) gives \(C_1=-1\), so \(x(t)=3t^2-1\).
For \(y\), integration gives \(y(t)=4t-t^2+C_2\). The condition \(y(0)=5\) gives \(C_2=5\), so \(y(t)=4t-t^2+5\).
For \(z\), integration gives \(z(t)=t^3+C_3\). The condition \(z(0)=2\) gives \(C_3=2\), so \(z(t)=t^3+2\). Therefore \(\mathbf{r}(t)=\langle3t^2-1,\ 4t-t^2+5,\ t^3+2\rangle\) metres.
At \(t=2\), the coordinates are \(x(2)=3(2^2)-1=11\), \(y(2)=4(2)-2^2+5=9\), and \(z(2)=2^3+2=10\). Thus \(\mathbf{r}(2)=\langle11,9,10\rangle\) metres. As a check, differentiating the position gives \(\langle6t,4-2t,3t^2\rangle\), exactly the specified velocity, and substituting \(t=0\) gives the initial position.
Initial Data at a Nonzero Time
The reference time does not have to be zero. When initial conditions are given at \(t=t_0\), the most direct approach is to integrate from \(t_0\), or to apply the conditions to the constants after finding antiderivatives. In either approach, use the same reference time consistently for all three coordinates.
Worked Example: Starting from a Later Reference Time
Suppose the coordinate accelerations are \(a_x(t)=2t\), \(a_y(t)=4\), and \(a_z(t)=-6t\). At \(t=2\), the position is \(\langle5,1,-3\rangle\) and the velocity is \(\langle1,-2,4\rangle\). Find \(\mathbf{r}(t)\).
For \(x\), integrate \(x''(t)=2t\) to obtain \(x'(t)=t^2+C_1\). The condition \(x'(2)=1\) gives \(4+C_1=1\), so \(C_1=-3\). Thus \(x'(t)=t^2-3\). A further integration gives \(x(t)=t^3/3-3t+C_2\). Using \(x(2)=5\), we get \(8/3-6+C_2=5\), so \(C_2=25/3\).
For \(y\), \(y''(t)=4\) gives \(y'(t)=4t+C_3\). Since \(y'(2)=-2\), \(8+C_3=-2\), hence \(C_3=-10\). Then \(y(t)=2t^2-10t+C_4\). From \(y(2)=1\), \(8-20+C_4=1\), so \(C_4=13\).
For \(z\), \(z''(t)=-6t\) gives \(z'(t)=-3t^2+C_5\). Since \(z'(2)=4\), \(-12+C_5=4\), hence \(C_5=16\). Then \(z(t)=-t^3+16t+C_6\). From \(z(2)=-3\), \(-8+32+C_6=-3\), so \(C_6=-27\).
Therefore \(\mathbf{r}(t)=\langle t^3/3-3t+25/3,\ 2t^2-10t+13,\ -t^3+16t-27\rangle\). Differentiating gives \(\mathbf{v}(t)=\langle t^2-3,\ 4t-10,\ -3t^2+16\rangle\), and differentiating again gives \(\mathbf{a}(t)=\langle2t,4,-6t\rangle\). At \(t=2\), the position is \(\langle5,1,-3\rangle\) and the velocity is \(\langle1,-2,4\rangle\), as required.
What Each Coordinate Contributes
Each coordinate function describes the particle’s position along one axis. Its first derivative gives the velocity component along that axis, and its second derivative gives the acceleration component along that axis. A coordinate with zero acceleration can still change with time: its velocity is constant, not necessarily zero. Likewise, zero velocity at one instant means that coordinate is momentarily not changing; it does not imply that the other coordinates are stationary.
The three coordinates must be combined at the end because the particle has one spatial position at each time. Keeping the component equations distinct during calculation helps prevent one coordinate’s initial data or integration constants from being assigned to another.
A Common Mistake: Stopping after One Integration
Suppose \(y''(t)=6t\), \(y'(0)=1\), and \(y(0)=-2\). A common error is to integrate the acceleration once, obtain \(y'(t)=3t^2+1\), and report this as the position. That expression is the velocity, because it is the first antiderivative of acceleration. Integrate once more to obtain \(y(t)=t^3+t+C\); then \(y(0)=-2\) gives \(C=-2\). The correct position is \(y(t)=t^3+t-2\). Differentiating it gives \(y'(t)=3t^2+1\) and \(y''(t)=6t\), confirming the distinction.
Check Your Understanding
Answer each question by considering the coordinate equations and their initial data.
- If \(x''(t)=0\), what can be concluded about \(x'(t)\), and what additional information determines \(x(t)\)?
- Why are both an initial coordinate and an initial velocity needed to recover a coordinate from acceleration?
- When initial data are specified at \(t=3\), what must be substituted into each coordinate solution?
- If \(z'(t)=0\) for all \(t\), does that require \(x'(t)\) and \(y'(t)\) to be zero? Explain.
- What two checks should be made after reconstructing position from acceleration and initial conditions?