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IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.11

Constant velocity: speed, position and closest approach Problem set A Worked solutions

Skills

  1. A drone moves with constant velocity \(\mathbf{v} = \begin{pmatrix} 7 \\ -3 \end{pmatrix}\) m s\(^{-1}\), where \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\) points east and \(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\) points north.

    2468−4−3−2−11eastnorthv

    1. (a) Find the direction of motion as a bearing.

      • Draw the components: \(7\) in the east/west direction and \(3\) in the north/south direction.
      • \(\theta = \tan^{-1}\frac{3}{7} = 23.2^\circ\) (below east)
      • \(\text{bearing} = 90^\circ + 23.2^\circ = 113^\circ\)

    2. (b) Find the speed.

      • \(\text{speed} = |\mathbf{v}| = \sqrt{49 + 9} = 7.62\)

    3. (c) A second object has velocity \(\begin{pmatrix} -5 \\ 4 \end{pmatrix}\) m s\(^{-1}\). Find its direction of motion as a bearing.

      • Sketch the components again; this vector points to the north-west.
      • \(\theta = \tan^{-1}\frac{5}{4} = 51.3^\circ\) (from north towards west)
      • \(\text{bearing} = 360^\circ - 51.3^\circ = 309^\circ\)

  2. A hot-air balloon moves with constant speed in a fixed direction. Distances are in metres and time in seconds.

    1. (a) The direction is \(\begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}\) and the speed is 15 m s\(^{-1}\). Find the velocity vector.

      • \(|\begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}| = \sqrt{9} = 3\)
      • \(\mathbf{v} = \frac{15}{3}\begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} = 5\begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}\)
      • \(\mathbf{v} = \begin{pmatrix} 5 \\ -10 \\ 10 \end{pmatrix}\)

    2. (b) The direction is \(\begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix}\) and the speed is 12 m s\(^{-1}\). Find the velocity vector, giving each component to 3 significant figures.

      • \(|\begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix}| = \sqrt{38}\)
      • Unit vector times speed: \(\mathbf{v} = \frac{12}{\sqrt{38}}\begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix}\)
      • Check the magnitude is 12: \(\mathbf{v} = \begin{pmatrix} 3.89 \\ 5.84 \\ 9.73 \end{pmatrix}\)

  3. An object moves with constant velocity \(\mathbf{v} = -\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}\) m s\(^{-1}\). At \(t = 0\) its position vector is \(\mathbf{r}_0 = 6\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}\) m, where \(t\) is in seconds.

    1. (a) Write down an equation for the position vector of the object at time \(t\).

      • Use \(\mathbf{r} = \mathbf{r}_0 + t\mathbf{v}\).
      • \(\mathbf{r} = (6\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}) + t(-\mathbf{i} + 3\mathbf{j} + 2\mathbf{k})\)

    2. (b) Find the distance of the object from the origin when \(t = 4\).

      • \(\mathbf{r} = 6\mathbf{i} - 4\mathbf{j} + 2\mathbf{k} + 4(-\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}) = 2\mathbf{i} + 8\mathbf{j} + 10\mathbf{k}\)
      • \(|\mathbf{r}| = \sqrt{4 + 64 + 100} = 13.0\) m

In context

  1. Two weather balloons move with constant velocities. Their position vectors (in km) at time \(t\) hours are given.

    1. (a) \(\mathbf{r}_A = \begin{pmatrix} 1 \\ -2 \\ 5 \end{pmatrix} + t\begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix}\) and \(\mathbf{r}_B = \begin{pmatrix} 7 \\ 2 \\ 2 \end{pmatrix} + t\begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix}\). Show that \(A\) and \(B\) do not collide.

      • Set \(\mathbf{r}_A = \mathbf{r}_B\) and compare: \(1 + 2t = 7 - t\) (1), \(-2 + 3t = 2 + t\) (2), \(5 - t = 2 + 2t\) (3)
      • Solve each for the same \(t\): \((1): t = 2\), \((2): t = 2\), \((3): t = 1\)
      • The values of \(t\) are not all the same, so the two never collide.

    2. (b) \(\mathbf{r}_C = \begin{pmatrix} 2 \\ 8 \\ -2 \end{pmatrix} + t\begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}\) and \(\mathbf{r}_D = \begin{pmatrix} 11 \\ 2 \\ 4 \end{pmatrix} + t\begin{pmatrix} -2 \\ 0 \\ 1 \end{pmatrix}\). Show that \(C\) and \(D\) collide, and find where.

      • Set \(\mathbf{r}_C = \mathbf{r}_D\) and compare: \(2 + t = 11 - 2t\) (1), \(8 - 2t = 2\) (2), \(-2 + 3t = 4 + t\) (3)
      • Solve each for the same \(t\): \((1): t = 3\), \((2): t = 3\), \((3): t = 3\)
      • All three give \(t = 3\), so they collide at that time, at \((5, 2, 7)\).

  2. Two drones have position vectors (in km) at time \(t\) hours: \(\mathbf{r}_A = \begin{pmatrix} 4 \\ 1 \\ 2 \end{pmatrix} + t\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\) and \(\mathbf{r}_B = \begin{pmatrix} -3 \\ 5 \\ -1 \end{pmatrix} + t\begin{pmatrix} 3 \\ -2 \\ 4 \end{pmatrix}\).

    1. (a) Find \(\overrightarrow{AB}\) in terms of \(t\).

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

    2. (b) Use your calculator to find the time at which they are closest together, and their distance apart at that time.

      • \(d = |\overrightarrow{AB}| = \sqrt{(-7 + t)^2 + (4 - t)^2 + (-3 + t)^2}\)
      • From the calculator: minimum of \(d\) against \(t\). \(t = 4.67\) hours, \(d = 2.94\) km

Stretch

  1. A survey drone starts at the point with position vector \(\begin{pmatrix} 40 \\ 25 \\ 10 \end{pmatrix}\) m, relative to its base O, and moves in the direction \(\begin{pmatrix} 3 \\ -1 \\ 5 \end{pmatrix}\) at a constant speed of 14 m s\(^{-1}\).

    1. (a) Find its velocity vector, giving each component to 3 significant figures.

      • \(|\begin{pmatrix} 3 \\ -1 \\ 5 \end{pmatrix}| = \sqrt{35}\)
      • \(\mathbf{v} = \frac{14}{\sqrt{35}}\begin{pmatrix} 3 \\ -1 \\ 5 \end{pmatrix} = \begin{pmatrix} 7.10 \\ -2.37 \\ 11.8 \end{pmatrix}\)

    2. (b) Write down its position vector at time \(t\) seconds.

      • Use \(\mathbf{r} = \mathbf{r}_0 + t\mathbf{v}\).
      • \(\mathbf{r} = \begin{pmatrix} 40 \\ 25 \\ 10 \end{pmatrix} + t\begin{pmatrix} 7.10 \\ -2.37 \\ 11.8 \end{pmatrix}\)

    3. (c) Find its distance from its base O after 20 seconds.

      • \(t = 20: \mathbf{r} = \begin{pmatrix} 182 \\ -22.3 \\ 247 \end{pmatrix}\)
      • \(|\mathbf{r}| = 307\) m

  2. Two aircraft have position vectors (in km) at time \(t\) hours: \(\mathbf{r}_A = \begin{pmatrix} 3 \\ 0 \\ -2 \end{pmatrix} + t\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}\) and \(\mathbf{r}_B = \begin{pmatrix} 9 \\ 2 \\ 7 \end{pmatrix} + t\begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix}\).

    1. (a) Show that they do not collide.

      • Set \(\mathbf{r}_A = \mathbf{r}_B\) and compare: \(3 + t = 9 - 2t\) (1), \(2t = 2 + t\) (2), \(-2 + 2t = 7 - t\) (3)
      • Solve each for the same \(t\): \((1): t = 2\), \((2): t = 2\), \((3): t = 3\)
      • The values of \(t\) are not all the same, so the two never collide.

    2. (b) Find the time at which they are closest together, and the distance between them then.

      • \(\overrightarrow{AB} = \mathbf{r}_B - \mathbf{r}_A = \begin{pmatrix} 6 - 3t \\ 2 - t \\ 9 - 3t \end{pmatrix}\)
      • \(d = |\overrightarrow{AB}|\)
      • From the calculator: minimum of \(d\). \(t = 2.47\) hours, \(d = 2.18\) km

    3. (c) Safety rules require at least 2.5 km between them at all times. Is the rule kept? Give a reason.

      • The smallest distance is 2.18 km.
      • Compare with 2.5 km: it is smaller, so the rule is broken.

jmaths.xyz · AI HL · AHL 4.11 Constant velocity: speed, position and closest approach