Putting Vector Motion Data Together
A vector-motion problem may give acceleration together with initial position and velocity, or give a position formula and ask for several quantities at a particular time. The calculations are connected: position changes according to velocity, and velocity changes according to acceleration.
A reliable solution keeps the vector components together, uses the given initial data to determine any constants, and checks the final quantities against their definitions. This matters because a correct component calculation can still be misinterpreted: velocity and displacement are vectors, while speed is a nonnegative scalar.
For planar motion, write \(\mathbf{r}(t)=\langle x(t),y(t)\rangle\). Then \(\mathbf{v}(t)=\langle x'(t),y'(t)\rangle\), \(\mathbf{a}(t)=\langle x''(t),y''(t)\rangle\), and speed is \(\sqrt{(x'(t))^2+(y'(t))^2}\). In three dimensions, include the third component inside the square root as well.
A Plan for Multi-Step Problems
Start by identifying the given quantity and the requested quantity. If acceleration is given, integrate its components to obtain velocity, then use the initial velocity. Integrate velocity to obtain position, then use the initial position. If position is already given as component formulas, differentiate instead. At the requested time, calculate speed from velocity; over a time interval, calculate displacement from endpoint positions or integrate velocity.
Record the initial time and the given position, velocity, acceleration, or component formulas. Keep the units attached to each quantity.
Differentiate known position components or integrate known acceleration and velocity components, applying initial conditions to determine constants.
Substitute the requested time, calculate speed as a magnitude, and describe displacement as the change in position.
Differentiate to verify derivative formulas, and compare interval displacement with the difference between the endpoint positions.
Why Displacement Is the Integral of Velocity
The relationship between velocity and displacement explains why integration is useful in these problems. The argument is componentwise, so it applies to motion in any finite number of coordinate directions.
To justify the result, write the position vector as \(\mathbf{r}(t)=\langle x_1(t),\ldots,x_n(t)\rangle\). Its velocity is \(\mathbf{v}(t)=\langle x_1'(t),\ldots,x_n'(t)\rangle\). By the Fundamental Theorem of Calculus, for each component \(i\),
Integrating each component therefore gives \(\langle x_1(b)-x_1(a),\ldots,x_n(b)-x_n(a)\rangle\), which is exactly \(\mathbf{r}(b)-\mathbf{r}(a)\). This proves the vector equation. In a calculation, it provides two equivalent ways to find displacement: subtract endpoint positions or integrate velocity over the interval.
Worked Example: Recovering Motion from Acceleration Data
Worked Example: Recovering Motion from Acceleration Data
A particle moves in the plane, with position measured in meters and time in seconds. Its acceleration is \(\mathbf{a}(t)=\langle 6t,4\rangle\ \text{m/s}^2\). At \(t=0\), its velocity is \(\mathbf{v}(0)=\langle 1,-2\rangle\ \text{m/s}\), and its position is \(\mathbf{r}(0)=\langle 3,5\rangle\ \text{m}\). Find its position, velocity, acceleration, and speed at \(t=2\), and its displacement from \(t=0\) to \(t=2\).
Integrate the acceleration components to obtain velocity:
Use \(\mathbf{v}(0)=\langle 1,-2\rangle\). Substitution gives \(\langle C_1,C_2\rangle=\langle 1,-2\rangle\), so \(C_1=1\) and \(C_2=-2\). Thus \(\mathbf{v}(t)=\langle 3t^2+1,4t-2\rangle\ \text{m/s}\).
Now integrate velocity to obtain position:
The condition \(\mathbf{r}(0)=\langle 3,5\rangle\) gives \(D_1=3\) and \(D_2=5\). Hence \(\mathbf{r}(t)=\langle t^3+t+3,2t^2-2t+5\rangle\ \text{m}\).
At \(t=2\), the position is \(\mathbf{r}(2)=\langle 8+2+3,8-4+5\rangle=\langle 13,9\rangle\ \text{m}\). The velocity is \(\mathbf{v}(2)=\langle 3(4)+1,8-2\rangle=\langle 13,6\rangle\ \text{m/s}\). The acceleration is \(\mathbf{a}(2)=\langle 6(2),4\rangle=\langle 12,4\rangle\ \text{m/s}^2\). Therefore, the instantaneous speed is \(\|\mathbf{v}(2)\|=\sqrt{13^2+6^2}=\sqrt{205}\ \text{m/s}\).
The displacement over the interval is \(\mathbf{r}(2)-\mathbf{r}(0)=\langle 13-3,9-5\rangle=\langle 10,4\rangle\ \text{m}\). As a check, differentiating the position gives \(\langle 3t^2+1,4t-2\rangle\), the velocity found above, and differentiating that velocity gives \(\langle 6t,4\rangle\), the stated acceleration. The endpoint subtraction confirms the displacement.
Worked Example: Position and Velocity from a Given Velocity
Worked Example: Position and Velocity from a Given Velocity
Suppose \(\mathbf{v}(t)=\langle 2t,3-t\rangle\ \text{m/s}\), and \(\mathbf{r}(1)=\langle 4,-1\rangle\ \text{m}\). Find the position at \(t=3\), the acceleration and speed at \(t=3\), and the displacement from \(t=1\) to \(t=3\).
Integrate the velocity components: \(\mathbf{r}(t)=\langle t^2+C_1,3t-\frac{t^2}{2}+C_2\rangle\). Use the initial position at \(t=1\). The first component gives \(1+C_1=4\), so \(C_1=3\). The second gives \(3-\frac12+C_2=-1\), or \(\frac52+C_2=-1\), so \(C_2=-\frac72\). Therefore,
At \(t=3\), \(\mathbf{r}(3)=\langle 9+3,9-\frac92-\frac72\rangle=\langle 12,1\rangle\ \text{m}\). Differentiating velocity gives \(\mathbf{a}(t)=\langle 2,-1\rangle\ \text{m/s}^2\), so \(\mathbf{a}(3)=\langle 2,-1\rangle\ \text{m/s}^2\). The velocity at that time is \(\mathbf{v}(3)=\langle 6,0\rangle\ \text{m/s}\), and the speed is \(\sqrt{6^2+0^2}=6\ \text{m/s}\).
The displacement is \(\mathbf{r}(3)-\mathbf{r}(1)=\langle 12-4,1-(-1)\rangle=\langle 8,2\rangle\ \text{m}\). Check by integrating the velocity: \(\int_1^3 2t\,dt=[t^2]_1^3=9-1=8\), and \(\int_1^3(3-t)\,dt=[3t-\frac{t^2}{2}]_1^3=(9-\frac92)-(3-\frac12)=\frac92-\frac52=2\). Thus the vector integral is \(\langle 8,2\rangle\ \text{m}\), agreeing with the endpoint subtraction. Differentiating the position also returns the given velocity.
Worked Example: Differentiating Component Formulas
Worked Example: Differentiating Component Formulas
A particle has position \(\mathbf{r}(t)=\langle 4\cos t,4\sin t\rangle\ \text{m}\), with \(t\) in seconds. Find its position, velocity, acceleration, and speed at \(t=\frac{\pi}{2}\), as well as its displacement from \(t=0\) to \(t=\frac{\pi}{2}\).
Differentiate each position component: \(\mathbf{v}(t)=\langle -4\sin t,4\cos t\rangle\ \text{m/s}\). Differentiate again: \(\mathbf{a}(t)=\langle -4\cos t,-4\sin t\rangle\ \text{m/s}^2\).
At \(t=\frac{\pi}{2}\), \(\cos(\frac{\pi}{2})=0\) and \(\sin(\frac{\pi}{2})=1\). Therefore the position is \(\mathbf{r}(\frac{\pi}{2})=\langle 0,4\rangle\ \text{m}\), the velocity is \(\mathbf{v}(\frac{\pi}{2})=\langle -4,0\rangle\ \text{m/s}\), and the acceleration is \(\mathbf{a}(\frac{\pi}{2})=\langle 0,-4\rangle\ \text{m/s}^2\). The speed is \(\sqrt{(-4)^2+0^2}=4\ \text{m/s}\).
At the initial time, \(\mathbf{r}(0)=\langle 4,0\rangle\ \text{m}\). Thus the displacement is \(\langle 0-4,4-0\rangle=\langle -4,4\rangle\ \text{m}\). The check is direct: subtracting the endpoint positions gives the same vector. Also, differentiating \(\mathbf{r}(t)\) once and twice gives precisely the velocity and acceleration stated above.
Interpret the Vectors Before Concluding
Each answer describes a different feature of the motion. Position specifies location relative to the chosen origin. Velocity gives the instantaneous direction of motion and the rate of change of position. Acceleration gives the rate of change of velocity. Speed is the magnitude of velocity, so it describes how quickly the particle moves without specifying direction.
Displacement compares two positions and points from the starting location to the ending location. It is not generally the total distance traveled along the path. For example, a particle can move and return to its starting point; its displacement is then the zero vector even though its speed may have been positive during the trip.
| Quantity | Definition or calculation | Units, if position is in meters and time in seconds |
|---|---|---|
| Position | \(\mathbf{r}(t)\) | m |
| Velocity | \(\mathbf{r}'(t)\) | m/s |
| Acceleration | \(\mathbf{v}'(t)=\mathbf{r}''(t)\) | m/s² |
| Speed | \(\|\mathbf{v}(t)\|\) | m/s |
| Displacement | \(\mathbf{r}(b)-\mathbf{r}(a)\) | m |
Common Pitfall: Treating Speed as a Vector
Suppose the velocity is \(\langle 3,4\rangle\ \text{m/s}\). A common incorrect response is to call the speed \(\langle 3,4\rangle\ \text{m/s}\). That is the velocity vector, not the speed. The correction is to take its magnitude: \(\sqrt{3^2+4^2}=\sqrt{9+16}=5\ \text{m/s}\). The result is a scalar, and it cannot be negative.
Another useful check is dimensional: position and displacement have units of length, while velocity and speed have units of length per time. If integrating acceleration, the first integration must produce velocity units, and the second must produce position units. Initial conditions then determine the constants in each component; they should not be applied to only one coordinate.
Check Your Understanding
For each question, state the relevant vector or scalar and keep its units clear.
- If \(\mathbf{v}(t)=\langle 5,12\rangle\ \text{m/s}\) at an instant, what is the instantaneous speed?
- A particle has \(\mathbf{a}(t)=\langle 2,6t\rangle\). What componentwise integration step gives a general velocity vector before an initial velocity is applied?
- If \(\mathbf{r}(2)=\langle 7,1\rangle\ \text{m}\) and \(\mathbf{r}(5)=\langle 3,4\rangle\ \text{m}\), what is the displacement from \(t=2\) to \(t=5\)?
- Why does integrating velocity from \(a\) to \(b\) give the displacement between those times?
- A student reports a displacement in meters per second. What distinction between motion quantities has likely been missed?
