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

Kinematics with constant velocity Problem set B Worked solutions

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.

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    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