Revise Vector applications: kinematics and motion for IB Mathematics: Applications and Interpretation (3.6)

Model answer — full working.\n\nPosition vector at time $t$. The initial position is $\mathbf{r}_0 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ and the velocity is $\mathbf{v} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$, so\n$\mathbf{r} = \mathbf{r}_0 + t\mathbf{v} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} + t\begin{pmatrix} 3 \\ 4 \end{pmatrix}.$\n\nPosition after 2 hours. Substitute $t = 2$:\n$\mathbf{r}(2) = \begin{pmatrix} 2 \\ 1 \end{pmatrix} + 2\begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} + \begin{pmatrix} 6 \\ 8 \end{pmatrix} = \begin{pmatrix} 8 \\ 9 \end{pmatrix} \text{ km}.$\n\nSpeed. The speed is the magnitude of the velocity vector:\n$\text{speed} = |\mathbf{v}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ km/h}.$\n\nFinal answer: $\mathbf{r} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} + t\begin{pmatrix} 3 \\ 4 \end{pmatrix}$; after 2 hours the drone is at $\begin{pmatrix} 8 \\ 9 \end{pmatrix}$ km; its speed is $5$ km/h.\n\n---\nHow our marking engine awards the 5 marks:\n\n- M1 — set up the model. A method mark for writing the position in the form $\mathbf{r} = \mathbf{r}_0 + t\mathbf{v}$, correctly identifying $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ as the starting vector and $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ as the velocity vector.\n- A1 — the position equation. The accuracy mark for the fully correct expression $\begin{pmatrix} 2 \\ 1 \end{pmatrix} + t\begin{pmatrix} 3 \\ 4 \end{pmatrix}$. It depends on the M1: the method must be there for this A-mark to stand.\n- M1 — substitute $t = 2$. A method mark for putting $t = 2$ into the candidate's own position equation, i.e. $\begin{pmatrix} 2 \\ 1 \end{pmatrix} + 2\begin{pmatrix} 3 \\ 4 \end{pmatrix}$. Follow-through (FT) applies — a student who wrote a slightly different equation but substitutes correctly still earns this.\n- A1 — the position $\begin{pmatrix} 8 \\ 9 \end{pmatrix}$. The accuracy mark for the evaluated position, FT on the candidate's equation.\n- A1 — the speed $5$ km/h. Awarded for $\sqrt{3^2 + 4^2} = 5$ with correct units. The engine accepts $\sqrt{25}$ or $5.0$ and any equivalent form; once the correct speed appears, ISW means later restatements do not lose the mark.\n\nBottom line: two of the five marks are method marks that survive an arithmetic slip, and the accuracy marks are shielded by follow-through — but only if the model and the substitution are actually written on the page.

(a) Initial position. At $t = 0$ the position is the vector not multiplied by $t$: $\mathbf{r}_0 = \begin{pmatrix} 2 \\ -5 \end{pmatrix}$, i.e. the point $(2, -5)$. [A1]\n\n**(b) Speed.** The velocity is $\mathbf{v} = \begin{pmatrix} 4 \\ -3 \end{pmatrix}$ m/s.\n$\text{speed} = |\mathbf{v}| = \sqrt{4^2 + (-3)^2}$ [M1]\n$= \sqrt{16 + 9} = \sqrt{25} = 5$ m/s. [A1]\n\n**(c) Position after 10 s.** Substitute $t = 10$:\n$\mathbf{r}(10) = \begin{pmatrix} 2 \\ -5 \end{pmatrix} + 10\begin{pmatrix} 4 \\ -3 \end{pmatrix} = \begin{pmatrix} 2 \\ -5 \end{pmatrix} + \begin{pmatrix} 40 \\ -30 \end{pmatrix} = \begin{pmatrix} 42 \\ -35 \end{pmatrix}$, the point $(42, -35)$. [M1][A1]\n\n**(d) Distance in the first 10 s.** Distance $=$ speed $\times$ time $= 5 \times 10 = 50$ m. [A1]\n\nNotice the direction of motion: both components of $\mathbf{v}$ tell the story — the particle drifts right ($+4$) and downward ($-3$) each second.

(a) Position vectors.\n$\mathbf{r}_A(t) = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + t\begin{pmatrix} 4 \\ 1 \end{pmatrix}$ [A1]\n$\mathbf{r}_B(t) = \begin{pmatrix} 3 \\ 1 \end{pmatrix} + t\begin{pmatrix} 2 \\ 2 \end{pmatrix}$ [A1]\n\n**(b) Test for a collision.** For a collision the positions must be equal at the same time, so set $\mathbf{r}_A(t) = \mathbf{r}_B(t)$:\n$\begin{pmatrix} 1 \\ 2 \end{pmatrix} + t\begin{pmatrix} 4 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} + t\begin{pmatrix} 2 \\ 2 \end{pmatrix}$ [M1]\nSeparating into components:\n$x$: $1 + 4t = 3 + 2t \;\Rightarrow\; 2t = 2 \;\Rightarrow\; t = 1.$ [A1]\n$y$: $2 + t = 1 + 2t \;\Rightarrow\; t = 1.$ [A1]\nBoth components give the same value $t = 1$, so the boats are at the same place at the same time: they collide 1 hour after noon. [R1]\n\n**(c) Collision point.** Substitute $t = 1$ into either position — using boat A:\n$\mathbf{r}_A(1) = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + 1\begin{pmatrix} 4 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$ km. [A1]\nThe boats collide at the point with position vector $\begin{pmatrix} 5 \\ 3 \end{pmatrix}$ km. (Checking with boat B: $\begin{pmatrix} 3 \\ 1 \end{pmatrix} + \begin{pmatrix} 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$ — agreement confirms the collision.)

For a collision the positions must be equal at the same time, so set $\mathbf{r}_P(t) = \mathbf{r}_Q(t)$:\n$\begin{pmatrix} 0 \\ 5 \end{pmatrix} + t\begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 1 \end{pmatrix} + t\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ [M1]\nComponents:\n$x$: $0 + 3t = 6 + t \;\Rightarrow\; 2t = 6 \;\Rightarrow\; t = 3.$ [A1]\n$y$: $5 + t = 1 + 2t \;\Rightarrow\; t = 4.$ [A1]\nThe $x$-components agree at $t = 3$ but the $y$-components agree only at $t = 4$. There is no single time satisfying both, so although the flight paths cross, the aircraft are never at that point together: they do NOT collide. [R1]