Kinematics with constant velocity Problem set A Answers

1. (a) 24°
2. (b) 293°

1. (a) √89 = 9.43 m s⁻¹
2. (b) 9 m s⁻¹
3. (c) √35 = 5.92 m s⁻¹

1. (a) \(\begin{pmatrix}24\\45\end{pmatrix}\) m s⁻¹
2. (b) \(3\mathbf{i} + 12\mathbf{j} + 24\mathbf{k}\) m s⁻¹
3. (c) \(\mathbf{r} = \begin{pmatrix}-4\\0\\9\end{pmatrix} + t\begin{pmatrix}30\\10\\15\end{pmatrix}\)

1. (a) √6.5 = 2.55 m s⁻¹
2. (b) \(\mathbf{r} = \begin{pmatrix}-3\\6\\2\end{pmatrix} + t\begin{pmatrix}2\\-1.5\\0.5\end{pmatrix}\)
3. (c) Yes: t = 4 gives the point in every component

1. (a) (−2, 7, 4)
2. (b) 7 m s⁻¹
3. (c) 29.7 m

7.21 km

1. (a) \(\mathbf{r}_P = (-12 + 6t)\mathbf{i} + (5 + 2t)\mathbf{j},\quad \mathbf{r}_Q = (3 + t)\mathbf{i} + (-8 - 3t)\mathbf{j}\)
2. (b) 23.5 km
3. (c) t = 3 hours

1. (a) They cross at (8, −3, 5)
2. (b) The first object is at (8, −3, 5) when t = 3 and the second when t = 5: different times

1. (a) √14 = 3.74 m s⁻¹ and 3 m s⁻¹
2. (b) They do not meet

1. (a) Shown
2. (b) 6.64 km
3. (c) t = 3

1. (a) They collide at t = 2 at (7, 2, 2)
2. (b) 1.73 km

1. (a) t = 4
2. (b) 3.5 m

1. (a) c = 1
2. (b) 43.3 km h⁻¹