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

Kinematics with constant velocity Problem set A 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 km h⁻¹.

    −112345246810eastnorthv

    1. (a) A ferry: \(\mathbf{v} = \begin{pmatrix}4\\9\end{pmatrix}\)

      • Sketch: east 4, north 9
      • Angle from the north–south line: \(\tan\alpha = \dfrac{4}{9}\), \(\alpha = 24.0^\circ\)
      • Bearing = 24.0°
      • 24° to the nearest degree

    2. (b) A glider: \(\mathbf{v} = -7\mathbf{i} + 3\mathbf{j}\)

      • Sketch: east −7, north 3
      • Angle from the north–south line: \(\tan\alpha = \dfrac{7}{3}\), \(\alpha = 66.8^\circ\)
      • Bearing = 360° − 66.8° = 293.2°
      • 293° 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 skater: \(\mathbf{v} = \begin{pmatrix}8\\-5\end{pmatrix}\)

      • \(|\mathbf{v}| = \sqrt{8^2 + (-5)^2} = \sqrt{89} = 9.43\)
      • Speed = √89 = 9.43 m s⁻¹

    2. (b) A drone: \(\mathbf{v} = 4\mathbf{i} + 7\mathbf{j} - 4\mathbf{k}\)

      • \(|\mathbf{v}| = \sqrt{4^2 + 7^2 + (-4)^2} = 9\)
      • Speed = 9 m s⁻¹

    3. (c) A kite: \(\mathbf{v} = \begin{pmatrix}3\\-1\\5\end{pmatrix}\)

      • \(|\mathbf{v}| = \sqrt{3^2 + (-1)^2 + 5^2} = \sqrt{35} = 5.92\)
      • Speed = √35 = 5.92 m s⁻¹

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

    1. (a) Direction \(\begin{pmatrix}8\\15\end{pmatrix}\), speed 51 m s⁻¹

      • \(|\mathbf{d}| = 17\)
      • \(\mathbf{v} = \dfrac{51}{17}\,\begin{pmatrix}8\\15\end{pmatrix} = \begin{pmatrix}24\\45\end{pmatrix}\)

    2. (b) Direction \(\mathbf{i} + 4\mathbf{j} + 8\mathbf{k}\), speed 27 m s⁻¹

      • \(|\mathbf{d}| = 9\)
      • \(\mathbf{v} = \dfrac{27}{9}\,(\mathbf{i} + 4\mathbf{j} + 8\mathbf{k}) = 3\mathbf{i} + 12\mathbf{j} + 24\mathbf{k}\)

    3. (c) A drone starts at the point (−4, 0, 9) and flies at 35 m s⁻¹ in the direction of \(6\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}\). Write down its position vector at time \(t\) seconds.

      • \(|\mathbf{d}| = 7\)
      • \(\mathbf{v} = 35\div 7\times \begin{pmatrix}6\\2\\3\end{pmatrix} = \begin{pmatrix}30\\10\\15\end{pmatrix}\)
      • \(\mathbf{r} = \begin{pmatrix}-4\\0\\9\end{pmatrix} + t\begin{pmatrix}30\\10\\15\end{pmatrix}\)

  4. A particle moves with constant velocity \(\mathbf{v} = \begin{pmatrix}2\\-1.5\\0.5\end{pmatrix}\) m s⁻¹. At \(t = 0\) it is at the point (−3, 6, 2) (distances in metres, time in seconds).

    1. (a) Find the speed.

      • \(|\mathbf{v}| = \sqrt{2^2 + (-1.5)^2 + 0.5^2} = \sqrt{6.5} = 2.55\)
      • Speed = √6.5 = 2.55 m s⁻¹

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

      • \(\mathbf{r} = \begin{pmatrix}-3\\6\\2\end{pmatrix} + t\begin{pmatrix}2\\-1.5\\0.5\end{pmatrix}\)

    3. (c) Does it pass through the point (5, 0, 4)? Give a reason.

      • \(x: -3 + 2t = 5 \Rightarrow t = 4\)
      • \(y: 6 - 1.5(4) = 0\)
      • \(z: 2 + 0.5(4) = 4\)
      • All three components match at the same t, so it passes through the point.

  5. A weather balloon moves so that its position vector at time \(t\) seconds is \(\mathbf{r} = \begin{pmatrix}-2\\7\\4\end{pmatrix} + t\begin{pmatrix}3\\-2\\6\end{pmatrix}\), where distances are in metres.

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

      • Put t = 0: \(\mathbf{r} = \begin{pmatrix}-2\\7\\4\end{pmatrix}\)
      • Initial position (−2, 7, 4)

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

      • \(|\mathbf{v}| = \sqrt{3^2 + (-2)^2 + 6^2} = 7\)
      • Speed = 7 m s⁻¹

    3. (c) Find the distance of the object from the origin after 4 seconds.

      • \(\mathbf{r} = \begin{pmatrix}-2\\7\\4\end{pmatrix} + 4\begin{pmatrix}3\\-2\\6\end{pmatrix} = \begin{pmatrix}10\\-1\\28\end{pmatrix}\)
      • \(|\mathbf{r}| = \sqrt{885} = 29.7\) m

In context

  1. Two balloons have position vectors \(\mathbf{r}_1 = \begin{pmatrix}2\\-4\\1\end{pmatrix} + t\begin{pmatrix}1\\3\\0.5\end{pmatrix}\) and \(\mathbf{r}_2 = \begin{pmatrix}-5\\2\\4\end{pmatrix} + t\begin{pmatrix}2\\-1\\-0.5\end{pmatrix}\) at time \(t\) hours, where distances are in km. Find the distance between the balloons when \(t = 3\).

    • \(\mathbf{r}_1 = \begin{pmatrix}2\\-4\\1\end{pmatrix} + 3\begin{pmatrix}1\\3\\0.5\end{pmatrix} = \begin{pmatrix}5\\5\\2.5\end{pmatrix}\)
    • \(\mathbf{r}_2 = \begin{pmatrix}-5\\2\\4\end{pmatrix} + 3\begin{pmatrix}2\\-1\\-0.5\end{pmatrix} = \begin{pmatrix}1\\-1\\2.5\end{pmatrix}\)
    • \(\mathbf{r}_1 - \mathbf{r}_2 = \begin{pmatrix}4\\6\\0\end{pmatrix}\)
    • \(\sqrt{52} = 7.21\) km

  2. Ship P starts at \(-12\mathbf{i} + 5\mathbf{j}\) km and moves with velocity \(6\mathbf{i} + 2\mathbf{j}\) km h⁻¹. Ship Q starts at \(3\mathbf{i} - 8\mathbf{j}\) km and moves with velocity \(\mathbf{i} - 3\mathbf{j}\) km h⁻¹. The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) point east and north.

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

      • \(\mathbf{r}_P = (-12\mathbf{i} + 5\mathbf{j}) + t(6\mathbf{i} + 2\mathbf{j})\)
      • \(\mathbf{r}_Q = (3\mathbf{i} - 8\mathbf{j}) + t(\mathbf{i} - 3\mathbf{j})\)

    2. (b) Find the distance between P and Q after 2 hours.

      • \(\mathbf{r}_P - \mathbf{r}_Q = \begin{pmatrix}0\\9\end{pmatrix} - \begin{pmatrix}5\\-14\end{pmatrix} = \begin{pmatrix}-5\\23\end{pmatrix}\)
      • \(d = \sqrt{554} = 23.5\) km

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

      • Due north: the i-components (east positions) are equal.
      • \(-12 + 6t = 3 + t \Rightarrow t = 3\)
      • Check: at t = 3 the north positions are 11 and −17, so P is north of Q.

  3. Two submarines move in straight lines. Their position vectors at time \(t\) minutes are \(\mathbf{r}_1 = \begin{pmatrix}2\\-6\\8\end{pmatrix} + t\begin{pmatrix}2\\1\\-1\end{pmatrix}\) and \(\mathbf{r}_2 = \begin{pmatrix}13\\-13\\0\end{pmatrix} + t\begin{pmatrix}-1\\2\\1\end{pmatrix}\).

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

      • \(\begin{pmatrix}2\\-6\\8\end{pmatrix} + \lambda\begin{pmatrix}2\\1\\-1\end{pmatrix} = \begin{pmatrix}13\\-13\\0\end{pmatrix} + \mu\begin{pmatrix}-1\\2\\1\end{pmatrix}\)
      • \(i: 2 + 2\lambda = 13 - \mu \qquad j: -6 + \lambda = -13 + 2\mu\)
      • \(\lambda = 3,\ \mu = 5\)
      • \(k: 8 - 1(3) = 5 \text{ and } 0 + 1(5) = 5\) Equal, so the paths cross at (8, −3, 5)

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

      • The first object is at (8, −3, 5) when t = 3. The second is there when t = 5.
      • 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 particles move so that their position vectors at time \(t\) seconds are \(\mathbf{r}_1 = (1 + 3t)\mathbf{i} + (-2 + t)\mathbf{j} + (7 + 2t)\mathbf{k}\) and \(\mathbf{r}_2 = (9 - t)\mathbf{i} + (-4 + 2t)\mathbf{j} + (4 + 2t)\mathbf{k}\), where distances are in metres.

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

      • First: \(\sqrt{3^2 + 1^2 + 2^2} = \sqrt{14} = 3.74\)
      • Second: \(\sqrt{(-1)^2 + 2^2 + 2^2} = 3\)

    2. (b) Determine whether the particles meet.

      • \(i: 1 + 3t = 9 - t \Rightarrow t = 2\)
      • \(j: -2 + t = -4 + 2t \Rightarrow t = 2\)
      • \(k: 7 + 2t = 4 + 2t\) has no solution (the t terms cancel and 7 ≠ 4).
      • The same t works for i and j but not k, so the particles do not meet.

Stretch

  1. Two helicopters search a coastline. Helicopter P starts at \(2\mathbf{i} - 8\mathbf{j}\) km and moves with velocity \(2\mathbf{i}\) km per minute. Helicopter Q starts at \(6\mathbf{i} + \mathbf{j}\) km and moves with velocity \(3\mathbf{i} - 3\mathbf{j}\) km per minute. The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) point east and north. Time \(t\) is in minutes.

    1. (a) Show that the distance between P and Q at time \(t\) is \(d = \sqrt{10t^2 - 46t + 97}\).

      • \(\mathbf{r}_Q - \mathbf{r}_P = \begin{pmatrix}4 + t\\9 - 3t\end{pmatrix}\)
      • \(d^2 = (4 + t)^2 + (9 - 3t)^2 = 10t^2 - 46t + 97\)
      • \(d = \sqrt{10t^2 - 46t + 97}\)

    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 = 2.30, d = 6.64 km

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

      • Due east: same j-component (north position).
      • \(-8 = 1 - 3t \Rightarrow t = 3\)
      • Check: at t = 3 the east positions are 8 (P) and 15 (Q), so Q is east of P.

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

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

      • \(i: 1 + 3t = 5 + t \Rightarrow t = 2\)
      • \(j: -2 + 2t = 4 - t \Rightarrow t = 2\)
      • \(k: 4 - t = -2 + 2t \Rightarrow t = 2\)
      • All three give t = 2, so they collide.
      • \(\mathbf{r}_1 = \begin{pmatrix}1\\-2\\4\end{pmatrix} + 2\begin{pmatrix}3\\2\\-1\end{pmatrix} = \begin{pmatrix}7\\2\\2\end{pmatrix}\)

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

      • \(\text{At } t = 1:\ \mathbf{r}_2 = \begin{pmatrix}5\\4\\-2\end{pmatrix} + 1\begin{pmatrix}1\\-1\\2\end{pmatrix} = \begin{pmatrix}6\\3\\0\end{pmatrix}\)
      • \(\text{By } t = 2:\ \mathbf{r}_2 = \begin{pmatrix}6\\3\\0\end{pmatrix} + 1\begin{pmatrix}2\\0\\1\end{pmatrix} = \begin{pmatrix}8\\3\\1\end{pmatrix}\)
      • \(\begin{pmatrix}8\\3\\1\end{pmatrix} - \begin{pmatrix}7\\2\\2\end{pmatrix} = \begin{pmatrix}1\\1\\-1\end{pmatrix}\)
      • \(\sqrt{3} = 1.73\) km

  3. Two hummingbirds move so that their position vectors at time \(t\) seconds are \(\mathbf{r}_1 = \begin{pmatrix}1\\2\\3.2\end{pmatrix} + t\begin{pmatrix}2\\1\\0.4\end{pmatrix}\) and \(\mathbf{r}_2 = \begin{pmatrix}7\\8\\0.5\end{pmatrix} + t\begin{pmatrix}0.5\\-0.5\\0.2\end{pmatrix}\), 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 hummingbird.

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

      • Vertically above: the i and j components are equal.
      • \(i: 1 + 2t = 7 + 0.5t \Rightarrow t = 4\)
      • \(j: 2 + t = 8 - 0.5t \Rightarrow t = 4\)
      • Both give t = 4. At this time the first hummingbird is above the second hummingbird: k-components 4.8 and 1.3.

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

      • \(4.8 - 1.3 = 3.5\)
      • Distance 3.5 m (the horizontal positions are the same).

  4. Distances are in kilometres and time in hours. A container ship moves with constant velocity \(36\mathbf{i}\) km h⁻¹ and at \(t = 0\) it is at the origin. A submarine is at the point (0, 0.6, −0.03) at \(t = 0\). It then moves with constant velocity in the direction of the vector \(30\mathbf{i} - 20\mathbf{j} + c\mathbf{k}\).

    1. (a) Given that the submarine reaches the container ship, find the value of \(c\).

      • Let the velocity be λ times the direction vector, and let T be the time of meeting.
      • \(\text{j-components: } 0.6 - 20\lambda T = 0 \Rightarrow \lambda T = 0.03\)
      • \(\text{k-components: } -0.03 + c\lambda T = 0 \Rightarrow c = 1\)

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

      • \(\text{i-components: } 30\lambda T = 36T \Rightarrow \lambda = 1.2\)
      • \(\text{Velocity} = 1.2\left(30\mathbf{i} - 20\mathbf{j} + \mathbf{k}\right)\)
      • \(\text{speed} = 1.2\sqrt{30^2 + (-20)^2 + (1)^2} = 43.3\) km h⁻¹

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