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IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.30

Vectors: exam review Exam review Worked solutions

Section A: short questions

  1. A drone is at the point P and a landing pad is at the point Q. Relative to an origin O, the position vectors are \(\overrightarrow{OP} = 4\mathbf{i} - \mathbf{j} + 2\mathbf{k}\) and \(\overrightarrow{OQ} = \mathbf{i} + 3\mathbf{j} - 4\mathbf{k}\), in metres. [6]

    1. (a) Write down \(\overrightarrow{PQ}\) as a column vector. [2]

      • M1 for \(\overrightarrow{OQ} - \overrightarrow{OP}\) (end point minus start point)
      • A1 for \(\begin{pmatrix} -3 \\ 4 \\ -6 \end{pmatrix}\)

    2. (b) Find the exact distance from P to Q. [2]

      • M1 for \(\sqrt{(-3)^2 + (4)^2 + (-6)^2}\)
      • A1 for \(\sqrt{61}\)

    3. (c) Find the vector of magnitude 10 in the same direction as \(\overrightarrow{PQ}\). Give the components to 3 significant figures. [2]

      • M1 for dividing \(\overrightarrow{PQ}\) by its magnitude and multiplying by 10, \(\dfrac{10}{\sqrt{61}}\begin{pmatrix} -3 \\ 4 \\ -6 \end{pmatrix}\)
      • A1 for \(\begin{pmatrix} -3.84 \\ 5.12 \\ -7.68 \end{pmatrix}\)

  2. In triangle OAB, \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). The point M is the midpoint of OA. The point P lies on AB with AP:PB = 1:2. The point Q is such that OAQB is a parallelogram. [6]

    OABQMP

    1. (a) Find \(\overrightarrow{AB}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\). [1]

      • A1 for \(\mathbf{b} - \mathbf{a}\)

    2. (b) Show that \(\overrightarrow{OP} = \dfrac{1}{3}(2\mathbf{a} + \mathbf{b})\). [2]

      • M1 for \(\overrightarrow{OP} = \overrightarrow{OA} + \dfrac{1}{3}\overrightarrow{AB}\)
      • A1 for \(\mathbf{a} + \dfrac{1}{3}(\mathbf{b} - \mathbf{a}) = \dfrac{1}{3}(2\mathbf{a} + \mathbf{b})\) (answer given, so the working must show the step)

    3. (c) Show that M, P and Q lie on a straight line. [3]

      • M1 for \(\overrightarrow{MP} = \overrightarrow{OP} - \overrightarrow{OM} = \dfrac{1}{6}(\mathbf{a} + 2\mathbf{b})\)
      • M1 for \(\overrightarrow{MQ} = \overrightarrow{OQ} - \overrightarrow{OM} = \mathbf{a} + \mathbf{b} - \overrightarrow{OM} = \dfrac{1}{2}(\mathbf{a} + 2\mathbf{b})\)
      • R1 for \(\overrightarrow{MQ} = 3\overrightarrow{MP}\), and M is a common point, so M, P, Q are collinear

  3. Let \(\mathbf{u} = \begin{pmatrix} 3 \\ p \\ -6 \end{pmatrix}\), \(\mathbf{v} = \begin{pmatrix} q \\ -10 \\ 4 \end{pmatrix}\) and \(\mathbf{w} = \begin{pmatrix} k \\ k + 2 \\ 2 \end{pmatrix}\), where \(p\), \(q\) and \(k\) are constants. [6]

    1. (a) Given that \(\mathbf{u}\) is parallel to \(\mathbf{v}\), find the value of \(p\) and the value of \(q\). [3]

      • M1 for \(\mathbf{u} = \lambda\mathbf{v}\) (or equal ratios of components)
      • M1 for \(-6 = \lambda(4)\) and so \(\lambda = -\frac{3}{2}\)
      • A1 for \(p = 15\) and \(q = -2\)

    2. (b) Given that \(|\mathbf{w}| = 6\), find the possible values of \(k\). Give your answers to 3 significant figures. [3]

      • M1 for \((k)^2 + (k + 2)^2 + (2)^2 = 36\)
      • A1 for the equation \(2 k^{2} + 4 k - 28 = 0\)
      • A1 for \(k = -4.87\) and \(k = 2.87\)

  4. The line \(l\) passes through the points A\((2, -1, 3)\) and B\((5, 1, -1)\). The point C has coordinates \((-1, -3, 7)\). [6]

    1. (a) Find a vector equation of \(l\). Give your answer in the form \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}\). [2]

      • M1 for \(\overrightarrow{AB} = \begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix}\) (or \(\overrightarrow{BA}\)) as the direction
      • A1 for \(\mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix}\) (any point of l and any multiple of the direction)

    2. (b) Determine whether C lies on \(l\). [2]

      • M1 for equating \(\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix} = \begin{pmatrix} -1 \\ -3 \\ 7 \end{pmatrix}\) and solving one component: \(-1 = 3 \lambda + 2 \Rightarrow \lambda = -1\)
      • A1 for checking the other two components with \(\lambda = -1\): they give \(-1, -3, 7\), which equal the coordinates of C, so C is on \(l\)

    3. (c) Write down parametric equations of \(l\). [2]

      • A1 for two correct equations, \(x = 3 \lambda + 2,\ y = 2 \lambda - 1\)
      • A1 for all three, with \(z = 3 - 4 \lambda\)

  5. A ship is sailing in open water with constant velocity. At time \(t\) hours after noon its position vector is \(\mathbf{r} = \begin{pmatrix} -4 \\ 7 \end{pmatrix} + t\begin{pmatrix} 9 \\ -4 \end{pmatrix}\), in km relative to the lighthouse L. The first component is the distance east and the second is the distance north. [7]

    1. (a) Find the speed of the ship. [2]

      • M1 for \(\sqrt{(9)^2 + (-4)^2}\)
      • A1 for \(\sqrt{97}\) = 9.85

    2. (b) Find the bearing on which the ship is moving. Give your answer to the nearest degree. [2]

      • M1 for a sketch or the angle with the north-south line: \(\tan\theta = \dfrac{9}{4}\), 66.0°
      • A1 for the bearing 114° (180° − 66.0°)

    3. (c) Find the distance of the ship from the lighthouse L when \(t = 2.5\). [3]

      • M1 for \(\mathbf{r} = \begin{pmatrix} -4 \\ 7 \end{pmatrix} + 2.5\begin{pmatrix} 9 \\ -4 \end{pmatrix}\)
      • A1 for \(\mathbf{r} = \begin{pmatrix} 18.5 \\ -3 \end{pmatrix}\)
      • A1 for \(\sqrt{(18.5)^2 + (-3)^2}\) = 18.7

  6. The scalar product (dot product) of two vectors is written \(\mathbf{a} \cdot \mathbf{b}\). [7]

    1. (a) The vectors \(\mathbf{a}\) and \(\mathbf{b}\) have \(|\mathbf{a}| = 4\) and \(|\mathbf{b}| = 7\). The angle between them is 140°. Find \(\mathbf{a} \cdot \mathbf{b}\). Give your answer to 3 significant figures. [2]

      • M1 for \(\mathbf{a} \cdot \mathbf{b} = 4 \times 7 \times \cos 140^\circ\)
      • A1 for \(-21.4\)

    2. (b) Let \(\mathbf{p} = \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix}\) and \(\mathbf{q} = \begin{pmatrix} -1 \\ 4 \\ 2 \end{pmatrix}\). Find the angle between \(\mathbf{p}\) and \(\mathbf{q}\). Give your answer to the nearest degree. [3]

      • M1 for \(\mathbf{p} \cdot \mathbf{q} = (3)(-1) + (-2)(4) + (1)(2) = -9\)
      • M1 for \(\cos\theta = \dfrac{-9}{\sqrt{14}\sqrt{21}}\)
      • A1 for 122° (the cosine is negative, so the angle is obtuse)

    3. (c) The vector \(\mathbf{s} = 5\mathbf{i} + t\mathbf{j} - \mathbf{k}\) is perpendicular to \(\mathbf{z} = 3\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}\). Find the value of \(t\). [2]

      • M1 for \(\mathbf{s} \cdot \mathbf{z} = 0:\ 13 - 4 t = 0\)
      • A1 for \(t = \frac{13}{4}\)

  7. The vector product (cross product) of two vectors is written \(\mathbf{a} \times \mathbf{b}\). [7]

    1. (a) The vectors \(\mathbf{a}\) and \(\mathbf{b}\) have \(|\mathbf{a}| = 6\) and \(|\mathbf{b}| = 7\). The angle between them is 125°. Find \(|\mathbf{a} \times \mathbf{b}|\). Give your answer to 3 significant figures. [2]

      • M1 for \(|\mathbf{a} \times \mathbf{b}| = 6 \times 7 \times \sin 125^\circ\)
      • A1 for 34.4

    2. (b) Let \(\mathbf{p} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}\) and \(\mathbf{q} = \mathbf{i} + 4\mathbf{j} - \mathbf{k}\). Find \(\mathbf{p} \times \mathbf{q}\). [2]

      • M1 for the component form \(\begin{pmatrix} (-1)(-1) - (3)(4) \\ (3)(1) - (2)(-1) \\ (2)(4) - (-1)(1) \end{pmatrix}\)
      • A1 for \(\begin{pmatrix} -11 \\ 5 \\ 9 \end{pmatrix}\)

    3. (c) Find the component of \(\mathbf{p}\) in the direction of \(\mathbf{q}\). Give your answer to 3 significant figures. [3]

      • M1 for the scalar product \(\mathbf{p} \cdot \mathbf{q} = -5\) (direction goes with the scalar product)
      • M1 for dividing by \(|\mathbf{q}| = \sqrt{18}\)
      • A1 for -1.18

  8. A model train runs on a straight track. Its velocity at time \(t\) seconds is \(v = 7\cos(0.8t)\) m s⁻¹. A second model train runs on a parallel straight track. When its displacement from the station is \(s\) metres, its velocity is \(v = 4 + 0.3s^2\) m s⁻¹. [6]

    1. (a) Find the acceleration of the first train when \(t = 2\). Give your answer to 3 significant figures. [2]

      • M1 for differentiating the velocity with respect to \(t\) (chain rule): \(a = -5.6\sin(0.8t)\)
      • A1 for \(a(2) = -5.60\) (calculator in radian mode)

    2. (b) Find the change in displacement of the first train between \(t = 0\) and \(t = 3\). Give your answer to 3 significant figures. [2]

      • M1 for the definite integral \(\displaystyle\int_0^{3} 7\cos(0.8t)\,dt\)
      • A1 for \(5.91\) metres

    3. (c) Find the acceleration of the second train when \(s = 2\). [2]

      • M1 for using \(a = v\dfrac{dv}{ds} = (4 + 0.3s^2)(0.6s)\)
      • A1 for \(a = 6.24\)

  9. A drone flies in a horizontal plane. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix} 3 t^{2} - 4 \\ 6 t - 5 \end{pmatrix}\) m s⁻¹, for \(t \geq 0\). When \(t = 0\) its position vector relative to a fixed origin O is \(\begin{pmatrix} 2 \\ -1 \end{pmatrix}\) metres. [7]

    1. (a) Find the acceleration of the drone when \(t = 2\). [2]

      • M1 for differentiating each component: \(\mathbf{a} = \begin{pmatrix} 6 t \\ 6 \end{pmatrix}\)
      • A1 for \(\mathbf{a}(2) = \begin{pmatrix} 12 \\ 6 \end{pmatrix}\)

    2. (b) Find an expression for the position vector of the drone at time \(t\). [3]

      • M1 for integrating each component of the velocity
      • A1 for the correct integrals, \(t^{3} - 4 t\) and \(3 t^{2} - 5 t\), plus constants
      • A1 for using \(\mathbf{r}(0) = \begin{pmatrix} 2 \\ -1 \end{pmatrix}\) to find the constants: \(\mathbf{r} = \begin{pmatrix} t^{3} - 4 t + 2 \\ 3 t^{2} - 5 t - 1 \end{pmatrix}\)

    3. (c) Find the speed of the drone when \(t = 2\). Give your answer to 3 significant figures. [2]

      • M1 for \(|\mathbf{v}(2)| = \sqrt{(8)^2 + (7)^2}\)
      • A1 for \(10.6\)

  10. A stone is thrown from a point O on level ground. Its position vector relative to O at time \(t\) seconds is \(\mathbf{r} = \begin{pmatrix} 9t \\ 14t - 4.9t^2 \end{pmatrix}\) metres, where the first component is horizontal and the second is vertically upward. [6]

    1. (a) Find the velocity vector of the stone at time \(t\). [2]

      • M1 for differentiating each component of \(\mathbf{r}\) with respect to \(t\)
      • A1 for \(\begin{pmatrix} 9 \\ 14 - 9.8t \end{pmatrix}\)

    2. (b) Find the greatest height of the stone above the ground. Give your answer to 3 significant figures. [2]

      • M1 for setting the vertical velocity to zero, \(14 - 9.8t = 0\), so \(t = 1.43\)
      • A1 for \(14(1.43) - 4.9(1.43)^2 = 10.0\)

    3. (c) Find the horizontal distance from O to the point where the stone lands. Give your answer to 3 significant figures. [2]

      • M1 for setting the vertical displacement to zero, \(14t - 4.9t^2 = 0\), so \(t = 2.86\) (reject \(t = 0\))
      • A1 for \(9(2.86) = 25.7\)

  11. A bead slides on a circular wire. Its position vector at time \(t\) seconds, relative to a fixed origin O, is \(\mathbf{r} = \begin{pmatrix} 6 + 4\cos(0.5t) \\ -3 + 4\sin(0.5t) \end{pmatrix}\) metres. A second bead starts at the highest point of the circle and goes round clockwise, taking 8 seconds for each complete turn, on the same circle. [8]

    1. (a) Show that the bead moves on a circle, and write down the centre and the radius of the circle. [3]

      • M1 for \(\mathbf{r} - \begin{pmatrix} 6 \\ -3 \end{pmatrix}\) having components 4\cos(0.5t) and 4\sin(0.5t)
      • M1 for \(\sqrt{4^2\cos^2(0.5t) + 4^2\sin^2(0.5t)} = 4\) using \(\sin^2\theta + \cos^2\theta = 1\)
      • A1 for centre \((6, -3)\) and radius \(4\) metres

    2. (b) State whether the bead goes round the circle clockwise or anticlockwise, and find the time it takes to go round once. Give your answer to 3 significant figures. [2]

      • M1 for the direction of motion at \(t = 0\), from the position and velocity (\(\mathbf{r}(0) - \text{centre} = \begin{pmatrix} 4 \\ 0 \end{pmatrix}, \ \mathbf{v}(0) = \begin{pmatrix} 0 \\ 2 \end{pmatrix}\)): from the point to the right of the centre the bead moves upward, so anticlockwise
      • A1 for anticlockwise and \(T = \dfrac{2\pi}{0.5} = 12.6\) seconds

    3. (c) Write down an expression for the position vector of the second bead at time \(t\) seconds. [3]

      • M1 for angle speed \(\dfrac{2\pi}{8}\) so the argument is \(\dfrac{\pi t}{4}\)
      • M1 for centre \(\begin{pmatrix} 6 \\ -3 \end{pmatrix}\) plus radius \(4\) with the correct start point and direction (at t=0 position (cx, cy+R) and velocity (Rw, 0) to the right: clockwise)
      • A1 for \(\mathbf{r} = \begin{pmatrix} 6 + 4\sin(\dfrac{\pi t}{4}) \\ -3 + 4\cos(\dfrac{\pi t}{4}) \end{pmatrix}\)

Section B: long questions

  1. Two straight cables in a stadium roof are modelled by lines. The lines \(l_1\) and \(l_2\) are given by \(l_1:\ \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}\) and \(l_2:\ \mathbf{r} = 4\mathbf{i} + 6\mathbf{j} + 3\mathbf{k} + \mu\left(\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}\right)\). [14]

    1. (a) Show that \(l_1\) and \(l_2\) intersect, and find the coordinates of the point of intersection. [4]

      • M1 for equating the \(x\) and \(y\) components of the two lines: \(2 \lambda + 1 = \mu + 4\) and \(\lambda + 2 = 6 - 2 \mu\)
      • A1 for \(\lambda = 2\) and \(\mu = 1\)
      • R1 for checking the \(z\) component: \(5 = 5\), so the lines intersect
      • A1 for \((5, 4, 5)\)

    2. (b) Find the acute angle between \(l_1\) and \(l_2\). [3]

      • M1 for the direction vectors \(\begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}\) (not the position vectors)
      • M1 for \(\cos\theta = \dfrac{|6|}{\sqrt{14}\sqrt{9}}\)
      • A1 for 57.7°

    3. (c) The point P has coordinates \((8, 3, 7)\). Find the shortest distance from P to \(l_2\). Give your answer to 3 significant figures. [4]

      • M1 for \(\overrightarrow{PB} = \begin{pmatrix} \mu - 4 \\ 3 - 2 \mu \\ 2 \mu - 4 \end{pmatrix}\) for a general point B on \(l_2\)
      • M1 for \(\overrightarrow{PB} \cdot \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} = 0\)
      • A1 for \(\mu = 2\) and \(\overrightarrow{PB} = \begin{pmatrix} -2 \\ -1 \\ 0 \end{pmatrix}\)
      • A1 for \(\sqrt{5}\) = 2.24

    4. (d) A third line \(l_3\) has parametric equations \(x = s + 2,\ y = 5 - s,\ z = 2 s - 1\). Show that \(l_1\) and \(l_3\) do not intersect. [3]

      • M1 for equating two components of \(l_1\) and \(l_3\): \(2 \lambda + 1 = s + 2\) and \(\lambda + 2 = 5 - s\)
      • A1 for \(\lambda = \frac{4}{3}\) and \(s = \frac{5}{3}\)
      • R1 for the \(z\) components: \(3 \neq \frac{7}{3}\), so the lines do not intersect

  2. Two drones, A and B, fly in a light show. Positions are measured in metres from a fixed point O, with \(\mathbf{i}\) east, \(\mathbf{j}\) north and \(\mathbf{k}\) vertically upwards, and \(t\) is the time in seconds. At \(t = 0\) A is at \((10, 20, 30)\) and moves with constant velocity \(\begin{pmatrix} 4 \\ -2 \\ 1 \end{pmatrix}\) m/s. At \(t = 0\) B is at \((-8, 32, 12)\) and moves with constant speed 6 m/s in the direction of \(\begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}\). [14]

    1. (a) Find the speed of A. Give your answer to 3 significant figures. [2]

      • M1 for \(\sqrt{(4)^2 + (-2)^2 + (1)^2}\)
      • A1 for 4.58

    2. (b) Show that the velocity of B is \(\begin{pmatrix} 2 \\ -4 \\ 4 \end{pmatrix}\). [3]

      • A1 for \(|\begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}| = 3\)
      • M1 for \(\mathbf{v} = 6 \times \dfrac{1}{3}\begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}\)
      • A1 for \(\begin{pmatrix} 2 \\ -4 \\ 4 \end{pmatrix}\) (answer given, so the scaling step must be seen)

    3. (c) Write down the position vector of B at time \(t\). [1]

      • A1 for \(\mathbf{r}_{B} = \begin{pmatrix} -8 \\ 32 \\ 12 \end{pmatrix} + t\begin{pmatrix} 2 \\ -4 \\ 4 \end{pmatrix}\)

    4. (d) Find the distance between A and B when \(t = 3\). Give your answer to 3 significant figures. [2]

      • M1 for finding both positions at the time, or the vector between them: \(\begin{pmatrix} -24 \\ 6 \\ -9 \end{pmatrix}\)
      • A1 for 26.3

    5. (e) Show that A and B do not collide. [3]

      • M1 for equating the \(x\) components: \(4 t + 10 = 2 t - 8\) gives \(t = -9\)
      • M1 for equating a second component: \(20 - 2 t = 32 - 4 t\) gives \(t = 6\)
      • R1 for the values of \(t\) being different, so the same \(t\) does not work in every component and they do not collide

    6. (f) Use your calculator to find the time at which they are closest together, and the distance between them at that time. Give your answers to 3 significant figures. [3]

      • M1 for an expression for the distance, \(d = |\mathbf{r}_{B} - \mathbf{r}_{A}| = \sqrt{17 t^{2} - 84 t + 792}\)
      • A1 for \(t = 2.47\)
      • A1 for \(d = 26.2\)

  3. A triangular sail ABC is held in the air by three poles. The corners are A\((2, 1, -1)\), B\((5, 3, 1)\) and C\((4, 5, 2)\), in metres. [14]

    1. (a) Find \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) as column vectors. [2]

      • A1 for \(\overrightarrow{AB} = \begin{pmatrix} 3 \\ 2 \\ 2 \end{pmatrix}\)
      • A1 for \(\overrightarrow{AC} = \begin{pmatrix} 2 \\ 4 \\ 3 \end{pmatrix}\)

    2. (b) Find the angle BAC. Give your answer to the nearest degree. [3]

      • M1 for the scalar product \(\overrightarrow{AB} \cdot \overrightarrow{AC} = (3)(2) + (2)(4) + (2)(3) = 20\)
      • M1 for \(\cos\theta = \dfrac{20}{\sqrt{17}\sqrt{29}}\)
      • A1 for 26°

    3. (c) Find \(\overrightarrow{AB} \times \overrightarrow{AC}\). [2]

      • M1 for the component form of the vector product, e.g. \((2)(3) - (2)(4)\)
      • A1 for \(\begin{pmatrix} -2 \\ -5 \\ 8 \end{pmatrix}\)

    4. (d) Hence find the area of triangle ABC. Give your answer to 3 significant figures. [2]

      • M1 for \(\tfrac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|\)
      • A1 for \(\tfrac{1}{2}\sqrt{93}\) = 4.82

    5. (e) Find a unit vector that is perpendicular to both \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\). [2]

      • M1 for dividing \(\begin{pmatrix} -2 \\ -5 \\ 8 \end{pmatrix}\) by its magnitude \(\sqrt{93}\)
      • A1 for \(\pm\dfrac{1}{\sqrt{93}}\begin{pmatrix} -2 \\ -5 \\ 8 \end{pmatrix}\) (either direction)

    6. (f) Find the shortest distance from C to the line AB. Give your answer to 3 significant figures. [3]

      • M1 for the perpendicular distance as the height of the triangle, \(\text{area} = \tfrac{1}{2}|\overrightarrow{AB}| \times h\), or the component \(\dfrac{|\overrightarrow{AB} \times \overrightarrow{AC}|}{|\overrightarrow{AB}|}\)
      • A1 for \(|\overrightarrow{AB}| = \sqrt{17}\)
      • A1 for 2.34

  4. C is a carriage on a circular track. Its position vector relative to the centre O of the track at time \(t\) seconds, where \(t \geq 0\), is \(\mathbf{r}_C = \begin{pmatrix} 8\cos(0.05t^2) \\ 8\sin(0.05t^2) \end{pmatrix}\) metres. A second carriage D starts from the same point as C, 5 seconds after C, and moves in exactly the same way. [12]

    1. (a) Show that C moves on a circle with centre O, and write down the radius of the circle. [2]

      • M1 for \(|\mathbf{r}_C|^2 = 64\cos^2(0.05t^2) + 64\sin^2(0.05t^2) = 64\) using \(\sin^2\theta + \cos^2\theta = 1\)
      • A1 for \(|\mathbf{r}_C| = 8\), constant, so C moves on a circle of radius 8 metres

    2. (b) Find an expression for the velocity of C at time \(t\). [3]

      • M1 for differentiating \(8\cos(0.05t^2)\) with the chain rule
      • A1 for \(-0.8t\sin(0.05t^2)\)
      • A1 for \(0.8t\cos(0.05t^2)\)

    3. (c) Find the speed of C when \(t = 4\). [2]

      • M1 for \(|\mathbf{v}_C| = \sqrt{(0.8t)^2\sin^2(0.05t^2) + (0.8t)^2\cos^2(0.05t^2)} = 0.8t\), or evaluating the magnitude on the calculator at \(t = 4\)
      • A1 for \(3.20\)

    4. (d) Find the number of complete revolutions that C makes in the first 20 seconds. [2]

      • M1 for the angle turned, \(0.05(20)^2 = 20.0\) radians, divided by \(2\pi\) to give \(3.18\) revolutions
      • A1 for 3 complete revolutions (whole number only)

    5. (e) Write down an expression for the position vector \(\mathbf{r}_D\) of D at time \(t\), for \(t \geq 5\). [1]

      • A1 for replacing \(t\) by \((t - 5)\) in \(\mathbf{r}_C\)

    6. (f) Find the distance between C and D when \(t = 9\). Give your answer to 3 significant figures. [2]

      • M1 for \(|\mathbf{r}_C - \mathbf{r}_D|\) with \(t = 9\), using \(\mathbf{r}_C\) at angle \(0.05(9)^2 = 4.05\) and \(\mathbf{r}_D\) at angle \(0.05(4)^2 = 0.800\)
      • A1 for 16.0

  5. A robot moves along a straight rail. The velocity of the robot at time \(t\) seconds is \(v = t^{2} - 9 t + 14\) m s⁻¹, for \(t \geq 0\). [7]

    1. (a) Find the times when the robot is at rest, and for each time say whether the robot changes direction. [3]

      • M1 for solving \(t^{2} - 9 t + 14 = 0\): \(t = 2\), \(t = 7\)
      • A1 for both times
      • R1 for \(v\) changing sign (positive, then negative, then positive) at each of these times, so the direction changes at both

    2. (b) Find the total distance travelled by the robot in the first 8 seconds. Give your answer to 3 significant figures. [2]

      • M1 for \(\displaystyle\int_0^{8} |t^{2} - 9 t + 14|\,dt\), using the calculator
      • A1 for \(36.3\) metres (the displacement is \(-5.33\) metres, which is not the distance)

    3. (c) The robot starts at the point where \(s = 0\). Find the first time after it starts at which it returns to this point. Give your answer to 3 significant figures. [2]

      • M1 for \(s = \displaystyle\int_0^t v\,dt = \frac{t^{3}}{3} - \frac{9 t^{2}}{2} + 14 t\), set equal to 0 and solved with technology
      • A1 for \(t = 4.86\) (reject \(t = 0\) and any negative time)

  6. A ball is thrown from a point 1.5 m above level ground, with speed 16 m s⁻¹ at an angle of 35° above the horizontal. Take O as the point on the ground directly below the point of release, with \(\mathbf{i}\) horizontal and \(\mathbf{j}\) vertically upward. Ignore air resistance and take \(g = 9.8\) m s⁻². [12]

    1. (a) Write down the position vector of the ball at time \(t\) seconds after it is thrown, in the form \(\mathbf{r} = \begin{pmatrix} x \\ y \end{pmatrix}\) where \(x\) and \(y\) are in terms of \(t\). [2]

      • M1 for horizontal component \(16\cos 35^\circ = 13.1\) and vertical component \(16\sin 35^\circ = 9.18\)
      • A1 for \(\mathbf{r} = \begin{pmatrix} 13.1t \\ 1.5 + 9.18t - 4.9t^2 \end{pmatrix}\) (with \(g = 9.8\))

    2. (b) Find the time that the ball takes to reach the ground, and the horizontal distance from O to the point where it lands. [3]

      • M1 for setting the vertical component equal to 0: \(1.5 + 9.18t - 4.9t^2 = 0\)
      • A1 for \(t = 2.02\) (the positive root)
      • A1 for \(x = 13.1 \times 2.02 = 26.5\) m

    3. (c) Find the greatest height of the ball above the ground. [2]

      • M1 for the top at \(9.18 - 9.8t = 0\), \(t = 0.936\), and substituting into \(y\)
      • A1 for \(5.80\) m

    4. (d) Find the speed of the ball when it lands, and the angle below the horizontal at which it lands. [3]

      • M1 for \(\mathbf{v}(t) = \begin{pmatrix} 13.1 \\ 9.18 - 9.8t \end{pmatrix}\) at \(t = 2.02\): \(\begin{pmatrix} 13.1 \\ -10.7 \end{pmatrix}\)
      • A1 for speed \(16.9\) m s⁻¹
      • A1 for \(\tan^{-1}\dfrac{10.7}{13.1} = 39.1^\circ\) below the horizontal

    5. (e) The ball passes a fence, of height 5.0 m, at a horizontal distance 15 m from O. Decide whether the ball clears the fence, and find the vertical distance between the ball and the top of the fence. [2]

      • M1 for \(t = \dfrac{15}{13.1} = 1.14\), then \(y = 5.58\) m at this time
      • A1 for comparing 5.58 with 5.0: the ball clears the fence, 0.585 m above the top

jmaths.xyz · AI HL · AHL 4.30 Unit review