Vector Kinematics

A boat crossing a lake, a drone climbing over a field, an aircraft on its approach: each one has a position that changes with time. If you write that position as a vector, the vector equation of a line becomes a model of the motion, and the parameter becomes time. This page shows how to read off position, velocity and speed, how to tell whether two moving objects collide, and how to find when they are closest together.

This page writes vectors as columns, as IB does. The Ontario vector pages on this site write the same vectors in square brackets, so $\begin{pmatrix} 3 \\ -2 \\ 6 \end{pmatrix}$ there is $[3, -2, 6]$.

Key ideas

Position at time t

An object that starts at position $\vec{r}_0$ and moves with constant velocity $\vec{v}$ is at

$$\vec{r} = \vec{r}_0 + t\vec{v}$$

after $t$ units of time. This is the line $\vec{r} = \vec{a} + \lambda\vec{b}$ with a meaning attached to each part:

In the line	In the motion	Meaning
$\vec{a}$	$\vec{r}_0$	position when $t = 0$
$\vec{b}$	$\vec{v}$	velocity: the displacement in one unit of time
$\lambda$	$t$	time since $t = 0$
$\lvert\vec{b}\rvert$	$\lvert\vec{v}\rvert$	speed

The path is the whole line; the position equation also tells you when the object is at each point. Units come from the question: if positions are in kilometres and $t$ is in hours, the velocity is in km/h.

The same equation works in two or three dimensions; in 3-D the third component is often a height.

Speed and distance

Speed is the magnitude of the velocity:

$$\text{speed} = \lvert\vec{v}\rvert = \sqrt{v_1^2 + v_2^2 + v_3^2}$$

With constant velocity, the distance travelled in time $t$ is $\text{speed} \times t$.

Starting at a different time

If an object only sets off at time $t = a$, replace $t$ by $t - a$:

$$\vec{r} = \vec{r}_0 + (t - a)\vec{v}, \qquad t \ge a$$

Now $\vec{r}_0$ is its position at $t = a$. Use the same clock for every object in a question.

Do two objects meet?

Two things can happen when the paths of objects $A$ and $B$ cross:

• They collide if they are at the same point at the same time: $\vec{r}_A(t) = \vec{r}_B(t)$ has a solution with one value of $t$ that works in every component.
• Their paths cross but they miss if they reach the crossing point at different times. To find the crossing point itself, give the two lines different parameters ($s$ and $t$) and solve, exactly as when you intersect two lines.

Relative position and closest approach

The position of $B$ relative to $A$ is

$$\overrightarrow{AB} = \vec{r}_B(t) - \vec{r}_A(t)$$

and the distance between them is its magnitude, $d(t) = \lvert\vec{r}_B(t) - \vec{r}_A(t)\rvert$. Because the velocities are constant, $d(t)^2$ is a quadratic in $t$, so it has a minimum. Three ways to find it:

1. Quadratic: expand $d(t)^2$ and find its vertex (or differentiate and set the derivative to $0$). Minimizing $d^2$ is easier than minimizing $d$, and it happens at the same $t$.
2. Dot product: at the closest moment, the relative position is perpendicular to the relative velocity $\vec{v}_B - \vec{v}_A$, so solve $\big(\vec{r}_B(t) - \vec{r}_A(t)\big) \cdot (\vec{v}_B - \vec{v}_A) = 0$. (See the dot product.)
3. Technology: graph $d(t)$ on your GDC and use its minimum feature.

Always check that the time makes sense (for example $t \ge 0$, or after both objects have set off).

Variable velocity in 2-D (AI HL)

In AI HL, the velocity can change. If the position is $\vec{r}(t) = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}$, differentiate each component:

$$\vec{v}(t) = \frac{d\vec{r}}{dt} = \begin{pmatrix} x'(t) \\ y'(t) \end{pmatrix}, \qquad \vec{a}(t) = \frac{d\vec{v}}{dt} = \begin{pmatrix} x''(t) \\ y''(t) \end{pmatrix}$$

Speed is still $\lvert\vec{v}(t)\rvert$, but now it changes with time. Going the other way, integrate each component of $\vec{v}(t)$ and use the starting position to find the constants. Two special cases come up often:

• Projectile motion: constant horizontal velocity and a vertical acceleration of $-9.8$ m s⁻², so $\vec{a} = \begin{pmatrix} 0 \\ -9.8 \end{pmatrix}$.
• Circular motion: $\vec{r}(t) = \begin{pmatrix} R\cos(\omega t) \\ R\sin(\omega t) \end{pmatrix}$ (radians), with constant speed $R\omega$ and acceleration pointing to the centre.

The AP-style page on motion in the plane with vectors has more practice with these ideas.

Worked examples

Example 1: A climbing drone

A drone takes off from the point $(2, 1, 0)$ and flies with constant velocity $\vec{v} = \begin{pmatrix} 3 \\ -2 \\ 6 \end{pmatrix}$ m s⁻¹. Distances are in metres and $z$ is the height above the ground.

• (a) Write the drone’s position $t$ seconds after take-off.
• (b) Find its speed.
• (c) Find its position after $4$ seconds.
• (d) When does it reach a height of $30$ m, and where is it then?

Solution.

(a) Start at $\vec{r}_0$ and add $t$ copies of the velocity:

$$\vec{r} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} + t\begin{pmatrix} 3 \\ -2 \\ 6 \end{pmatrix}$$

(b) $\lvert\vec{v}\rvert = \sqrt{3^2 + (-2)^2 + 6^2} = \sqrt{49} = 7$ m s⁻¹.

(c) Substitute $t = 4$:

$$\vec{r} = \begin{pmatrix} 2 + 12 \\ 1 - 8 \\ 0 + 24 \end{pmatrix} = \begin{pmatrix} 14 \\ -7 \\ 24 \end{pmatrix}$$

so the drone is at $(14, -7, 24)$.

(d) The height is the $z$-component, $6t$. Set $6t = 30$, so $t = 5$ s. Then

$$\vec{r} = \begin{pmatrix} 2 + 15 \\ 1 - 10 \\ 30 \end{pmatrix} = \begin{pmatrix} 17 \\ -9 \\ 30 \end{pmatrix}$$

Check: in $5$ s at $7$ m s⁻¹ the drone flies $35$ m, and the distance from $(2, 1, 0)$ to $(17, -9, 30)$ is $\sqrt{15^2 + 10^2 + 30^2} = \sqrt{1225} = 35$. ✓

Example 2: Paths that cross

Two kayaks are on a lake. Positions are in kilometres from a dock at the origin, and $t$ is the time in hours after 09:00:

$$\vec{r}_A = t\begin{pmatrix} 1 \\ 2 \end{pmatrix}, \qquad \vec{r}_B = \begin{pmatrix} 10 \\ 0 \end{pmatrix} + t\begin{pmatrix} -2 \\ 1 \end{pmatrix}$$

• (a) Find where their paths cross.
• (b) Do the kayaks collide?

Solution.

(a) The paths are lines, and each kayak may pass the crossing point at a different time, so use different parameters: $s$ for $A$ and $t$ for $B$.

$$\begin{aligned} s &= 10 - 2t && x\text{-components} \\ 2s &= t && y\text{-components} \end{aligned}$$

Substitute $t = 2s$ into the first equation: $s = 10 - 4s$, so $s = 2$ and $t = 4$. The paths cross at $\vec{r}_A(2) = \begin{pmatrix} 2 \\ 4 \end{pmatrix}$, the point $(2, 4)$. Check with $B$: $\begin{pmatrix} 10 - 8 \\ 4 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}$. ✓

(b) Kayak $A$ is at $(2, 4)$ at $t = 2$ (11:00), but kayak $B$ only gets there at $t = 4$ (13:00). They are never at that point at the same time, so they don’t collide. At 11:00, kayak $B$ is at $\begin{pmatrix} 6 \\ 2 \end{pmatrix}$, which is $\sqrt{4^2 + 2^2} = 2\sqrt{5} \approx 4.47$ km from $A$.

The paths cross at $(2, 4)$, but $A$ is there at $t = 2$ and $B$ at $t = 4$. They are closest at $t = 3$ (green), $\sqrt{10}$ km apart.

Example 3: Closest approach

For the kayaks in Example 2, find the time when they are closest together and the shortest distance between them.

Solution. The position of $B$ relative to $A$ is

$$\vec{r}_B - \vec{r}_A = \begin{pmatrix} 10 - 2t \\ t \end{pmatrix} - \begin{pmatrix} t \\ 2t \end{pmatrix} = \begin{pmatrix} 10 - 3t \\ -t \end{pmatrix}$$

Method 1: minimize the squared distance.

$$\begin{aligned} d^2 &= (10 - 3t)^2 + (-t)^2 \\ &= 100 - 60t + 9t^2 + t^2 \\ &= 10t^2 - 60t + 100 \\ &= 10(t - 3)^2 + 10 && \text{completing the square} \end{aligned}$$

The smallest value of $d^2$ is $10$, when $t = 3$. (Or: $\dfrac{d}{dt}(d^2) = 20t - 60 = 0$ gives $t = 3$.)

Method 2: the dot product. The relative velocity is $\vec{v}_B - \vec{v}_A = \begin{pmatrix} -3 \\ -1 \end{pmatrix}$. At the closest moment it is perpendicular to the relative position:

$$\begin{pmatrix} 10 - 3t \\ -t \end{pmatrix} \cdot \begin{pmatrix} -3 \\ -1 \end{pmatrix} = -30 + 9t + t = 10t - 30 = 0 \quad\Rightarrow\quad t = 3$$

Either way, the kayaks are closest at $t = 3$ (12:00). Then $A$ is at $(3, 6)$ and $B$ is at $(4, 3)$, so the shortest distance is

$$d = \sqrt{(4 - 3)^2 + (3 - 6)^2} = \sqrt{10} \approx 3.16 \text{ km (3 s.f.)}$$

Check: at $t = 0$ they are $10$ km apart and at $t = 2$ they are $4.47$ km apart, so $3.16$ km at $t = 3$ fits the pattern of getting closer, then moving apart.

Example 4: A projectile (AI HL)

A ball is thrown from a point $1.2$ m above the ground. Its position, in metres, $t$ seconds later is

$$\vec{r}(t) = \begin{pmatrix} 8t \\ 1.2 + 10t - 4.9t^2 \end{pmatrix}$$

where the first component is the horizontal distance and the second is the height.

• (a) Find the velocity and acceleration vectors.
• (b) Find the speed at which the ball is thrown.
• (c) Find the greatest height of the ball.
• (d) Find how far the ball travels horizontally before it hits the ground.

Solution.

(a) Differentiate each component:

$$\vec{v}(t) = \begin{pmatrix} 8 \\ 10 - 9.8t \end{pmatrix}, \qquad \vec{a}(t) = \begin{pmatrix} 0 \\ -9.8 \end{pmatrix}$$

The acceleration is constant and straight down: that’s gravity.

(b) At $t = 0$, $\vec{v} = \begin{pmatrix} 8 \\ 10 \end{pmatrix}$, so the speed is $\sqrt{8^2 + 10^2} = \sqrt{164} \approx 12.8$ m s⁻¹.

(c) At the top, the ball is moving horizontally, so the vertical velocity is $0$: $10 - 9.8t = 0$ gives $t = \dfrac{10}{9.8} \approx 1.0204$ s. The height then is

$$1.2 + 10(1.0204) - 4.9(1.0204)^2 \approx 6.30 \text{ m (3 s.f.)}$$

(d) The ball lands when the height is $0$: $1.2 + 10t - 4.9t^2 = 0$. Your GDC’s solver (or the quadratic formula) gives $t \approx 2.1545$ s; the other root is negative and is rejected. The horizontal distance is

$$8(2.1545) \approx 17.2 \text{ m (3 s.f.)}$$

Common mistakes

Using the same parameter for both paths when looking for a crossing point. Writing $\vec{r}_A(t) = \vec{r}_B(t)$ only finds a collision. If there’s no solution, the paths may still cross, just at different times. To find where the paths cross, give each line its own parameter.

Using different times when testing for a collision. The reverse mistake: if you solve with $s$ and $t$ and find that the paths cross, that does not mean the objects collide. They collide only if $s = t$.

Treating the velocity vector as the speed. Speed is a number, the magnitude $\lvert\vec{v}\rvert$. A velocity of $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ km/h means a speed of $5$ km/h, not ”$3$ and $4$”.

Using a point on the path as the starting position. In $\vec{r} = \vec{r}_0 + t\vec{v}$, the vector $\vec{r}_0$ must be the position at $t = 0$. If you are told where a ship is at 14:00 and $t$ counts hours from noon, either find its noon position first or write the equation with $t - 2$.

Minimizing the wrong quantity. The closest approach comes from the distance between the objects, $\lvert\vec{r}_B - \vec{r}_A\rvert$, not from either object’s distance from the origin.

Forgetting to finish the question. Finding $t$ at the closest approach is only halfway. Substitute it back to get the distance (and the positions, if asked), and give the time in the form the question wants, such as a clock time.

Practice

1. (Warm-up) A boat’s position in kilometres, $t$ hours after it leaves port, is $\vec{r} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} + t\begin{pmatrix} -1 \\ 4 \end{pmatrix}$.

• (a) Where is the port?
• (b) Find the boat’s position after $3$ hours.
• (c) Find its speed.

Solution

(a) At $t = 0$ the boat is at $(3, -2)$, so that’s the port.

(b) $\vec{r} = \begin{pmatrix} 3 - 3 \\ -2 + 12 \end{pmatrix} = \begin{pmatrix} 0 \\ 10 \end{pmatrix}$, the point $(0, 10)$.

(c) $\sqrt{(-1)^2 + 4^2} = \sqrt{17} \approx 4.12$ km/h (3 s.f.).

2. (Warm-up) A particle starts at $(1, 0, 2)$ and moves with constant velocity $\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ m s⁻¹. Find its position after $5$ seconds, its speed, and the distance it travels in those $5$ seconds.

Solution

$$\vec{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + 5\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 11 \\ -5 \\ 12 \end{pmatrix}$$

Speed: $\sqrt{4 + 1 + 4} = 3$ m s⁻¹. Distance in $5$ s: $3 \times 5 = 15$ m.

Check: $\lvert(11, -5, 12) - (1, 0, 2)\rvert = \sqrt{100 + 25 + 100} = 15$. ✓

3. (Core) A ship moves with constant velocity. At noon it is at $(-4, 6)$, and at 14:00 it is at $(8, 1)$, with distances in kilometres. Let $t$ be the time in hours after noon.

• (a) Find the ship’s velocity vector and its speed.
• (b) Write an equation for its position at time $t$.
• (c) Where is the ship at 15:00?

Solution

(a) In $2$ hours the displacement is $\begin{pmatrix} 8 - (-4) \\ 1 - 6 \end{pmatrix} = \begin{pmatrix} 12 \\ -5 \end{pmatrix}$, so the velocity is

$$\vec{v} = \frac{1}{2}\begin{pmatrix} 12 \\ -5 \end{pmatrix} = \begin{pmatrix} 6 \\ -2.5 \end{pmatrix} \text{ km/h}$$

Speed: $\sqrt{6^2 + 2.5^2} = \sqrt{42.25} = 6.5$ km/h.

(b) At $t = 0$ (noon) the ship is at $(-4, 6)$:

$$\vec{r} = \begin{pmatrix} -4 \\ 6 \end{pmatrix} + t\begin{pmatrix} 6 \\ -2.5 \end{pmatrix}$$

(c) At $t = 3$: $\vec{r} = \begin{pmatrix} -4 + 18 \\ 6 - 7.5 \end{pmatrix} = \begin{pmatrix} 14 \\ -1.5 \end{pmatrix}$, the point $(14, -1.5)$.

4. (Core) Two aircraft are tracked by radar, with positions in kilometres and $t$ in minutes:

$$\vec{r}_P = \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} + t\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}, \qquad \vec{r}_Q = \begin{pmatrix} 7 \\ 0 \\ 10 \end{pmatrix} + t\begin{pmatrix} 0 \\ 1 \\ -3 \end{pmatrix}$$

(The $z$-coordinate is measured from a reference level, so it can be negative.) Show that their paths cross, and decide whether the aircraft collide.

Solution

Give the paths different parameters, $s$ for $P$ and $u$ for $Q$:

$$\begin{aligned} 1 + 2s &= 7 && \Rightarrow\ s = 3 \\ 4 - s &= u && \Rightarrow\ u = 1 \\ -2 + 3s &= 10 - 3u && \text{check: } 7 = 7 \end{aligned}$$

All three equations agree, so the paths cross at $\vec{r}_P(3) = \begin{pmatrix} 7 \\ 1 \\ 7 \end{pmatrix}$, the point $(7, 1, 7)$.

But $P$ is there at $t = 3$ and $Q$ is there at $t = 1$. They reach the point at different times, so they don’t collide. (At $t = 3$, $Q$ is at $(7, 3, 1)$, a distance $\sqrt{0 + 4 + 36} = 2\sqrt{10} \approx 6.32$ km from $P$.)

5. (Core) Cyclist $A$ leaves a junction at the origin at $t = 0$ with velocity $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ m s⁻¹. Cyclist $B$ leaves the same junction one second later, at $t = 1$, with velocity $\begin{pmatrix} 6 \\ 8 \end{pmatrix}$ m s⁻¹.

• (a) Write the position of each cyclist at time $t$ (for $t \ge 1$).
• (b) When and where does $B$ catch up with $A$?

Solution

(a) $\vec{r}_A = t\begin{pmatrix} 3 \\ 4 \end{pmatrix}$. $B$ starts at $t = 1$, so replace $t$ by $t - 1$: $\vec{r}_B = (t - 1)\begin{pmatrix} 6 \\ 8 \end{pmatrix}$.

(b) Set them equal. The $x$-components give $3t = 6(t - 1)$, so $3t = 6$ and $t = 2$. The $y$-components agree: $4(2) = 8$ and $8(2 - 1) = 8$. ✓

$B$ catches $A$ at $t = 2$ s, at the point $(6, 8)$, which is $10$ m from the junction.

6. (Core) Two ships have positions, in kilometres, $t$ hours after midnight:

$$\vec{r}_A = t\begin{pmatrix} 4 \\ 3 \end{pmatrix}, \qquad \vec{r}_B = \begin{pmatrix} 14 \\ 0 \end{pmatrix} + t\begin{pmatrix} 0 \\ 5 \end{pmatrix}$$

• (a) Show that the distance between them is given by $d^2 = 20t^2 - 112t + 196$.
• (b) Find the time when they are closest, and the shortest distance.
• (c) Their radar can detect a ship within $5$ km. Does either ship ever detect the other?

Solution

(a) $\vec{r}_B - \vec{r}_A = \begin{pmatrix} 14 - 4t \\ 5t - 3t \end{pmatrix} = \begin{pmatrix} 14 - 4t \\ 2t \end{pmatrix}$, so

$$d^2 = (14 - 4t)^2 + (2t)^2 = 196 - 112t + 16t^2 + 4t^2 = 20t^2 - 112t + 196$$

(b) The vertex of the quadratic is at $t = \dfrac{112}{2(20)} = 2.8$ hours, that is, 02:48. Then

$$d^2 = 20(2.8)^2 - 112(2.8) + 196 = 156.8 - 313.6 + 196 = 39.2$$

so $d = \sqrt{39.2} \approx 6.26$ km (3 s.f.).

(c) The shortest distance is $6.26$ km, which is more than $5$ km, so neither ship ever detects the other.

7. (Core, AI HL) A particle moves in the plane with position $\vec{r}(t) = \begin{pmatrix} t^2 - 4t \\ 3t - t^2 \end{pmatrix}$ metres, for $t \ge 0$ seconds.

• (a) Find $\vec{v}(t)$ and $\vec{a}(t)$.
• (b) Find the speed at $t = 1$.
• (c) At what time is the particle moving parallel to the $x$-axis? Is it ever at rest?

Solution

(a) $\vec{v}(t) = \begin{pmatrix} 2t - 4 \\ 3 - 2t \end{pmatrix}$ and $\vec{a}(t) = \begin{pmatrix} 2 \\ -2 \end{pmatrix}$ (constant).

(b) $\vec{v}(1) = \begin{pmatrix} -2 \\ 1 \end{pmatrix}$, so the speed is $\sqrt{4 + 1} = \sqrt{5} \approx 2.24$ m s⁻¹.

(c) Moving parallel to the $x$-axis means the $y$-component of velocity is $0$ (and the $x$-component isn’t): $3 - 2t = 0$, so $t = 1.5$ s. Then $\vec{v} = \begin{pmatrix} -1 \\ 0 \end{pmatrix}$. ✓

At rest needs both components to be $0$ at once: $2t - 4 = 0$ gives $t = 2$, but $3 - 2t = 0$ gives $t = 1.5$. There’s no common time, so the particle is never at rest.

8. (Challenge) Two drones fly with constant velocities. Positions are in metres and $t$ is in seconds:

$$\vec{r}_A = t\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \qquad \vec{r}_B = \begin{pmatrix} 6 \\ 0 \\ 3 \end{pmatrix} + t\begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}$$

Use the dot product to find when the drones are closest, and find the shortest distance between them.

Solution

Relative position and relative velocity:

$$\vec{r}_B - \vec{r}_A = \begin{pmatrix} 6 - t \\ t \\ 3 \end{pmatrix}, \qquad \vec{v}_B - \vec{v}_A = \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}$$

At the closest moment these are perpendicular:

$$-(6 - t) + t + 0 = 2t - 6 = 0 \quad\Rightarrow\quad t = 3 \text{ s}$$

Then $\vec{r}_B - \vec{r}_A = \begin{pmatrix} 3 \\ 3 \\ 3 \end{pmatrix}$, so the shortest distance is $\sqrt{27} = 3\sqrt{3} \approx 5.20$ m (3 s.f.). (Drone $A$ is at $(3, 3, 3)$ and drone $B$ at $(6, 6, 6)$.)

Check with the quadratic: $d^2 = (6 - t)^2 + t^2 + 9 = 2t^2 - 12t + 45$, whose vertex is at $t = \dfrac{12}{4} = 3$. ✓

9. (Challenge, AI HL) A seat on a fairground ride moves in a horizontal circle. Its position, in metres, after $t$ seconds is

$$\vec{r}(t) = \begin{pmatrix} 5\cos(0.4t) \\ 5\sin(0.4t) \end{pmatrix}$$

with the angle in radians.

• (a) Find $\vec{v}(t)$ and show that the speed is constant.
• (b) Show that $\vec{a}(t) = -0.16\,\vec{r}(t)$, and explain what this says about the direction of the acceleration.
• (c) How long does one revolution take?

Solution

(a) Differentiate each component (chain rule):

$$\vec{v}(t) = \begin{pmatrix} -2\sin(0.4t) \\ 2\cos(0.4t) \end{pmatrix}$$$$\lvert\vec{v}\rvert = \sqrt{4\sin^2(0.4t) + 4\cos^2(0.4t)} = \sqrt{4} = 2 \text{ m s}^{-1}$$

for every $t$, so the speed is constant.

(b) Differentiate again:

$$\vec{a}(t) = \begin{pmatrix} -0.8\cos(0.4t) \\ -0.8\sin(0.4t) \end{pmatrix} = -0.16\begin{pmatrix} 5\cos(0.4t) \\ 5\sin(0.4t) \end{pmatrix} = -0.16\,\vec{r}(t)$$

The acceleration is a negative multiple of the position vector, so it always points from the seat back towards the centre of the circle. Its magnitude is $0.16 \times 5 = 0.8$ m s⁻².

(c) One revolution is when $0.4t$ increases by $2\pi$: $t = \dfrac{2\pi}{0.4} = 5\pi \approx 15.7$ s. (Check: the circumference is $10\pi$ m and the speed is $2$ m s⁻¹, giving $5\pi$ s. ✓)