Q1
Answer
The position vector of a particle, $\mathbf{r}$, at time $t$ is given by $$\mathbf{r} = 3t\mathbf{i} + qt^2\mathbf{j}$$ When $t = 2$, the distance of the particle from the origin is $10$.
  1. Given $q$ is positive, find $q$.
  2. Find the speed of the particle when $t = 1$.
  3. Find the direction of the acceleration of the particle.
  1. $q = 2$
  2. $5$
  3. $\mathbf{a} = 4\mathbf{j}$, North
Q2
Answer
The position vector of a particle, $\mathbf{r}$, at time $t$ is given by $$\mathbf{r} = \begin{pmatrix}pt^2 \\ \sqrt{t}\end{pmatrix}$$ When $t = 1$, the particle is accelerating on a bearing of $135^{\circ}$. Find $p$.
$p = \dfrac{1}{8}$
Q3
Answer
The velocity of a particle, $\mathbf{v}$, at time $t$ is given by $$\mathbf{v} = 2t^2\mathbf{i} + (3t-5)\mathbf{j}$$
  1. Find the speed of the particle when $t = 3$ to 3 significant figures.
  2. Find $t$ when the particle is moving in the south east direction.
  3. Find $t$ when the particle is accelerating parallel to $2\mathbf{i} + \mathbf{j}$
  1. $18.4$
  2. $t = 1$
  3. $t = 1.5$
Q4
Answer
The velocity of a particle, $\mathbf{v}$, at time $t$ is given by $$\mathbf{v} = 5\mathbf{i} + (1 - 2t)\mathbf{j}$$ At $t = 0$, the particle has position vector $3\mathbf{i} + 6\mathbf{j}$.
  1. Find the position vector of the particle.
  2. Find $t$ when the particle has a position vector parallel to $2\mathbf{i} - \mathbf{j}$.
  1. $(5t +3)\mathbf{i} + (t-t^2+6)\mathbf{j}$
  2. $t = 5$
Q5
Answer
The velocity of a particle, $\mathbf{v}$, at time $t$ is given by $$\mathbf{v} = \begin{pmatrix}5t\sqrt{t} \\ 4t\end{pmatrix}$$
  1. Find the speed of the particle when $t = 4$ correct to 3 significant figures.
  2. Find the acceleration of the particle when $t = 4$.
  3. When $t = 1$, the particle has a position vector of $\begin{pmatrix}1 \\ 5\end{pmatrix}$. Find the position vector of the particle at time $t$.
  1. $43.1$
  2. $\begin{pmatrix}15 \\ 4\end{pmatrix}$
  3. $\begin{pmatrix}2t^{\frac{5}{2}} -1 \\ 2t^2 + 3\end{pmatrix}$
Q6
Answer
The velocity of a particle, $\mathbf{v}$, at time $t$ is given by $$\mathbf{v} = \begin{pmatrix}(3+2t)^5 \\ t^2(1+t)^{\frac{1}{2}}\end{pmatrix}$$
  1. Find the acceleration of the particle at time $t$.
  2. Find the direction the particle is accelerating in when $t = 0$.
  1. $\begin{pmatrix}10(3+2t)^4 \\ 2t(1+t)^{\frac{1}{2}} + \frac{1}{2}t^2(1+t)^{-\frac{1}{2}}\end{pmatrix}$
  2. $0^{\circ}$ or East or a bearing of $090^{\circ}$
Q7
Answer
The acceleration of a particle, $\mathbf{a}$, at time $t$ is given by $$\mathbf{a} = \begin{pmatrix}2 \\ 1\end{pmatrix}$$
  1. When $t = 0$, the particle has velocity $\mathbf{v} = \begin{pmatrix}1 \\ 3\end{pmatrix}$. Find $\mathbf{v}$ in terms of $t$.
  2. When $t = 0$, the particle has position $\mathbf{r} = \begin{pmatrix}4 \\ -1\end{pmatrix}$. Find $\mathbf{r}$ in terms of $t$.
  1. $\begin{pmatrix}2t + 1 \\ t + 3\end{pmatrix}$
  2. $\begin{pmatrix}t^2 + t + 4 \\ \dfrac{1}{2}t^2 + 3t - 1\end{pmatrix}$
Q8
Answer
The acceleration of a particle, $\mathbf{a}$, at time $t$ is given by $$\mathbf{a} = \begin{pmatrix}2t \\ 3t^2\end{pmatrix}$$
  1. When $t = 0$, the particle has velocity $\mathbf{v} = \begin{pmatrix}1 \\ -2\end{pmatrix}$. Find $\mathbf{v}$ in terms of $t$.
  2. When $t = 0$, the particle has position $\mathbf{r} = \begin{pmatrix}4 \\ 3\end{pmatrix}$. Find $\mathbf{r}$ in terms of $t$.
  1. $\begin{pmatrix}t^2 + 1 \\ t^3 - 2\end{pmatrix}$
  2. $\begin{pmatrix}\dfrac{1}{3}t^3 + t + 4 \\ \dfrac{1}{4}t^4 - 2t + 3\end{pmatrix}$
Q9
Answer
The acceleration of a particle, $\mathbf{a}$, at time $t$ is given by $$\mathbf{a} = \begin{pmatrix}6t^2 \\ 4t + 1\end{pmatrix}$$ When $t = 0$, the particle has position $\begin{pmatrix}3\\1\end{pmatrix}$ and when $t = 1$ it has position $\begin{pmatrix}0\\4\end{pmatrix}$. Find the position of the particle at time $t$.
$\begin{pmatrix}\dfrac{1}{2}t^4 - \dfrac{7}{2}t + 3 \\ \dfrac{2}{3}t^3 + \dfrac{1}{2}t^2 + \dfrac{11}{6}t + 1\end{pmatrix}$
Q10
Answer
The acceleration of a particle, $\mathbf{a}$, at time $t$ is given by $$\mathbf{a} = \begin{pmatrix}3t^{-\frac{1}{2}} \\ 1 - 6t\end{pmatrix}$$ When $t = 1$, the particle has position $\begin{pmatrix}4\\2\end{pmatrix}$ and when $t = 4$ it has position $\begin{pmatrix}2\\8\end{pmatrix}$. Find the position of the particle at time $t$.
$\begin{pmatrix} 4t^{\frac{3}{2}} - 10t + 10 \\ \dfrac{1}{2}t^2 - t^3 + \dfrac{41}{2}t - 18 \end{pmatrix}$
Q11
Answer
A particle of mass $2$ kg is being acted on by a force $\mathbf{F} = 4\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}$. Initially, the particle has velocity $2\mathbf{i} + \mathbf{k}$ and is at the point with position vector $-3\mathbf{i} + \mathbf{j} - 4\mathbf{k}$. Find the position vector of the particle when $t = 3$.
    $12\mathbf{i} - \dfrac{7}{2}\mathbf{j} + \dfrac{25}{2}\mathbf{k}$
    Q12
    Answer
    A particle of mass $3$ kg is acted on by a force $\mathbf{F} = 6\mathbf{i} + 9\mathbf{j} - 3\mathbf{k}$. Initially the particle has velocity $-\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}$ and position vector $2\mathbf{i} - \mathbf{j} + 5\mathbf{k}$. Find the position vector of the particle when $t = 2$.
      $4\mathbf{i} + 9\mathbf{j} + 11\mathbf{k}$ m
      Q13
      Answer
      A particle of mass $4$ kg is acted on by a constant force $\mathbf{F}$. At time $t = 0$ the particle has velocity $3\mathbf{i} - 2\mathbf{j} + \mathbf{k}$ ms-1 and position vector $\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}$ m. At time $t = 2$ s the velocity of the particle is $7\mathbf{i} - 6\mathbf{j} + 5\mathbf{k}$ ms-1.
      1. Find the force $\mathbf{F}$.
        1. Find the position vector of the particle when $t = 4$.
          1. $\mathbf{F} = 8\mathbf{i} - 8\mathbf{j} + 8\mathbf{k}$
          2. $29\mathbf{i} - 20\mathbf{j} + 18\mathbf{k}$ m
          Q14
          Answer
          Two forces act on a particle of mass $5$ kg: $\mathbf{F}_1 = 10\mathbf{i} - 5\mathbf{j} + 15\mathbf{k}$ N and $\mathbf{F}_2 = -5\mathbf{i} + 10\mathbf{j} - 5\mathbf{k}$ N. Initially the particle is at rest at the origin.
          1. Find the acceleration of the particle.
            1. Find the position vector of the particle at time $t$.
              1. Find the distance of the particle from the origin when $t = 2$.
                1. $\mathbf{a} = \mathbf{i} + \mathbf{j} + 2\mathbf{k}$ ms-2
                2. $\mathbf{r} = \dfrac{1}{2}t^2\mathbf{i} + \dfrac{1}{2}t^2\mathbf{j} + t^2\mathbf{k}$
                3. $|\mathbf{r}| = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$ m
                Q15
                Answer
                Initially, ship A is at the origin and ship B is $8$ km north of A. A travels on a course of $060°$ at $12$ km h-1. Ship B travels on a course of $150°$ at $v$ km h-1.
                1. Find the position vector of A at time $t$.
                  1. Find the position vector of B at time $t$.
                    1. Given that the ships collide, find $v$.
                      1. $\mathbf{r}_A = t\begin{pmatrix}6\sqrt{3}\\6\end{pmatrix}$
                      2. $\mathbf{r}_B = \begin{pmatrix}\frac{v}{2}t\\8-\frac{v\sqrt{3}}{2}t\end{pmatrix}$
                      3. $v = 12\sqrt{3}$
                      Q16
                      Answer
                      At noon, ship A is at position vector $\begin{pmatrix}3\\-2\end{pmatrix}$ km and ship B is at position vector $\begin{pmatrix}-1\\4\end{pmatrix}$ km, relative to a fixed origin. Ship A has velocity $\begin{pmatrix}2\\1\end{pmatrix}$ km h-1 and ship B has velocity $\begin{pmatrix}-1\\-2\end{pmatrix}$ km h-1.
                      1. Find the position vectors of A and B at time $t$ hours after noon.
                        1. Find the displacement vector $\overrightarrow{AB}$ at time $t$.
                          1. Find the minimum distance between the two ships and the time at which it occurs.
                            1. $\mathbf{r}_A = \begin{pmatrix}3+2t\\-2+t\end{pmatrix}$, $\quad\mathbf{r}_B = \begin{pmatrix}-1-t\\4-2t\end{pmatrix}$
                            2. $\overrightarrow{AB} = \begin{pmatrix}-4-3t\\6-3t\end{pmatrix}$
                            3. Minimum distance $= 5\sqrt{2}$ km at $t = \dfrac{1}{3}$ hours
                            Q17
                            Answer
                            A lighthouse is at the origin. A ship leaves a port at position vector $\begin{pmatrix}-12\\5\end{pmatrix}$ km at 09:00, travelling with constant velocity $\begin{pmatrix}4\\-1\end{pmatrix}$ km h-1.
                            1. Find the position vector of the ship at time $t$ hours after 09:00.
                              1. Find the time at which the ship is closest to the lighthouse to the nearest minute.
                                1. Find the closest distance between the ship and the lighthouse.
                                  1. $\mathbf{r} = \begin{pmatrix}-12+4t\\5-t\end{pmatrix}$
                                  2. $t = \dfrac{53}{17}$ approximately 12:07
                                  3. Minimum distance $= \dfrac{8}{\sqrt{17}} \approx 1.94$ km
                                  Q18
                                  Answer
                                  Ship A sets off from a port located at the origin at 08:00 with velocity $\begin{pmatrix}6\\8\end{pmatrix}$ km h-1. Ship B sets off from the same port at 09:00 with velocity $\begin{pmatrix}p\\q\end{pmatrix}$ km h-1, where $p > 0$.
                                  1. Find the position vector of each ship at time $t$ hours after 09:00.
                                    1. Given that B has speed $15$ km h-1 and intercepts A, find $p$ and $q$ and the time of interception.
                                      1. $\mathbf{r}_A = \begin{pmatrix}6+6t\\8+8t\end{pmatrix}$, $\quad\mathbf{r}_B = \begin{pmatrix}pt\\qt\end{pmatrix}$
                                      2. $p = 9$, $q = 12$; interception at $t = 2$ hours after 09:00 (i.e. 11:00)
                                      Q19
                                      Answer
                                      At time $t = 0$, ship A is at position vector $\begin{pmatrix}10\\0\end{pmatrix}$ km and ship B is at position vector $\begin{pmatrix}0\\10\end{pmatrix}$ km. Ship A travels with velocity $\begin{pmatrix}-2\\3\end{pmatrix}$ km h-1 and ship B travels with velocity $\begin{pmatrix}4\\-1\end{pmatrix}$ km h-1.
                                      1. Write down the position vectors of A and B at time $t$.
                                        1. Show the ships do not collide.
                                          1. Find the time and distance of closest approach.
                                            1. $\mathbf{r}_A = \begin{pmatrix}10-2t\\3t\end{pmatrix}$, $\quad\mathbf{r}_B = \begin{pmatrix}4t\\10-t\end{pmatrix}$
                                            2. $\mathbf{i}$: $10 - 2t = 4t \Rightarrow t = \dfrac{5}{3}$; $\mathbf{j}$: $3t = 10 - t \Rightarrow t = \dfrac{5}{2}$; inconsistent so no collision
                                            3. $t = \dfrac{25}{13}$ hours; distance $= \dfrac{10\sqrt{13}}{13}$ km
                                            Q20
                                            Answer
                                            Initially, ship A is at position vector $\begin{pmatrix}0\\6\end{pmatrix}$ km and ship B is at position vector $\begin{pmatrix}4\\0\end{pmatrix}$ km. Ship A has velocity $\begin{pmatrix}3\\-2\end{pmatrix}$ km h-1. Ship B wishes to intercept ship A and travels at a constant speed of $5$ km h-1.
                                            1. Find the position vector of ship A at time $t$ hours.
                                              1. Find the time of interception.
                                                1. Find the bearing on which ship B must travel.
                                                  1. $\mathbf{r}_A = \begin{pmatrix}3t\\6-2t\end{pmatrix}$
                                                  2. $3t^2 + 12t - 13 = 0$ and $t = 0.887$
                                                  3. $342^{\circ}$