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AHL 4.11

Constant velocity: speed, position and closest approach

AHL 4.11

Constant velocity: speed, position and closest approach

Target: I can find the speed from a velocity vector, and the direction of motion as a bearing. (+4 more)

By the end of this section

Key points

Velocity and speed

  • An object with constant velocity \(\mathbf{v}\) moves parallel to \(\mathbf{v}\) with speed \(|\mathbf{v}|\). Velocity is a vector; speed is a number.
  • To find \(\mathbf{v}\) from a direction \(\mathbf{d}\) and a speed \(s\): \(\mathbf{v} = s \times \dfrac{\mathbf{d}}{|\mathbf{d}|}\). Check that \(|\mathbf{v}| = s\).

Bearings

  • A bearing is the angle clockwise from north, written with three figures.
  • Sketch the east and north components first. The quadrant tells you what to add: north-east is \(\theta\), south-east is \(90^\circ + \theta\), south-west is \(180^\circ + \theta\), north-west is \(360^\circ - \theta\), where \(\theta\) is the acute angle in your sketch.

Position at time t

  • For constant velocity \(\mathbf{v}\) from the starting position \(\mathbf{r}_0\): \(\mathbf{r} = \mathbf{r}_0 + t\mathbf{v}\).
  • This is the equation of a line with direction \(\mathbf{v}\), and \(t\) plays the part of \(\lambda\).
  • The distance from the origin is \(|\mathbf{r}|\).

Collisions and closest approach

  • Two objects collide only if all three coordinates are equal at the same time \(t\). Solve each equation for \(t\): if the values differ, they never collide.
  • Their paths can cross without a collision, because they pass the crossing point at different times.
  • The distance between them is \(|\mathbf{r}_B - \mathbf{r}_A|\). Use the calculator to find the minimum of this distance against \(t\).
  • Alternatively, \(d^2\) is a quadratic in \(t\) and its minimum is at the vertex.

On the calculator

Minimum of a graph

  • Enter the distance \(d(t)\) as a function of \(t\).
  • Graph it, with a window that shows the lowest point.
  • Use the calculator's minimum tool and read \(t\) and \(d\).

Screenshot and video to come

Checking with a table

  • Make a table of \(d(t)\) for values of \(t\).
  • Look for the smallest value of \(d\) and the \(t\) next to it, to check the graph.

Screenshot and video to come

Explore: two ships and their closest approach

Each ship has r = r₀ + tv. Drag a ship to change where it starts. Drag the orange tip to change its velocity (the arrow is one hour of travel, in km). Move the time and watch the distance. Predict the closest approach, then check.

Distance apart at this time

Click a ship or an orange tip, then use the arrow keys to move it one unit. Start positions are in km east and north. Velocities are in km per hour.

Move the time to see where the ships are.

  1. Where do the two paths cross? Are both ships at that point at the same time? What would have to be true for a collision?
  2. Find the time when you think the ships are closest. Check it. Is it the time when either ship passes the crossing point?
  3. Give the two ships the same velocity. What happens to the distance between them, and why?
  4. Change one velocity so the ships collide. What do you have to check, and what is the least distance then?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Bearing and speed from a velocity

An object moves with velocity \(\mathbf{v} = \begin{pmatrix} 5 \\ -2 \end{pmatrix}\) m s\(^{-1}\), where \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\) points east and \(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\) points north. Find (a) the direction of motion as a bearing, (b) the speed.

5 east2 southvθOPV
Working
  1. Draw it

    East 5, south 2

    Always sketch the components. The vector points right and down.

  2. The angle

    \(\theta = \tan^{-1}\frac{2}{5} = 21.8^\circ\) (below east)

    θ is measured below the east line.

  3. The bearing

    \(\text{bearing} = 90^\circ + 21.8^\circ = 112^\circ\)

    North is 0° and east is 90°. The vector is past east, so add θ to 90°.

  4. Speed

    \(|\mathbf{v}| = \sqrt{25 + 4} = 5.39\) m s\(^{-1}\)

    Speed is the magnitude of the velocity.

Video walkthrough coming
Velocity from a direction and a speed

Find the velocity vector of an object moving in the direction \(\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\) with speed 12 m s\(^{-1}\).

Working
  1. Magnitude

    \(\left|\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\right| = \sqrt{4 + 1 + 9} = \sqrt{14}\)

    This is the length of the direction vector.

  2. Unit vector

    \(\frac{1}{\sqrt{14}}\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\)

    Dividing by the length gives a vector of length 1 in the same direction.

  3. Times the speed

    \(\mathbf{v} = \frac{12}{\sqrt{14}}\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 6.41 \\ -3.21 \\ 9.62 \end{pmatrix}\)

    Multiply by 12 so that the speed is 12.

  4. Check

    \(|\mathbf{v}| = \frac{12}{\sqrt{14}} \times \sqrt{14} = 12\)

    The magnitude must equal the speed.

Video walkthrough coming
Position at time t, and distance from the origin

An object moves with constant velocity \(\mathbf{v} = 3\mathbf{i} + 2\mathbf{j} - 6\mathbf{k}\) m s\(^{-1}\). At \(t = 0\) its position vector is \(-2\mathbf{i} + 5\mathbf{j} + 4\mathbf{k}\) m. (a) Write down its position vector at time \(t\) seconds. (b) Find its distance from the origin when \(t = 4\).

Working
  1. Position at t

    \(\mathbf{r} = (-2\mathbf{i} + 5\mathbf{j} + 4\mathbf{k}) + t(3\mathbf{i} + 2\mathbf{j} - 6\mathbf{k})\)

    \(\mathbf{r} = \mathbf{r}_0 + t\mathbf{v}\): start plus time times velocity.

  2. At t = 4

    \(\mathbf{r} = -2\mathbf{i} + 5\mathbf{j} + 4\mathbf{k} + 4(3\mathbf{i} + 2\mathbf{j} - 6\mathbf{k}) = 10\mathbf{i} + 13\mathbf{j} - 20\mathbf{k}\)

    Work out each component.

  3. Distance

    \(|\mathbf{r}| = \sqrt{100 + 169 + 400} = 25.9\) m

    The distance from the origin is the magnitude of the position vector.

Video walkthrough coming
Two objects that never collide

Two drones have position vectors (in km) at time \(t\) hours: \(\mathbf{r}_A = \begin{pmatrix} -3 \\ 2 \\ 4 \end{pmatrix} + t\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}\) and \(\mathbf{r}_B = \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix} + t\begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix}\). Show that they do not collide.

Working
  1. Equate

    \(-3 + 2t = 3 - t\) (1), \(2 + t = -2 + 3t\) (2), \(4 - t = 1 + 2t\) (3)

    Same position needs all three coordinates equal. Both drones use the same \(t\), the same time.

  2. Each equation

    \((1): t = 2\), \((2): t = 2\), \((3): t = 1\)

    Unlike two crossing lines, there is one parameter, not two.

  3. Conclude

    (3) gives a different t

    No time makes all three equal, so they never collide. Their paths may still cross.

Video walkthrough coming
Closest approach with the calculator

Two aircraft have position vectors (in km) at time \(t\) hours: \(\mathbf{r}_A = \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix} + t\begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix}\) and \(\mathbf{r}_B = \begin{pmatrix} -3 \\ 9 \\ 1 \end{pmatrix} + t\begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}\). Find the time at which they are closest together, and their distance apart then.

Working
  1. Vector AB

    \(\overrightarrow{AB} = \mathbf{r}_B - \mathbf{r}_A = \begin{pmatrix} -4 + t \\ 7 - 2t \\ -3 + 2t \end{pmatrix}\)

    End minus start: the vector from A to B at time \(t\).

  2. Distance

    \(d = |\overrightarrow{AB}| = \sqrt{(-4 + t)^2 + (7 - 2t)^2 + (-3 + 2t)^2}\)

    This is the distance between them at time \(t\).

  3. The calculator

    From the calculator: minimum of \(d\) against \(t\).

    Graph \(d\) and find the minimum point.

  4. Read the result

    \(t = 2.67\) hours, \(d = 3.16\) km

    Another way: \(d^2 = 9t^2 - 48t + 74\) has its minimum at \(t = \frac{8}{3}\), but the calculator is quicker.

Video walkthrough coming