Projectiles, circular motion and delayed objects Problem set A Answers

1. (a) (i) 58.5° above the horizontal (ii) 39.2° below the horizontal
2. (b) 2 s
3. (c) \(\mathbf{r} = \begin{pmatrix}12t\\19.6t − 4.9t^2\end{pmatrix}\); greatest height 19.6 m

39 m

1. (a) \(|\mathbf{r}| = 3\) for all \(t\), so a circle of radius 3 m
2. (b) \(\mathbf{v} = \begin{pmatrix}-18\sin 6t\\18\cos 6t\end{pmatrix}\) m s⁻¹
3. (c) \(\mathbf{v}\cdot\mathbf{r} = 0\)
4. (d) 18 m s⁻¹

1. (a) 1.5 m s⁻¹
2. (b) 8 m s⁻¹

1. (a) \(\mathbf{v} = \begin{pmatrix}-6\sin 3t\\6\cos 3t\end{pmatrix}\), \(\mathbf{a} = \begin{pmatrix}-18\cos 3t\\−18\sin 3t\end{pmatrix}\)
2. (b) \(k = −9\)
3. (c) Towards the centre of the circle (the origin)
4. (d) 18 m s⁻²

1. (a) \(\mathbf{r} = \begin{pmatrix}2\sin 1.5t\\−2\cos 1.5t\end{pmatrix}\); \(|\mathbf{r}| = 2\), so a circle of radius 2 m
2. (b) Initial position \(\begin{pmatrix}0\\−2\end{pmatrix}\) m, radius 2 m
3. (c) 4.19 s
4. (d) \(\dfrac{2\pi\times 2}{4.19} = 3\) m s⁻¹, the speed

1. (a) \(\mathbf{r} = \begin{pmatrix}5t + t^2\\9t − t^3\end{pmatrix}\) m
2. (b) 17.2 m

25.7 m

\(b = 12.25\)

1. (a) \(\mathbf{r} = \begin{pmatrix}7t\\21t − 4.9t^2\end{pmatrix}\)
2. (b) \(\mathbf{r} = \begin{pmatrix}7(t − 2)\\21(t − 2) − 4.9(t − 2)^2\end{pmatrix}\)

1. (a) \(b = 20.5\)
2. (b) 48.1 m

1. (a) \(d = \sqrt{150^2 + (372.5 − 49t)^2}\)
2. (b) 7.60 s

1. (a) \(|\mathbf{r}| = r\), constant: a circle of radius r
2. (b) \(|\mathbf{v}| = r\omega\)
3. (c) \(T = \dfrac{2\pi}{\omega}\)
4. (d) \(|\mathbf{a}| = r\omega^2 = \dfrac{v^2}{r}\)
5. (e) 3 m s⁻¹, 2.51 s, 7.5 m s⁻²

1. (a) \(\mathbf{v}\cdot\mathbf{r} = 0\)
2. (b) \(\mathbf{a} = −\omega^2\mathbf{r}\)