This note shows how to describe motion with vectors: position, velocity and acceleration as functions of time, found by differentiating and integrating each component. It also covers speed, distance travelled and how to decide whether two moving objects meet or how close they come.
Key ideas
Consider an object moving in two or three dimensions. Its position vector at time \( t \) is \( \mathbf{r}(t) \). Everything else comes from calculus applied to each component separately.
- Velocity is the rate of change of position: \( \mathbf{v} = \dfrac{d\mathbf{r}}{dt} \). It points in the direction of motion.
- Acceleration is the rate of change of velocity: \( \mathbf{a} = \dfrac{d\mathbf{v}}{dt} = \dfrac{d^2\mathbf{r}}{dt^2} \).
- Going backwards. \( \mathbf{v} = \int \mathbf{a}\,dt \) and \( \mathbf{r} = \int \mathbf{v}\,dt \). Each integration of a component adds a constant, so a vector constant \( \mathbf{c} \) is added. Use a known starting value, such as \( \mathbf{r}(0) \), to find it.
- Speed is a number: \( |\mathbf{v}| = \sqrt{v_x^2 + v_y^2} \). Velocity is a vector, so two objects can have the same speed but different velocities.
- Displacement and distance. Displacement over an interval is \( \mathbf{r}(t_2) - \mathbf{r}(t_1) \), and it is a vector. Distance travelled along the path is \( \displaystyle\int_{t_1}^{t_2} |\mathbf{v}|\,dt \), which is always at least as large as the size of the displacement.
- Constant velocity. If \( \mathbf{v} \) does not change, then \( \mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v} \), which is the vector equation of a line with time as the parameter.
- Meeting and closest approach. Two objects are at the same place at the same time only if their position vectors are equal for the same value of \( t \). Equal paths crossing at different times does not mean a collision. To find the closest approach, write the distance between them as a function of \( t \) and find its minimum on your GDC.
- Direction and angle. The direction of motion is given by \( \mathbf{v} \). For example, the object moves parallel to the \( x \)-axis when \( v_y = 0 \), and is momentarily stationary when \( \mathbf{v} = \mathbf{0} \).
Trigonometric functions in kinematics use radians. Set your GDC to radian mode whenever \( \sin t \) or \( \cos t \) appears.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \( \mathbf{v} = \dfrac{d\mathbf{r}}{dt} \) | Velocity from position: differentiate each component. |
| \( \mathbf{a} = \dfrac{d\mathbf{v}}{dt} \) | Acceleration from velocity. |
| \( \mathbf{r} = \int \mathbf{v}\,dt \) | Position from velocity. Include the constant vector from the starting condition. |
| \( \text{speed} = |\mathbf{v}| \) | Magnitude of velocity. Not a vector. |
| \( \text{distance} = \displaystyle\int_{t_1}^{t_2}|\mathbf{v}|\,dt \) | Length of path travelled. Use the definite integral function on your GDC. |
| \( \text{displacement} = \mathbf{r}(t_2)-\mathbf{r}(t_1) \) | Overall change in position. |
| \( \mathbf{r} = \mathbf{r}_0 + t\mathbf{v} \) | Motion with constant velocity. |
| \( d(t) = |\mathbf{r}_A(t) - \mathbf{r}_B(t)| \) | Separation of two moving objects. Minimise to find closest approach. |
Worked examples
Example 1 A skater's position (in metres) at time \( t \) seconds is \( \mathbf{r} = \begin{pmatrix} 2t^2 - 3t \\ 5 + 4t - t^2 \end{pmatrix} \). Find the velocity, speed and acceleration when \( t = 2 \), and say what the velocity tells you about the direction of motion.
Differentiate each component: \( \mathbf{v} = \begin{pmatrix} 4t - 3 \\ 4 - 2t \end{pmatrix} \).
At \( t = 2 \): \( \mathbf{v} = \begin{pmatrix} 5 \\ 0 \end{pmatrix} \). Speed \( = \sqrt{25 + 0} = 5 \).
Differentiate again: \( \mathbf{a} = \begin{pmatrix} 4 \\ -2 \end{pmatrix} \). This is constant, so it is the same at \( t = 2 \).
Since \( v_y = 0 \), the skater is moving parallel to the \( x \)-axis at that instant.
Answer: \( \mathbf{v} = (5, 0)\ \text{m s}^{-1} \), speed \( 5\ \text{m s}^{-1} \), \( \mathbf{a} = (4, -2)\ \text{m s}^{-2} \), moving in the positive \( x \) direction.
Example 2 A model rocket has velocity \( \mathbf{v} = \begin{pmatrix} 6t \\ 4 - 2t \end{pmatrix} \) m s-1 and starts at \( (1, 3) \) m. Find (a) its position at \( t = 3 \), (b) the distance it travels in the first 3 seconds.
(a) Integrate: \( \mathbf{r} = \begin{pmatrix} 3t^2 \\ 4t - t^2 \end{pmatrix} + \mathbf{c} \). At \( t = 0 \), \( \mathbf{r} = (1, 3) \), so \( \mathbf{c} = (1, 3) \).
\( \mathbf{r} = \begin{pmatrix} 3t^2 + 1 \\ 4t - t^2 + 3 \end{pmatrix} \). At \( t = 3 \): \( \mathbf{r} = (27 + 1,\ 12 - 9 + 3) = (28, 6) \).
(b) Speed \( = \sqrt{(6t)^2 + (4 - 2t)^2} \). Distance travelled is \( \displaystyle\int_0^3 \sqrt{36t^2 + (4-2t)^2}\,dt \).
Evaluate with the definite integral function on your GDC: \( \approx 28.7 \) m.
Compare with the displacement: \( (28,6) - (1,3) = (27, 3) \), of magnitude \( \sqrt{738} \approx 27.2 \) m. The distance is larger because the path is curved.
Answer: (a) \( (28, 6) \) m; (b) \( 28.7 \) m (3 s.f.).
Example 3 Two drones fly at constant velocity. Drone A starts at \( (2, 9) \) with velocity \( (3, -2) \). Drone B starts at \( (14, 1) \) with velocity \( (-1, 1) \). Distances are in metres, times in seconds. Do they collide? Find their closest approach.
Positions: \( \mathbf{r}_A = (2 + 3t,\ 9 - 2t) \), \( \mathbf{r}_B = (14 - t,\ 1 + t) \).
Collision needs equal \( x \) and equal \( y \) at the same time. \( 2 + 3t = 14 - t \) gives \( t = 3 \). \( 9 - 2t = 1 + t \) gives \( t = \tfrac83 \). The times differ, so the drones never occupy the same point at the same moment: no collision.
Separation: \( \mathbf{r}_A - \mathbf{r}_B = (-12 + 4t,\ 8 - 3t) \), so \( d(t) = \sqrt{(4t - 12)^2 + (8 - 3t)^2} \).
Plot \( d(t) \) on your GDC and use the minimum function. The minimum occurs at \( t = 2.88 \), where \( d = 0.800 \).
Check: at \( t = 2.88 \), the separation vector is \( (-0.48, -0.64) \), whose magnitude is \( \sqrt{0.2304 + 0.4096} = 0.8 \).
Answer: no collision; closest approach is \( 0.800 \) m at \( t = 2.88 \) s.
Common errors
- Common error: forgetting the constant of integration. Without the starting position or velocity, the position function is wrong by a constant vector.
- Common error: calling \( \mathbf{v} \) "the speed". Speed is the magnitude \( |\mathbf{v}| \); give a single number with units.
- Common error: integrating the velocity vector (or its components) when asked for distance travelled. You must integrate the speed \( |\mathbf{v}| \).
- Common error: concluding two objects collide because their paths cross. They must be at the same point at the same time.
- Common error: leaving the GDC in degree mode for expressions containing \( \sin t \) and \( \cos t \).
Exam tip: Write which quantity you are finding (velocity, speed, displacement or distance) before you calculate. State units and give the integral you are evaluating, so that the method is clear even if you only quote the GDC value.
Check yourself
- A particle has position \( \mathbf{r} = (4\cos t,\ 4\sin t) \). Find its velocity and show that its speed is constant.
Answer
\( \mathbf{v} = (-4\sin t,\ 4\cos t) \). Speed \( = \sqrt{16\sin^2 t + 16\cos^2 t} = \sqrt{16} = 4 \), which does not depend on \( t \). - A cyclist has \( \mathbf{v} = (2t, 3) \) and starts at \( (1, -2) \). Find the position at \( t = 4 \).
Answer
\( \mathbf{r} = (t^2, 3t) + \mathbf{c} \) with \( \mathbf{c} = (1, -2) \). At \( t = 4 \): \( (16 + 1,\ 12 - 2) = (17, 10) \). - A particle has \( \mathbf{r} = (t^3,\ 2t^2) \). Find its velocity, speed and acceleration at \( t = 2 \).
Answer
\( \mathbf{v} = (3t^2, 4t) = (12, 8) \). Speed \( = \sqrt{144 + 64} = \sqrt{208} \approx 14.4 \). \( \mathbf{a} = (6t, 4) = (12, 4) \). - A buoy starts at \( (1, 4) \) and drifts with constant velocity \( (2, -1) \). Does it pass through \( (7, 1) \)?
Answer
\( \mathbf{r} = (1 + 2t,\ 4 - t) \). Setting \( 1 + 2t = 7 \) gives \( t = 3 \); then \( 4 - t = 1 \). Both coordinates agree at \( t = 3 \), so yes.