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IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.28

Projectiles, circular motion and delayed objects 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. A ball is kicked from the origin. Take \(\mathbf{i}\) as horizontal and \(\mathbf{j}\) as vertically up. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix}12\\19.6 − 9.8t\end{pmatrix}\) m s⁻¹. It starts at the origin.

    1. (a) Find the angle that the velocity makes with the horizontal (i) at the start (ii) when \(t = 3\).

      • Initially: \(\tan\theta = \dfrac{19.6}{12}\), so \(\theta = 58.5^\circ\)
      • At \(t = 3\): \(v_y = 19.6 − 9.8 \times 3 = −9.8\)
      • \(\tan\theta = \dfrac{−9.8}{12}\), so \(\theta = −39.2^\circ\), which is below the horizontal.

    2. (b) Find the time at which it reaches its greatest height.

      • At the top the vertical velocity is zero.
      • \(19.6 − 9.8t = 0\)
      • \(t = \dfrac{19.6}{9.8} = 2\)

    3. (c) Find an expression for the displacement from the origin at time \(t\). Hence find the greatest height.

      • Integrate with \(\mathbf{r}(0) = \mathbf{0}\): \(\mathbf{r} = \begin{pmatrix}12t\\19.6t − 4.9t^2\end{pmatrix}\)
      • Vertical displacement at \(t = 2\): \(y = 19.6\times2 − 4.9\times2^2 = 19.6\)
      • Greatest height 19.6 m

  2. A golf ball is hit. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix}13\\14.7 − 9.8t\end{pmatrix}\) m s⁻¹, where the origin is the launch point on level ground. Find the distance from the origin to the point where it lands on the same level ground.

    • \(\mathbf{r} = \begin{pmatrix}13t\\14.7t − 4.9t^2\end{pmatrix}\)
    • On the ground \(y = 0\): \(t(14.7 − 4.9t) = 0\), so \(t = 3\)
    • \(x = 13\times3 = 39\)
    • The point is on the ground, so the distance from the origin is 39 m

  3. A bead slides on a wire. Its position vector relative to the origin at time \(t\) seconds is \(\mathbf{r} = \begin{pmatrix}3\cos 6t\\3\sin 6t\end{pmatrix}\) m.

    1. (a) By finding \(|\mathbf{r}|\), show that the object moves in a circle. State the radius.

      • \(|\mathbf{r}| = \sqrt{3^2\cos^2 6t + 3^2\sin^2 6t}\)
      • \(= 3\sqrt{\cos^2 6t + \sin^2 6t}\)\( = 3\), using \(\sin^2\theta + \cos^2\theta = 1\)
      • The distance from the origin is constant, so the object moves in a circle of radius 3 m.

    2. (b) Find the velocity vector.

      • Differentiate each component: \(\mathbf{v} = \begin{pmatrix}-18\sin 6t\\18\cos 6t\end{pmatrix}\)
      • The derivative of \(\cos 6t\) is \(−6\sin 6t\): the \(6\) comes out in front.

    3. (c) Show that the velocity is always perpendicular to the position vector.

      • \(\mathbf{v}\cdot\mathbf{r} = (3\cos 6t)(−18\sin 6t) + (3\sin 6t)(18\cos 6t)\)
      • \(= −54\sin 6t\cos 6t + 54\sin 6t\cos 6t = 0\)
      • The scalar product is zero, so \(\mathbf{v}\) is perpendicular to \(\mathbf{r}\).

    4. (d) Show that the speed is constant and find its value.

      • \(|\mathbf{v}| = \sqrt{18^2\sin^2 6t + 18^2\cos^2 6t} = 18\)
      • The speed does not depend on \(t\), so it is constant: 18 m s⁻¹.

  4. Two objects move in circles about the origin. Find the speed of each, in m s⁻¹.

    1. (a) A fan blade tip: \(\mathbf{v} = \begin{pmatrix}-1.5\sin 4t\\1.5\cos 4t\end{pmatrix}\)

      • \(|\mathbf{v}| = \sqrt{1.5^2\sin^2 4t + 1.5^2\cos^2 4t}\)
      • \(= 1.5\) since \(\sin^2 + \cos^2 = 1\)

    2. (b) A pebble on a turntable: \(\mathbf{v} = \begin{pmatrix}8\cos 0.25t\\8\sin 0.25t\end{pmatrix}\)

      • \(|\mathbf{v}| = \sqrt{8^2\cos^2 0.25t + 8^2\sin^2 0.25t}\)
      • \(= 8\) since \(\sin^2 + \cos^2 = 1\)

  5. A weight on a spring-driven arm moves in a circle. Its position vector at time \(t\) seconds is \(\mathbf{r} = \begin{pmatrix}2\cos 3t\\2\sin 3t\end{pmatrix}\) m.

    xyrOP

    1. (a) Find the velocity and then the acceleration.

      • \(\mathbf{v} = \begin{pmatrix}-6\sin 3t\\6\cos 3t\end{pmatrix}\)
      • \(\mathbf{a} = \begin{pmatrix}-18\cos 3t\\−18\sin 3t\end{pmatrix}\)

    2. (b) Show that \(\mathbf{a} = k\mathbf{r}\) and find \(k\).

      • \(\mathbf{a} = \begin{pmatrix}-18\cos 3t\\−18\sin 3t\end{pmatrix} = −9\begin{pmatrix}2\cos 3t\\2\sin 3t\end{pmatrix}\)
      • \(\mathbf{a} = −9\mathbf{r}\), so \(k = −9\)

    3. (c) State the direction of the acceleration.

      • \(k\) is negative, so \(\mathbf{a}\) points the opposite way to \(\mathbf{r}\).
      • The position vector points away from the centre, so the acceleration points towards the centre.

    4. (d) Show that the magnitude of the acceleration is constant and find its value.

      • \(|\mathbf{a}| = |−9|\times|\mathbf{r}| = 9\times 2\)
      • \(= 18\) m s⁻², which does not depend on \(t\).

  6. A marker on a roundabout moves in a circle. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix}3\cos 1.5t\\3\sin 1.5t\end{pmatrix}\) m s⁻¹. At \(t = 0\) the position vector is \(\begin{pmatrix}0\\−2\end{pmatrix}\) m.

    1. (a) Find the position vector at time \(t\), show that the object moves in a circle about the origin and state the radius.

      • Integrate: \(\mathbf{r} = \begin{pmatrix}2\sin 1.5t + c_1\\−2\cos 1.5t + c_2\end{pmatrix}\)
      • At \(t = 0\): \(\begin{pmatrix}c_1\\−2 + c_2\end{pmatrix}\) = \(\begin{pmatrix}0\\−2\end{pmatrix}\), so \(c_1 = 0\) and \(c_2 = 0\)
      • \(\mathbf{r} = \begin{pmatrix}2\sin 1.5t\\−2\cos 1.5t\end{pmatrix}\)
      • \(|\mathbf{r}| = 2\sqrt{\sin^2 1.5t + \cos^2 1.5t} = 2\): a circle about the origin, radius 2 m.

    2. (b) Write down the initial position and the radius of the circle.

      • At \(t = 0\): \(\mathbf{r} = \begin{pmatrix}0\\−2\end{pmatrix}\)
      • Radius \(2\) m from the constant \(|\mathbf{r}|\).

    3. (c) Find the time \(T\) when the object first returns to its starting position.

      • The angle \(1.5t\) must reach \(2\pi\).
      • \(1.5T = 2\pi\)
      • \(T = \dfrac{2\pi}{1.5} = 4.19\) s

    4. (d) Verify that the speed equals \(\dfrac{2\pi r}{T}\).

      • \(\dfrac{2\pi\times 2}{4.19} = \dfrac{2\pi\times 2}{2\pi / 1.5} = 2\times 1.5 = 3\)
      • \(|\mathbf{v}| = 3\), the same value: the speed equals \(\dfrac{2\pi r}{T}\).

  7. A remote-control car moves on a flat floor. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix}5 + 2t\\9 − 3t^2\end{pmatrix}\) m s⁻¹. It starts at the origin.

    1. (a) Find an expression for the displacement from the origin at time \(t\).

      • Integrate each component and use \(\mathbf{r}(0) = \mathbf{0}\) so the constants are zero.
      • \(\mathbf{r} = \begin{pmatrix}5t + t^2\\9t − t^3\end{pmatrix}\)

    2. (b) Find the distance from the origin when \(t = 2\).

      • \(\mathbf{r}(2) = \begin{pmatrix}14\\10\end{pmatrix}\)
      • \(|\mathbf{r}| = \sqrt{14^2 + 10^2} = 17.2\) m

In context

  1. A drone starts at the origin when \(t = 0\). Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix}7 − 2t\\3 + 0.6t^2\end{pmatrix}\) m s⁻¹. A second drone follows the same path with the same velocity but leaves the origin 2 seconds later. Find the distance between the two drones 3 seconds after the second one leaves.

    • \(\mathbf{r} = \begin{pmatrix}7t − t^2\\3t + 0.2t^3\end{pmatrix}\)
    • First drone: \(t = 5\), \(\mathbf{r} = \begin{pmatrix}10\\40\end{pmatrix}\)
    • Second drone has travelled 3 s: \(\mathbf{r} = \begin{pmatrix}12\\14.4\end{pmatrix}\)
    • \(d = \sqrt{(−2)^2 + (25.6)^2} = 25.7\) m

  2. A cannon fires a ball. The velocity of the first at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix}20\\14.7 − 9.8t\end{pmatrix}\) m s⁻¹. A second cannon fires another ball. The velocity of the second is \(\mathbf{v} = \begin{pmatrix}24\\b − 9.8t\end{pmatrix}\) m s⁻¹. Both start at the origin and land on level ground in the same place. Find the value of \(b\).

    • First: lands when \(y = 0\), at \(t = 3\), where \(x = 20\times3 = 60\)
    • Second: \(y = bt − 4.9t^2 = 0\) gives \(t = \dfrac{b}{4.9}\)
    • Then \(x = 24\times\dfrac{b}{4.9} = 60\)
    • \(b = \dfrac{60\times 4.9}{24} = 12.25\)

  3. A basketball is thrown from the origin. The acceleration is \(\begin{pmatrix}0\\−9.8\end{pmatrix}\) m s⁻² and the initial velocity is \(\begin{pmatrix}7\\21\end{pmatrix}\) m s⁻¹.

    1. (a) Find the displacement from the starting point at time \(t\) seconds after the launch.

      • Integrate the acceleration: \(\mathbf{v} = \begin{pmatrix}7\\21 − 9.8t\end{pmatrix}\) (using the initial velocity).
      • Integrate again with \(\mathbf{r}(0) = \mathbf{0}\): \(\mathbf{r} = \begin{pmatrix}7t\\21t − 4.9t^2\end{pmatrix}\)

    2. (b) A second identical launch is made 2 seconds after the first, from the same point. Find the displacement of the second object from the starting point at time \(t\) seconds after the first launch, for \(t \geq 2\).

      • The second object has been moving for \(t − 2\) seconds.
      • Replace \(t\) by \(t − 2\): \(\mathbf{r} = \begin{pmatrix}7(t − 2)\\21(t − 2) − 4.9(t − 2)^2\end{pmatrix}\)

  4. Two fireworks are launched. The first is launched from the origin at time \(t = 0\) with initial velocity \(\begin{pmatrix}8\\30\end{pmatrix}\) m s⁻¹. The second is launched from the origin 2 seconds later with initial velocity \(\begin{pmatrix}12\\b\end{pmatrix}\) m s⁻¹. Both move with acceleration \(\begin{pmatrix}0\\−9.8\end{pmatrix}\) m s⁻². The two collide.

    1. (a) Find the value of \(b\).

      • First: \(\mathbf{r} = \begin{pmatrix}8t\\30t − 4.9t^2\end{pmatrix}\)
      • Second: \(\mathbf{r} = \begin{pmatrix}12(t − 2)\\b(t − 2) − 4.9(t − 2)^2\end{pmatrix}\)
      • Horizontal: \(8t = 12(t − 2)\), so \(t = 6\)
      • Vertical at \(t = 6\): first \(y = 30\times6 − 4.9\times6^2 = 3.6\)
      • Second: \(b(4) − 4.9(4)^2 = 3.6\), so \(b = 20.5\)

    2. (b) Find the distance from the origin of the point where they collide.

      • They meet at \(x = 8\times6 = 48\) and \(y = 3.6\)
      • \(|\mathbf{r}| = \sqrt{48^2 + 3.6^2} = 48.1\) m

  5. A rocket is launched from the origin with initial velocity \(\begin{pmatrix}30\\50\end{pmatrix}\) m s⁻¹ and moves with acceleration \(\begin{pmatrix}0\\−9.8\end{pmatrix}\) m s⁻². A second rocket is launched from the origin with the same initial velocity 5 seconds later.

    1. (a) Find an expression for the distance between the two rockets \(t\) seconds after the first launch, for \(t \geq 5\).

      • First: \(\mathbf{r}_1 = \begin{pmatrix}30t\\50t − 4.9t^2\end{pmatrix}\). Second: \(\mathbf{r}_2 = \begin{pmatrix}30(t − 5)\\50(t − 5) − 4.9(t − 5)^2\end{pmatrix}\)
      • Horizontal difference: \(30t − 30(t − 5) = 150\)
      • Vertical difference: \(372.5 − 49t\)
      • \(d = \sqrt{150^2 + (372.5 − 49t)^2}\)

    2. (b) Use your calculator to find the time, after the first launch, at which the distance is least while both are in flight.

      • Graph \(d(t)\) for \(5 \leq t \leq 10.2\) (both are in flight until the first lands at 10.2 s).
      • From the calculator: the minimum is at \(t = 7.60\) s, where \(d = 150\) m.

Stretch

  1. A sensor on a rotating arm moves in a circle about the origin. Its position vector is \(\mathbf{r} = \begin{pmatrix}r\cos \omega t\\r\sin \omega t\end{pmatrix}\), where \(r\) and \(\omega\) are positive constants.

    1. (a) Show that it moves in a circle and state its radius.

      • \(|\mathbf{r}| = \sqrt{r^2\cos^2 \omega t + r^2\sin^2 \omega t} = r\)
      • Using \(\sin^2\theta + \cos^2\theta = 1\). The distance from the origin is constant, so the path is a circle of radius \(r\).

    2. (b) Show that its speed is \(v = r\omega\).

      • \(\mathbf{v} = \begin{pmatrix}-r\omega\sin \omega t\\r\omega\cos \omega t\end{pmatrix}\)
      • \(|\mathbf{v}| = \sqrt{r^2\omega^2\sin^2 \omega t + r^2\omega^2\cos^2 \omega t} = r\omega\)

    3. (c) Show that the time for one full circle is \(T = \dfrac{2\pi}{\omega}\).

      • One full circle: the angle \(\omega t\) increases by \(2\pi\).
      • \(\omega T = 2\pi\)
      • \(T = \dfrac{2\pi}{\omega}\)

    4. (d) Show that the magnitude of the acceleration is \(\dfrac{v^2}{r}\).

      • \(\mathbf{a} = \begin{pmatrix}-r\omega^2\cos \omega t\\-r\omega^2\sin \omega t\end{pmatrix}\)
      • \(|\mathbf{a}| = r\omega^2\)
      • Since \(v = r\omega\), \(\omega = \dfrac{v}{r}\), so \(|\mathbf{a}| = r\times\dfrac{v^2}{r^2} = \dfrac{v^2}{r}\)

    5. (e) For \(r = 1.2\) m and \(\omega = 2.5\) rad s⁻¹, find the speed, the time for one full circle and the magnitude of the acceleration.

      • \(v = r\omega = 1.2\times2.5 = 3\) m s⁻¹
      • \(T = \dfrac{2\pi}{2.5} = 2.51\) s
      • \(|\mathbf{a}| = \dfrac{3^2}{1.2} = 7.5\) m s⁻²

  2. A planet is modelled as moving in a circle about the origin. Its position vector is \(\mathbf{r} = \begin{pmatrix}r\cos \omega t\\r\sin \omega t\end{pmatrix}\), where \(r\) and \(\omega\) are positive constants.

    xyrOP

    1. (a) Show that the velocity is perpendicular to the position vector.

      • \(\mathbf{v} = \begin{pmatrix}-r\omega\sin \omega t\\r\omega\cos \omega t\end{pmatrix}\)
      • \(\mathbf{v}\cdot\mathbf{r} = -r^2\omega\sin \omega t\cos \omega t + r^2\omega\sin \omega t\cos \omega t = 0\)
      • The scalar product is zero, so the vectors are perpendicular.

    2. (b) Show that the acceleration is directed towards the centre of the circle.

      • \(\mathbf{a} = \begin{pmatrix}-r\omega^2\cos \omega t\\-r\omega^2\sin \omega t\end{pmatrix} = −\omega^2\begin{pmatrix}r\cos \omega t\\r\sin \omega t\end{pmatrix}\)
      • \(\mathbf{a} = −\omega^2\mathbf{r}\)
      • This is a negative multiple of \(\mathbf{r}\), so it points from the object towards the centre.

jmaths.xyz · AI HL · AHL 4.28 Mixed practice: projectiles, circular motion and delayed objects