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IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.12

Kinematics with constant velocity Problem set B One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Skills

  1. Find the bearing on which each object moves, to the nearest degree. The first component is east, the second north. Velocities are in m s⁻¹.

    2468−11234eastnorthv

    1. (a) A kayak: \(\mathbf{v} = \begin{pmatrix}8\\3\end{pmatrix}\)

      • Sketch: east 8, north 3
      • Angle from the north–south line: \(\tan\alpha = \dfrac{8}{3}\), \(\alpha = 69.4^\circ\)
      • Bearing = 69.4°
      • 69° to the nearest degree

    2. (b) A rescue boat: \(\mathbf{v} = 5\mathbf{i} - 6\mathbf{j}\)

      • Sketch: east 5, north −6
      • Angle from the north–south line: \(\tan\alpha = \dfrac{5}{6}\), \(\alpha = 39.8^\circ\)
      • Bearing = 180° − 39.8° = 140.2°
      • 140° to the nearest degree

  2. Find the speed of each object. Give each answer exactly and to 3 significant figures. Velocities are in m s⁻¹.

    1. (a) A cyclist: \(\mathbf{v} = \begin{pmatrix}-6\\11\end{pmatrix}\)

      • \(|\mathbf{v}| = \sqrt{(-6)^2 + 11^2} = \sqrt{157} = 12.5\)
      • Speed = √157 = 12.5 m s⁻¹

    2. (b) A rocket: \(\mathbf{v} = 2\mathbf{i} - 6\mathbf{j} + 9\mathbf{k}\)

      • \(|\mathbf{v}| = \sqrt{2^2 + (-6)^2 + 9^2} = 11\)
      • Speed = 11 m s⁻¹

    3. (c) A bird: \(\mathbf{v} = \begin{pmatrix}7\\2\\-4\end{pmatrix}\)

      • \(|\mathbf{v}| = \sqrt{7^2 + 2^2 + (-4)^2} = \sqrt{69} = 8.31\)
      • Speed = √69 = 8.31 m s⁻¹

  3. Find the velocity vector of an object moving in the given direction at the given speed.

    1. (a) Direction \(\begin{pmatrix}20\\21\end{pmatrix}\), speed 58 m s⁻¹

      • \(|\mathbf{d}| = 29\)
      • \(\mathbf{v} = \dfrac{58}{29}\,\begin{pmatrix}20\\21\end{pmatrix} = \begin{pmatrix}40\\42\end{pmatrix}\)

    2. (b) Direction \(4\mathbf{i} + 4\mathbf{j} + 7\mathbf{k}\), speed 18 m s⁻¹

      • \(|\mathbf{d}| = 9\)
      • \(\mathbf{v} = \dfrac{18}{9}\,(4\mathbf{i} + 4\mathbf{j} + 7\mathbf{k}) = 8\mathbf{i} + 8\mathbf{j} + 14\mathbf{k}\)

    3. (c) A diving robot starts at the point (5, −2, 1) and moves at 14 m s⁻¹ in the direction of \(3\mathbf{i} + 6\mathbf{j} + 2\mathbf{k}\). Write down its position vector at time \(t\) seconds.

      • \(|\mathbf{d}| = 7\)
      • \(\mathbf{v} = 14\div 7\times \begin{pmatrix}3\\6\\2\end{pmatrix} = \begin{pmatrix}6\\12\\4\end{pmatrix}\)
      • \(\mathbf{r} = \begin{pmatrix}5\\-2\\1\end{pmatrix} + t\begin{pmatrix}6\\12\\4\end{pmatrix}\)

  4. A model train moves with constant velocity \(\mathbf{v} = -\mathbf{i} + 3\mathbf{j} + 2.5\mathbf{k}\) m s⁻¹. At \(t = 0\) it is at the point (7, −1, −4) (distances in metres, time in seconds).

    1. (a) Find the speed.

      • \(|\mathbf{v}| = \sqrt{(-1)^2 + 3^2 + 2.5^2} = \sqrt{16.25} = 4.03\)
      • Speed = √16.25 = 4.03 m s⁻¹

    2. (b) Write down an equation for the position vector at time t.

      • \(\mathbf{r} = (7\mathbf{i} - \mathbf{j} - 4\mathbf{k}) + t(-\mathbf{i} + 3\mathbf{j} + 2.5\mathbf{k})\)

    3. (c) Does it pass through the point (5, 5, 2)? Give a reason.

      • \(x: 7 - t = 5 \Rightarrow t = 2\)
      • \(y: -1 + 3(2) = 5\)
      • \(z: -4 + 2.5(2) = 1\), but the point has z = 2
      • The z-component does not match, so it does not pass through the point.

  5. A submarine moves so that its position vector at time \(t\) seconds is \(\mathbf{r} = \begin{pmatrix}1\\-5\\2\end{pmatrix} + t\begin{pmatrix}2\\4\\4\end{pmatrix}\), where distances are in metres.

    1. (a) Find the initial position of the object.

      • Put t = 0: \(\mathbf{r} = \begin{pmatrix}1\\-5\\2\end{pmatrix}\)
      • Initial position (1, −5, 2)

    2. (b) Find the speed of the object.

      • \(|\mathbf{v}| = \sqrt{2^2 + 4^2 + 4^2} = 6\)
      • Speed = 6 m s⁻¹

    3. (c) Find the distance of the object from the origin after 3 seconds.

      • \(\mathbf{r} = \begin{pmatrix}1\\-5\\2\end{pmatrix} + 3\begin{pmatrix}2\\4\\4\end{pmatrix} = \begin{pmatrix}7\\7\\14\end{pmatrix}\)
      • \(|\mathbf{r}| = \sqrt{294} = 17.1\) m

In context

  1. Two birds have position vectors \(\mathbf{r}_1 = \begin{pmatrix}7\\1\\5\end{pmatrix} + t\begin{pmatrix}-2\\0.5\\1\end{pmatrix}\) and \(\mathbf{r}_2 = \begin{pmatrix}-1\\6\\2\end{pmatrix} + t\begin{pmatrix}1\\-1.5\\2\end{pmatrix}\) at time \(t\) seconds, where distances are in m. Find the distance between the birds when \(t = 4\).

    • \(\mathbf{r}_1 = \begin{pmatrix}7\\1\\5\end{pmatrix} + 4\begin{pmatrix}-2\\0.5\\1\end{pmatrix} = \begin{pmatrix}-1\\3\\9\end{pmatrix}\)
    • \(\mathbf{r}_2 = \begin{pmatrix}-1\\6\\2\end{pmatrix} + 4\begin{pmatrix}1\\-1.5\\2\end{pmatrix} = \begin{pmatrix}3\\0\\10\end{pmatrix}\)
    • \(\mathbf{r}_1 - \mathbf{r}_2 = \begin{pmatrix}-4\\3\\-1\end{pmatrix}\)
    • \(\sqrt{26} = 5.10\) m

  2. Hiker R starts at \(9\mathbf{i} - 2\mathbf{j}\) km from a hut and walks with velocity \(-3\mathbf{i} + 4\mathbf{j}\) km h⁻¹. Hiker S starts at \(-6\mathbf{i} + 10\mathbf{j}\) km and walks with velocity \(2\mathbf{i} - \mathbf{j}\) km h⁻¹. The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) point east and north.

    1. (a) Find expressions for the position vectors of R and S after \(t\) hours.

      • \(\mathbf{r}_R = (9\mathbf{i} - 2\mathbf{j}) + t(-3\mathbf{i} + 4\mathbf{j})\)
      • \(\mathbf{r}_S = (-6\mathbf{i} + 10\mathbf{j}) + t(2\mathbf{i} - \mathbf{j})\)

    2. (b) Find the distance between R and S after 2 hours.

      • \(\mathbf{r}_R - \mathbf{r}_S = \begin{pmatrix}3\\6\end{pmatrix} - \begin{pmatrix}-2\\8\end{pmatrix} = \begin{pmatrix}5\\-2\end{pmatrix}\)
      • \(d = \sqrt{29} = 5.39\) km

    3. (c) Find the time at which R is due north of S.

      • Due north: the i-components (east positions) are equal.
      • \(9 - 3t = -6 + 2t \Rightarrow t = 3\)
      • Check: at t = 3 the north positions are 10 and 7, so R is north of S.

  3. Two seagulls move in straight lines. Their position vectors at time \(t\) seconds are \(\mathbf{r}_1 = \begin{pmatrix}-10\\8\\-1\end{pmatrix} + t\begin{pmatrix}3\\-1\\2\end{pmatrix}\) and \(\mathbf{r}_2 = \begin{pmatrix}-8\\-2\\7\end{pmatrix} + t\begin{pmatrix}1\\2\\-1\end{pmatrix}\).

    1. (a) Show that the paths cross, and find the point where they cross.

      • \(\begin{pmatrix}-10\\8\\-1\end{pmatrix} + \lambda\begin{pmatrix}3\\-1\\2\end{pmatrix} = \begin{pmatrix}-8\\-2\\7\end{pmatrix} + \mu\begin{pmatrix}1\\2\\-1\end{pmatrix}\)
      • \(i: -10 + 3\lambda = -8 + \mu \qquad j: 8 - \lambda = -2 + 2\mu\)
      • \(\lambda = 2,\ \mu = 4\)
      • \(k: -1 + 2(2) = 3 \text{ and } 7 - 1(4) = 3\) Equal, so the paths cross at (−4, 6, 3)

    2. (b) Show that the objects do not collide.

      • The first object is at (−4, 6, 3) when t = 2. The second is there when t = 4.
      • The times differ, so they are never at the same point at the same time. No collision. (Equating with the same t gives different t from the i and j components.)

  4. Two robots move so that their position vectors at time \(t\) seconds are \(\mathbf{r}_1 = (1 + 2t)\mathbf{i} + (7 - 3t)\mathbf{j} + (-8 + 6t)\mathbf{k}\) and \(\mathbf{r}_2 = (-3 + 4t)\mathbf{i} + 1\mathbf{j} + (10 - 3t)\mathbf{k}\), where distances are in metres.

    1. (a) Find the speed of each robot.

      • First: \(\sqrt{2^2 + (-3)^2 + 6^2} = 7\)
      • Second: \(\sqrt{4^2 + 0^2 + (-3)^2} = 5\)

    2. (b) Determine whether the robots meet.

      • \(i: 1 + 2t = -3 + 4t \Rightarrow t = 2\)
      • \(j: 7 - 3t = 1 \Rightarrow t = 2\)
      • \(k: -8 + 6t = 10 - 3t \Rightarrow t = 2\)
      • The same t = 2 in all three components, so the robots meet at (5, 1, 4).

Stretch

  1. Two joggers run across a park. Jogger P starts at \(7\mathbf{i}\) km from a gate and runs with velocity \(-\mathbf{i} - 2\mathbf{j}\) km h⁻¹. Jogger Q starts at \(\mathbf{i} - 6\mathbf{j}\) km and runs with velocity \(3\mathbf{i}\) km h⁻¹. The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) point east and north. Time \(t\) is in hours.

    1. (a) Show that the distance between P and Q at time \(t\) is \(d = \sqrt{20t^2 - 72t + 72}\).

      • \(\mathbf{r}_Q - \mathbf{r}_P = \begin{pmatrix}-6 + 4t\\-6 + 2t\end{pmatrix}\)
      • \(d^2 = (-6 + 4t)^2 + (-6 + 2t)^2 = 20t^2 - 72t + 72\)
      • \(d = \sqrt{20t^2 - 72t + 72}\)

    2. (b) Use your calculator to find the least distance between them.

      • From the calculator: graph d(t) and find the minimum for t ≥ 0.
      • Minimum at t = 1.80, d = 2.68 km

    3. (c) Find the time at which Q is due east of P.

      • Due east: same j-component (north position).
      • \(-2t = -6 \Rightarrow t = 3\)
      • Check: at t = 3 the east positions are 4 (P) and 10 (Q), so Q is east of P.

  2. The position vectors of two paragliders at time \(t\) minutes are \(\mathbf{r}_1 = \begin{pmatrix}-3\\5\\1\end{pmatrix} + t\begin{pmatrix}2\\-1\\4\end{pmatrix}\) and \(\mathbf{r}_2 = \begin{pmatrix}6\\-4\\10\end{pmatrix} + t\begin{pmatrix}-1\\2\\1\end{pmatrix}\), with distances in m.

    1. (a) Show that the paragliders would collide, and find the position vector of the point of collision.

      • \(i: -3 + 2t = 6 - t \Rightarrow t = 3\)
      • \(j: 5 - t = -4 + 2t \Rightarrow t = 3\)
      • \(k: 1 + 4t = 10 + t \Rightarrow t = 3\)
      • All three give t = 3, so they collide.
      • \(\mathbf{r}_1 = \begin{pmatrix}-3\\5\\1\end{pmatrix} + 3\begin{pmatrix}2\\-1\\4\end{pmatrix} = \begin{pmatrix}3\\2\\13\end{pmatrix}\)

    2. (b) After 2 minutes, the second paraglider changes its velocity to \(\begin{pmatrix}0\\3\\1\end{pmatrix}\) to avoid the collision. Find the distance between the two paragliders at the time they would have collided.

      • \(\text{At } t = 2:\ \mathbf{r}_2 = \begin{pmatrix}6\\-4\\10\end{pmatrix} + 2\begin{pmatrix}-1\\2\\1\end{pmatrix} = \begin{pmatrix}4\\0\\12\end{pmatrix}\)
      • \(\text{By } t = 3:\ \mathbf{r}_2 = \begin{pmatrix}4\\0\\12\end{pmatrix} + 1\begin{pmatrix}0\\3\\1\end{pmatrix} = \begin{pmatrix}4\\3\\13\end{pmatrix}\)
      • \(\begin{pmatrix}4\\3\\13\end{pmatrix} - \begin{pmatrix}3\\2\\13\end{pmatrix} = \begin{pmatrix}1\\1\\0\end{pmatrix}\)
      • \(\sqrt{2} = 1.41\) m

  3. Two moths move so that their position vectors at time \(t\) seconds are \(\mathbf{r}_1 = \begin{pmatrix}-2\\5\\2.5\end{pmatrix} + t\begin{pmatrix}1\\-2\\0.5\end{pmatrix}\) and \(\mathbf{r}_2 = \begin{pmatrix}4\\0.5\\1.2\end{pmatrix} + t\begin{pmatrix}-1\\-0.5\\0.4\end{pmatrix}\), in metres. The base vectors \(\mathbf{i}\) and \(\mathbf{j}\) are in the horizontal plane and \(\mathbf{k}\) points upwards. Call them the first and second moth.

    1. (a) Show that there is a time when the first moth is vertically above the second moth, and find this time.

      • Vertically above: the i and j components are equal.
      • \(i: -2 + t = 4 - t \Rightarrow t = 3\)
      • \(j: 5 - 2t = 0.5 - 0.5t \Rightarrow t = 3\)
      • Both give t = 3. At this time the first moth is above the second moth: k-components 4 and 2.4.

    2. (b) Find the distance between them at that time.

      • \(4 - 2.4 = 1.6\)
      • Distance 1.6 m (the horizontal positions are the same).

  4. Distances are in kilometres and time in hours. A hot-air balloon moves with constant velocity \(15\mathbf{i}\) km h⁻¹ and at \(t = 0\) it is at the origin. A drone is at the point (0, 0.9, −0.15) at \(t = 0\). It then moves with constant velocity in the direction of the vector \(20\mathbf{i} - 12\mathbf{j} + c\mathbf{k}\).

    1. (a) Given that the drone reaches the hot-air balloon, find the value of \(c\).

      • Let the velocity be λ times the direction vector, and let T be the time of meeting.
      • \(\text{j-components: } 0.9 - 12\lambda T = 0 \Rightarrow \lambda T = 0.075\)
      • \(\text{k-components: } -0.15 + c\lambda T = 0 \Rightarrow c = 2\)

    2. (b) Find the speed of the drone.

      • \(\text{i-components: } 20\lambda T = 15T \Rightarrow \lambda = 0.75\)
      • \(\text{Velocity} = 0.75\left(20\mathbf{i} - 12\mathbf{j} + 2\mathbf{k}\right)\)
      • \(\text{speed} = 0.75\sqrt{20^2 + (-12)^2 + (2)^2} = 17.6\) km h⁻¹

jmaths.xyz · AI HL · AHL 4.12 Mixed practice: kinematics with constant velocity