IB Mathematics AHL 5.13 Kinematic problems involving displacement-AI HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
Speed = magnitude of velocity vector \( \begin{pmatrix} -0.40 \\ -0.35 \\ 0.1 \end{pmatrix} \):
\( \sqrt{(-0.40)^2 + (-0.35)^2 + (0.1)^2} = \sqrt{0.16 + 0.1225 + 0.01} = \sqrt{0.2925} \approx 0.541 \text{ km/min}. \)
\( \boxed{0.541 \text{ km/min}} \)
(b)
Two valid statements:
1. The airplane is flying in a straight line in a north-west direction (since velocity components are negative in \(x\) (east) and positive in \(y\) (north)).
2. It is flying at a constant altitude of 7 km (since \(z\)-component of velocity is 0).
\( \boxed{\text{Flying in a straight NW direction at constant altitude 7 km.}} \)
(c)(i)
Set \(z\)-coordinates equal: \( 7 = 6 + 0.1t_1 \implies 0.1t_1 = 1 \implies t_1 = 10 \text{ minutes}. \)
\( \boxed{t_1 = 10 \text{ minutes after 10:00 am}} \)
(c)(ii)
At \(t_1 = 10\):
Airplane position: \( \begin{pmatrix} 5 \\ -2 \\ 7 \end{pmatrix} + 10 \begin{pmatrix} -1.4 \\ 1.65 \\ 0 \end{pmatrix} = \begin{pmatrix} -9 \\ 14.5 \\ 7 \end{pmatrix}. \)
Birds position: \( \begin{pmatrix} -11 \\ 30 \\ 6 \end{pmatrix} + 10 \begin{pmatrix} -0.40 \\ -0.35 \\ 0.1 \end{pmatrix} = \begin{pmatrix} -15 \\ 26.5 \\ 7 \end{pmatrix}. \)
The \(x\) and \(y\) coordinates differ (\(-9 \ne -15\) and \(14.5 \ne 26.5\)), so they are not at the same horizontal position when at the same height. Hence, no collision.
\( \boxed{\text{At } t_1=10 \text{, their (x,y) positions differ, so they do not collide.}} \)
(d)
At 10:20, \(t_1 = 20\) for the airplane. Its position then:
\( \mathbf{r}_{A}(20) = \begin{pmatrix} 5 \\ -2 \\ 7 \end{pmatrix} + 20 \begin{pmatrix} -1.4 \\ 1.65 \\ 0 \end{pmatrix} = \begin{pmatrix} -23 \\ 31 \\ 7 \end{pmatrix}. \)
For \(t_2 \ge 0\), airplane equation: \( \mathbf{r}_{A}(t_2) = \begin{pmatrix} -23 \\ 31 \\ 7 \end{pmatrix} + t_2 \begin{pmatrix} -1.4 \\ 1.65 \\ 0 \end{pmatrix}. \)
\( \boxed{\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -23 \\ 31 \\ 7 \end{pmatrix} + t_2 \begin{pmatrix} -1.4 \\ 1.65 \\ 0 \end{pmatrix}} \)
(e)
Let \( \mathbf{r}_J(t_2) = \begin{pmatrix} -95 \\ 32 \\ 1.5 \end{pmatrix} + t_2 \begin{pmatrix} 1.3 \\ 1.45 \\ 0.2 \end{pmatrix} \)
Let \( \mathbf{r}_A(t_2) = \begin{pmatrix} -23 \\ 31 \\ 7 \end{pmatrix} + t_2 \begin{pmatrix} -1.4 \\ 1.65 \\ 0 \end{pmatrix} \)
Difference: \( \mathbf{d}(t_2) = \mathbf{r}_J – \mathbf{r}_A = \begin{pmatrix} -72 \\ 1 \\ -5.5 \end{pmatrix} + t_2 \begin{pmatrix} 2.7 \\ -0.2 \\ 0.2 \end{pmatrix}. \)
Distance: \( D(t_2) = \sqrt{(2.7t_2 – 72)^2 + (-0.2t_2 + 1)^2 + (0.2t_2 – 5.5)^2}. \)
Using GDC to minimise: minimum distance \( \approx 4.33 \text{ km} \) (occurs at some \(t_2 > 0\)).
Since \(4.33 < 5\), they will break the 5 km separation rule.
\( \boxed{\text{Yes, minimum distance } \approx 4.33 \text{ km} < 5 \text{ km, so law is broken.}} \)