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IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.23

Variable velocity in two dimensions Problem set A Worked solutions

Skills

  1. Find the acceleration vector at the given time. Velocities are in m s⁻¹, time in seconds.

    1. (a) \(\mathbf{v} = \begin{pmatrix}5t^{2} - 4t\\9 - 2t^{3}\end{pmatrix}\), at \(t = 2\)

      • \(\mathbf{a} = \dfrac{d\mathbf{v}}{dt} = \begin{pmatrix}10t - 4\\-6t^{2}\end{pmatrix}\)
      • \(\mathbf{a}(2) = \begin{pmatrix}16\\-24\end{pmatrix}\) m s⁻²

    2. (b) \(\mathbf{v} = \begin{pmatrix}7e^{3t}\\12e^{-2t}\end{pmatrix}\), at \(t = 0.5\)

      • \(\mathbf{a} = \dfrac{d\mathbf{v}}{dt} = \begin{pmatrix}21e^{3t}\\-24e^{-2t}\end{pmatrix}\)
      • \(\mathbf{a}(0.5) = \begin{pmatrix}94.1\\-8.83\end{pmatrix}\) m s⁻²

    3. (c) \(\mathbf{v} = \begin{pmatrix}8\sin 4t\\6\cos 4t\end{pmatrix}\), at \(t = 1.2\)

      • \(\mathbf{a} = \dfrac{d\mathbf{v}}{dt} = \begin{pmatrix}32\cos 4t\\-24\sin 4t\end{pmatrix}\)
      • From the calculator, in radians: \(\mathbf{a}(1.2) = \begin{pmatrix}2.80\\23.9\end{pmatrix}\) m s⁻²

  2. Find an expression for the position vector \(\mathbf{r}\) in metres. Velocities are in m s⁻¹, and \(\mathbf{r}_0\) is the position at \(t = 0\).

    1. (a) \(\mathbf{v} = \begin{pmatrix}-t^{2} + 6t\\4t^{3} + 2\end{pmatrix},\ \mathbf{r}_0 = \begin{pmatrix}1\\-3\end{pmatrix}\)

      • \(\mathbf{r} = \int \mathbf{v}\,dt = \begin{pmatrix}-\frac{t^{3}}{3} + 3t^{2} + c_1\\t^{4} + 2t + c_2\end{pmatrix}\)
      • \(c_1 = 1,\ c_2 = −3\)
      • \(\mathbf{r} = \begin{pmatrix}-\frac{t^{3}}{3} + 3t^{2} + 1\\t^{4} + 2t - 3\end{pmatrix}\)

    2. (b) \(\mathbf{v} = \begin{pmatrix}9e^{3t}\\8e^{-2t}\end{pmatrix},\ \mathbf{r}_0 = \begin{pmatrix}2\\5\end{pmatrix}\)

      • \(\mathbf{r} = \int \mathbf{v}\,dt = \begin{pmatrix}3e^{3t} + c_1\\-4e^{-2t} + c_2\end{pmatrix}\)
      • \(c_1 = −1,\ c_2 = 9\)
      • \(\mathbf{r} = \begin{pmatrix}3e^{3t} - 1\\9 - 4e^{-2t}\end{pmatrix}\)

    3. (c) \(\mathbf{v} = \begin{pmatrix}6\cos 3t\\8\sin 2t\end{pmatrix},\ \mathbf{r}_0 = \begin{pmatrix}-1\\3\end{pmatrix}\)

      • \(\mathbf{r} = \int \mathbf{v}\,dt = \begin{pmatrix}2\sin 3t + c_1\\-4\cos 2t + c_2\end{pmatrix}\)
      • \(c_1 = −1,\ c_2 = 7\)
      • \(\mathbf{r} = \begin{pmatrix}2\sin 3t - 1\\7 - 4\cos 2t\end{pmatrix}\)

  3. Find the speed at the given time. Velocities are in m s⁻¹, time in seconds. Your calculator should be in radians.

    1. (a) \(\mathbf{v} = \begin{pmatrix}3t^{2} - 8t\\2t + 5\end{pmatrix}\), at \(t = 4\)

      • \(\mathbf{v}(4) = \begin{pmatrix}16\\13\end{pmatrix}\)
      • \(|\mathbf{v}| = \sqrt{16^2 + 13^2} = 20.6\) m s⁻¹

    2. (b) \(\mathbf{v} = \begin{pmatrix}6e^{2t}\\9e^{-t}\end{pmatrix}\), at \(t = 0.7\)

      • \(\mathbf{v}(0.7) = \begin{pmatrix}24.3\\4.47\end{pmatrix}\)
      • \(|\mathbf{v}| = \sqrt{24.3^2 + 4.47^2} = 24.7\) m s⁻¹

    3. (c) \(\mathbf{v} = \begin{pmatrix}10\cos 3t\\7\sin 3t\end{pmatrix}\), at \(t = 2\)

      • From the calculator, in radians: \(\mathbf{v}(2) = \begin{pmatrix}9.60\\-1.96\end{pmatrix}\)
      • \(|\mathbf{v}| = \sqrt{9.60^2 + (−1.96)^2} = 9.80\) m s⁻¹

  4. A diver's centre of mass moves in two dimensions. Its position vector is \(\mathbf{r} = \begin{pmatrix}t^{3} - 6t\\5\sin 2t\end{pmatrix}\) metres at time \(t\) seconds. Find its speed when \(t = 1.5\).

    • \(\mathbf{v} = \dfrac{d\mathbf{r}}{dt} = \begin{pmatrix}3t^{2} - 6\\10\cos 2t\end{pmatrix}\)
    • From the calculator, in radians: \(\mathbf{v}(1.5) = \begin{pmatrix}0.750\\-9.90\end{pmatrix}\)
    • \(|\mathbf{v}| = 9.93\) m s⁻¹

  5. A model rocket moves in a vertical plane. Its velocity at time \(t\) seconds is \(\mathbf{v} = (6 + 3t)\mathbf{i} + (14 - 5t)\mathbf{j}\) m s⁻¹.

    1. (a) Find the angle the initial velocity makes with \(\mathbf{i}\).

      • \(\mathbf{v}(0) = 6\mathbf{i} + 14\mathbf{j}\)
      • \(\tan\theta = \dfrac{14}{6}\)
      • \(\theta = 66.8^\circ\)

    2. (b) Show that the acceleration is constant and find its magnitude.

      • \(\mathbf{a} = \dfrac{d\mathbf{v}}{dt} = 3\mathbf{i} - 5\mathbf{j}\)
      • The acceleration has no t, so it is constant.
      • \(|\mathbf{a}| = \sqrt{3^2 + (−5)^2} = 5.83\) m s⁻²

  6. Find the distance from the origin at the given time. Displacements are in metres, time in seconds.

    1. (a) \(\mathbf{r} = \begin{pmatrix}2t^{2} - t\\9 - t^{3}\end{pmatrix}\), at \(t = 3\)

      • \(\mathbf{r}(3) = \begin{pmatrix}15\\-18\end{pmatrix}\)
      • \(|\mathbf{r}| = \sqrt{15^2 + (−18)^2} = 23.4\) m

    2. (b) \(\mathbf{r} = \begin{pmatrix}-t^{2} + 4t\\t^{3} - 12\end{pmatrix}\), at \(t = 2\)

      • Substitute: \(\mathbf{r}(2) = \begin{pmatrix}4\\-4\end{pmatrix}\)
      • \(|\mathbf{r}| = 5.66\) m

In context

  1. A boat leaves a harbour region and its position is measured in metres east and north of a marker at the origin. Its velocity is \(\mathbf{v} = \begin{pmatrix}2t - 5\\6e^{-t}\end{pmatrix}\) m s⁻¹, and at \(t = 0\) it is at \(\mathbf{r}_0 = \begin{pmatrix}1\\-2\end{pmatrix}\) m. Find, when \(t = 2\):

    1. (a) the acceleration

      • \(\mathbf{a} = \begin{pmatrix}2\\-6e^{-t}\end{pmatrix}\)
      • \(\mathbf{a}(2) = \begin{pmatrix}2\\-0.812\end{pmatrix}\)

    2. (b) the position vector

      • \(\mathbf{r} = \int \mathbf{v}\,dt = \begin{pmatrix}t^{2} - 5t + c_1\\-6e^{-t} + c_2\end{pmatrix}\)
      • \(c_1 = 1,\ c_2 = 4\)
      • \(\mathbf{r} = \begin{pmatrix}t^{2} - 5t + 1\\4 - 6e^{-t}\end{pmatrix}\)
      • \(\mathbf{r}(2) = \begin{pmatrix}-5\\3.19\end{pmatrix}\)

    3. (c) the distance from the origin

      • \(|\mathbf{r}(2)| = \sqrt{(−5)^2 + 3.19^2} = 5.93\) m

    4. (d) the speed

      • \(\mathbf{v}(2) = \begin{pmatrix}-1\\0.812\end{pmatrix}\)
      • \(|\mathbf{v}| = 1.29\) m s⁻¹

  2. A model plane flies on a control line. Its velocity is \(\mathbf{v} = \begin{pmatrix}12\cos 2t\\-12\sin 2t\end{pmatrix}\) m s⁻¹, and at \(t = 0\) it is at \(\mathbf{r}_0 = \begin{pmatrix}0\\6\end{pmatrix}\) m. Find, when \(t = 0.7\):

    1. (a) the speed

      • From the calculator, in radians: \(\mathbf{v}(0.7) = \begin{pmatrix}2.04\\-11.8\end{pmatrix}\)
      • \(|\mathbf{v}| = 12\) m s⁻¹

    2. (b) the position vector

      • \(\mathbf{r} = \int \mathbf{v}\,dt = \begin{pmatrix}6\sin 2t + c_1\\6\cos 2t + c_2\end{pmatrix}\)
      • \(c_1 = 0,\ c_2 = 0\)
      • \(\mathbf{r} = \begin{pmatrix}6\sin 2t\\6\cos 2t\end{pmatrix}\)
      • From the calculator, in radians: \(\mathbf{r}(0.7) = \begin{pmatrix}5.91\\1.02\end{pmatrix}\)

    3. (c) the distance from the origin. What do you notice about your answers and about the shape of the path?

      • \(|\mathbf{r}| = 6\) m
      • Check another time: both stay the same.
      • Constant speed and constant distance from the origin: motion in a circle.

Stretch

  1. A shuttlecock's horizontal and vertical velocity are modelled by two components. Its velocity is \(\mathbf{v} = \begin{pmatrix}t^{2} - 4t\\3t - 6\end{pmatrix}\) m s⁻¹ for \(t \ge 0\). Use your calculator to find the least speed and the time at which it happens.

    • \(\text{speed} = \sqrt{(t^{2} - 4t)^2 + (3t - 6)^2}\)
    • Graph the speed against t for t ≥ 0 and use the minimum tool.
    • Minimum: \(t = 2.00\) s, speed 4 m s⁻¹

  2. A particle moves in the plane. Its velocity is \(\mathbf{v} = \begin{pmatrix}6\cos 2t\\-8\sin 2t\end{pmatrix}\) m s⁻¹. Find the first time \(t > 0\) at which the particle moves parallel to \(\mathbf{j}\), and its speed then.

    • \(\mathbf{v}\) is parallel to \(\mathbf{j}\) when the i-component is zero.
    • \(6\cos 2t = 0\) gives \(2t = \dfrac{\pi}{2}\), so \(t = 0.785\) s.
    • From the calculator, in radians: \(|\mathbf{v}(0.785)| = 8\) m s⁻¹

jmaths.xyz · AI HL · AHL 4.23 Variable velocity in two dimensions