Variable velocity in two dimensions Problem set B One question at a time

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

1. (a) \(\mathbf{v} = \begin{pmatrix}4t^{2} - 7t\\5 - 3t^{3}\end{pmatrix}\), at \(t = 3\)

   • \(\mathbf{a} = \dfrac{d\mathbf{v}}{dt} = \begin{pmatrix}8t - 7\\-9t^{2}\end{pmatrix}\)
   • \(\mathbf{a}(3) = \begin{pmatrix}17\\-81\end{pmatrix}\) m s⁻²

2. (b) \(\mathbf{v} = \begin{pmatrix}5e^{4t}\\9e^{-3t}\end{pmatrix}\), at \(t = 0.4\)

   • \(\mathbf{a} = \dfrac{d\mathbf{v}}{dt} = \begin{pmatrix}20e^{4t}\\-27e^{-3t}\end{pmatrix}\)
   • \(\mathbf{a}(0.4) = \begin{pmatrix}99.1\\-8.13\end{pmatrix}\) m s⁻²

3. (c) \(\mathbf{v} = \begin{pmatrix}5\cos 3t\\9\sin 3t\end{pmatrix}\), at \(t = 1.5\)

   • \(\mathbf{a} = \dfrac{d\mathbf{v}}{dt} = \begin{pmatrix}-15\sin 3t\\27\cos 3t\end{pmatrix}\)
   • From the calculator, in radians: \(\mathbf{a}(1.5) = \begin{pmatrix}14.7\\-5.69\end{pmatrix}\) m s⁻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) \(\mathbf{v} = \begin{pmatrix}-t^{2} + 8t\\3t^{2} + 4\end{pmatrix},\ \mathbf{r}_0 = \begin{pmatrix}-2\\5\end{pmatrix}\)

   • \(\mathbf{r} = \int \mathbf{v}\,dt = \begin{pmatrix}-\frac{t^{3}}{3} + 4t^{2} + c_1\\t^{3} + 4t + c_2\end{pmatrix}\)
   • \(c_1 = −2,\ c_2 = 5\)
   • \(\mathbf{r} = \begin{pmatrix}-\frac{t^{3}}{3} + 4t^{2} - 2\\t^{3} + 4t + 5\end{pmatrix}\)

2. (b) \(\mathbf{v} = \begin{pmatrix}8e^{2t}\\9e^{-3t}\end{pmatrix},\ \mathbf{r}_0 = \begin{pmatrix}1\\4\end{pmatrix}\)

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

3. (c) \(\mathbf{v} = \begin{pmatrix}10\sin 5t\\6\cos 2t\end{pmatrix},\ \mathbf{r}_0 = \begin{pmatrix}3\\-1\end{pmatrix}\)

   • \(\mathbf{r} = \int \mathbf{v}\,dt = \begin{pmatrix}-2\cos 5t + c_1\\3\sin 2t + c_2\end{pmatrix}\)
   • \(c_1 = 5,\ c_2 = −1\)
   • \(\mathbf{r} = \begin{pmatrix}5 - 2\cos 5t\\3\sin 2t - 1\end{pmatrix}\)

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

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

   • \(\mathbf{v}(5) = \begin{pmatrix}5\\23\end{pmatrix}\)
   • \(|\mathbf{v}| = \sqrt{5^2 + 23^2} = 23.5\) m s⁻¹

2. (b) \(\mathbf{v} = \begin{pmatrix}5e^{3t}\\11e^{-2t}\end{pmatrix}\), at \(t = 0.6\)

   • \(\mathbf{v}(0.6) = \begin{pmatrix}30.2\\3.31\end{pmatrix}\)
   • \(|\mathbf{v}| = \sqrt{30.2^2 + 3.31^2} = 30.4\) m s⁻¹

3. (c) \(\mathbf{v} = \begin{pmatrix}12\sin 2t\\5\cos 2t\end{pmatrix}\), at \(t = 1\)

   • From the calculator, in radians: \(\mathbf{v}(1) = \begin{pmatrix}10.9\\-2.08\end{pmatrix}\)
   • \(|\mathbf{v}| = \sqrt{10.9^2 + (−2.08)^2} = 11.1\) m s⁻¹

A bee flies in two dimensions in a greenhouse. Its position vector is \(\mathbf{r} = \begin{pmatrix}2t^{2} - 5t\\8\cos 3t\end{pmatrix}\) metres at time \(t\) seconds. Find its speed when \(t = 2\).

• \(\mathbf{v} = \dfrac{d\mathbf{r}}{dt} = \begin{pmatrix}4t - 5\\-24\sin 3t\end{pmatrix}\)
• From the calculator, in radians: \(\mathbf{v}(2) = \begin{pmatrix}3\\6.71\end{pmatrix}\)
• \(|\mathbf{v}| = 7.35\) m s⁻¹

A hot-air balloon is modelled in a vertical plane. Its velocity at time \(t\) seconds is \(\mathbf{v} = (8 + 2t)\mathbf{i} + (9 - 6t)\mathbf{j}\) m s⁻¹.

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

   • \(\mathbf{v}(0) = 8\mathbf{i} + 9\mathbf{j}\)
   • \(\tan\theta = \dfrac{9}{8}\)
   • \(\theta = 48.4^\circ\)

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

   • \(\mathbf{a} = \dfrac{d\mathbf{v}}{dt} = 2\mathbf{i} - 6\mathbf{j}\)
   • The acceleration has no t, so it is constant.
   • \(|\mathbf{a}| = \sqrt{2^2 + (−6)^2} = 6.32\) m s⁻²

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

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

   • \(\mathbf{r}(2) = \begin{pmatrix}14\\-8\end{pmatrix}\)
   • \(|\mathbf{r}| = \sqrt{14^2 + (−8)^2} = 16.1\) m

2. (b) \(\mathbf{r} = \begin{pmatrix}t^{2} - 7\\-t^{3} + 3t\end{pmatrix}\), at \(t = 3\)

   • Substitute: \(\mathbf{r}(3) = \begin{pmatrix}2\\-18\end{pmatrix}\)
   • \(|\mathbf{r}| = 18.1\) m

A robot vacuum cleaner moves on a floor. Its position is measured in metres east and north of the charging point at the origin. Its velocity is \(\mathbf{v} = \begin{pmatrix}3t - 4\\8e^{-2t}\end{pmatrix}\) m s⁻¹, and at \(t = 0\) it is at \(\mathbf{r}_0 = \begin{pmatrix}-1\\3\end{pmatrix}\) m. Find, when \(t = 1.5\):

1. (a) the acceleration

   • \(\mathbf{a} = \begin{pmatrix}3\\-16e^{-2t}\end{pmatrix}\)
   • \(\mathbf{a}(1.5) = \begin{pmatrix}3\\-0.797\end{pmatrix}\)

2. (b) the position vector

   • \(\mathbf{r} = \int \mathbf{v}\,dt = \begin{pmatrix}\frac{3t^{2}}{2} - 4t + c_1\\-4e^{-2t} + c_2\end{pmatrix}\)
   • \(c_1 = −1,\ c_2 = 7\)
   • \(\mathbf{r} = \begin{pmatrix}\frac{3t^{2}}{2} - 4t - 1\\7 - 4e^{-2t}\end{pmatrix}\)
   • \(\mathbf{r}(1.5) = \begin{pmatrix}-3.63\\6.80\end{pmatrix}\)

3. (c) the distance from the origin

   • \(|\mathbf{r}(1.5)| = \sqrt{(−3.63)^2 + 6.80^2} = 7.71\) m

4. (d) the speed

   • \(\mathbf{v}(1.5) = \begin{pmatrix}0.500\\0.398\end{pmatrix}\)
   • \(|\mathbf{v}| = 0.639\) m s⁻¹

A model train runs on a track. Its velocity is \(\mathbf{v} = \begin{pmatrix}-15\sin 3t\\15\cos 3t\end{pmatrix}\) m s⁻¹, and at \(t = 0\) it is at \(\mathbf{r}_0 = \begin{pmatrix}5\\0\end{pmatrix}\) m. Find, when \(t = 0.9\):

1. (a) the speed

   • From the calculator, in radians: \(\mathbf{v}(0.9) = \begin{pmatrix}-6.41\\-13.6\end{pmatrix}\)
   • \(|\mathbf{v}| = 15\) m s⁻¹

2. (b) the position vector

   • \(\mathbf{r} = \int \mathbf{v}\,dt = \begin{pmatrix}5\cos 3t + c_1\\5\sin 3t + c_2\end{pmatrix}\)
   • \(c_1 = 0,\ c_2 = 0\)
   • \(\mathbf{r} = \begin{pmatrix}5\cos 3t\\5\sin 3t\end{pmatrix}\)
   • From the calculator, in radians: \(\mathbf{r}(0.9) = \begin{pmatrix}-4.52\\2.14\end{pmatrix}\)

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

   • \(|\mathbf{r}| = 5\) m
   • Check another time: both stay the same.
   • Constant speed and constant distance from the origin: motion in a circle.

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

• \(\text{speed} = \sqrt{(t^{2} - 6t + 4)^2 + (2t - 3)^2}\)
• Graph the speed against t for t ≥ 0 and use the minimum tool.
• Minimum: \(t = 0.896\) s, speed 1.34 m s⁻¹

A particle moves in the plane. Its velocity is \(\mathbf{v} = \begin{pmatrix}9\cos 3t\\5\sin 3t\end{pmatrix}\) 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.
• \(9\cos 3t = 0\) gives \(3t = \dfrac{\pi}{2}\), so \(t = 0.524\) s.
• From the calculator, in radians: \(|\mathbf{v}(0.524)| = 5\) m s⁻¹