Variable velocity in two dimensions Problem set A

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 \(t = 2\)
2. (b) \(\mathbf{v} = \begin{pmatrix}7e^{3t}\\12e^{-2t}\end{pmatrix}\), at \(t = 0.5\)
3. (c) \(\mathbf{v} = \begin{pmatrix}8\sin 4t\\6\cos 4t\end{pmatrix}\), at \(t = 1.2\)

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) \(\mathbf{v} = \begin{pmatrix}-t^{2} + 6t\\4t^{3} + 2\end{pmatrix},\ \mathbf{r}_0 = \begin{pmatrix}1\\-3\end{pmatrix}\)
2. (b) \(\mathbf{v} = \begin{pmatrix}9e^{3t}\\8e^{-2t}\end{pmatrix},\ \mathbf{r}_0 = \begin{pmatrix}2\\5\end{pmatrix}\)
3. (c) \(\mathbf{v} = \begin{pmatrix}6\cos 3t\\8\sin 2t\end{pmatrix},\ \mathbf{r}_0 = \begin{pmatrix}-1\\3\end{pmatrix}\)

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 \(t = 4\)
2. (b) \(\mathbf{v} = \begin{pmatrix}6e^{2t}\\9e^{-t}\end{pmatrix}\), at \(t = 0.7\)
3. (c) \(\mathbf{v} = \begin{pmatrix}10\cos 3t\\7\sin 3t\end{pmatrix}\), at \(t = 2\)

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 \(t = 1.5\).

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 \(\mathbf{i}\).
2. (b) Show that the acceleration is constant and find its magnitude.

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

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
2. (b) the position vector
3. (c) the distance from the origin
4. (d) the speed

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

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.

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.