Constant Acceleration in 2D: past paper questions & worked solutions

The velocity changes from $-\mathbf{i}+4\mathbf{j}$ at $t=1$ to $5\mathbf{i}-8\mathbf{j}$ at $t=4$. The time difference is $3$ s.

$\mathbf{a}=\dfrac{(5\mathbf{i}-8\mathbf{j})-(-\mathbf{i}+4\mathbf{j})}{3}=\dfrac{6\mathbf{i}-12\mathbf{j}}{3}=2\mathbf{i}-4\mathbf{j}$.

From $t=1$ to $t=3.5$ is $2.5$ s, so $\mathbf{v}=(-\mathbf{i}+4\mathbf{j})+2.5(2\mathbf{i}-4\mathbf{j})$.

$\mathbf{v}=-\mathbf{i}+4\mathbf{j}+5\mathbf{i}-10\mathbf{j}=4\mathbf{i}-6\mathbf{j}$.

Speed $=\sqrt{4^2+(-6)^2}=\sqrt{52}=2\sqrt{13}\ \mathrm{m\,s^{-1}}$, about $7.2\ \mathrm{m\,s^{-1}}$.

Given positions for particle A: ; At $t = 2$: $\mathbf{r} = -5\mathbf{i} + 16\mathbf{j}$ ; At $t = 5$: $\mathbf{r} = 10\mathbf{i} + 4\mathbf{j}$ $\mathbf{v} = \frac{\Delta \mathbf{r}}{\Delta t}$ $\mathbf{v}_A = \frac{(10\mathbf{i} + 4\mathbf{j}) - (-5\mathbf{i} + 16\mathbf{j})}{5 - 2}$ $= \frac{(10 + 5)\mathbf{i} + (4 - 16)\mathbf{j}}{3}$ $= \frac{15\mathbf{i} - 12\mathbf{j}}{3}$ $= 5\mathbf{i} - 4\mathbf{j} \text{ m s}^{-1}$

$\mathbf{r}(t) = \mathbf{r}(t_0) + \mathbf{v}(t - t_0)$ Using $t_0 = 2$ with $\mathbf{r}(2) = -5\mathbf{i} + 16\mathbf{j}$: $\mathbf{r}(t) = (-5\mathbf{i} + 16\mathbf{j}) + (5\mathbf{i} - 4\mathbf{j})(t - 2)$ $= -5\mathbf{i} + 16\mathbf{j} + 5(t-2)\mathbf{i} - 4(t-2)\mathbf{j}$ $= -5\mathbf{i} + 16\mathbf{j} + (5t - 10)\mathbf{i} + (-4t + 8)\mathbf{j}$ $= (-5 + 5t - 10)\mathbf{i} + (16 - 4t + 8)\mathbf{j}$ $= (5t - 15)\mathbf{i} + (24 - 4t)\mathbf{j}$

For A at $t = 5$: $\mathbf{r}_A = 10\mathbf{i} + 4\mathbf{j}$ (given) For B at $t = 5$: $\mathbf{s}_B = -5\mathbf{i} + 7\mathbf{j} + 5(2\mathbf{i} - 3\mathbf{j})$ $= -5\mathbf{i} + 7\mathbf{j} + 10\mathbf{i} - 15\mathbf{j}$ $= 5\mathbf{i} - 8\mathbf{j}$

$\overrightarrow{AB} = \mathbf{s}_B - \mathbf{r}_A$ $= (5\mathbf{i} - 8\mathbf{j}) - (10\mathbf{i} + 4\mathbf{j})$ $= -5\mathbf{i} - 12\mathbf{j}$

Angle from South (negative j direction): $\tan \alpha = \frac{5}{12}$ $\alpha = \tan^{-1}\left(\frac{5}{12}\right)$ $\alpha = 22.619...°$ Bearing = $180° + \alpha = 180° + 22.619...° = 202.619...°$

From $t=1$ to $t=9$, the displacement is $(10\mathbf{i}-3\mathbf{j})-(-2\mathbf{i}+5\mathbf{j})=12\mathbf{i}-8\mathbf{j}$.

The time interval is $8$ seconds, so $\mathbf{v}=\dfrac{12\mathbf{i}-8\mathbf{j}}{8}=1.5\mathbf{i}-\mathbf{j}\ \mathrm{m\,s^{-1}}$.

Time $t=7$ is 6 seconds after $t=1$. Therefore $\mathbf{r}_B(7)=(-2\mathbf{i}+5\mathbf{j})+6(1.5\mathbf{i}-\mathbf{j})=7\mathbf{i}-\mathbf{j}$.

The target has position vector $13\mathbf{i}-5\mathbf{j}$. The vector from the ball to the target is $(13\mathbf{i}-5\mathbf{j})-(7\mathbf{i}-\mathbf{j})=6\mathbf{i}-4\mathbf{j}$.

The distance is $\sqrt{6^2+(-4)^2}=\sqrt{52}=2\sqrt{13}\ \mathrm{m}$, which is about $7.2\ \mathrm{m}$.

$\mathbf{r}=(2-3)\mathbf{i}+(1-2(2))\mathbf{j}=-\mathbf{i}-3\mathbf{j}$.

The vector is in the third quadrant. The angle between $\mathbf{r}$ and the positive $\mathbf{j}$ direction satisfies $\tan\alpha=\dfrac{1}{3}$ from the vertical comparison.

$\alpha=18.4^\circ$ from the negative $\mathbf{j}$ direction, so the size of the angle with positive $\mathbf{j}$ is $180^\circ-18.4^\circ=161.6^\circ$, which is $162^\circ$.

Distance from O is $2.5$, so $(t-3)^2+(1-2t)^2=2.5^2$.

$t^2-6t+9+1-4t+4t^2=6.25$, so $5t^2-10t+3.75=0$.

Multiply by $4$: $20t^2-40t+15=0$, or $4t^2-8t+3=0$. Hence $(2t-1)(2t-3)=0$, so $t=\dfrac12$ or $t=\dfrac32$.

At maximum height, velocity = 0 ; $u = 5$ m s$^{-1}$ (initial velocity) ; $v = 0$ m s$^{-1}$ (at maximum height) ; $a = -9.8$ m s$^{-2}$ (gravity, downward) ; $s = ?$ (displacement above A)

$v^2 = u^2 + 2as$ $0^2 = 5^2 + 2(-9.8)s$ $0 = 25 - 19.6s$ $19.6s = 25$ $s = \frac{25}{19.6}$ $s = 1.276...$ $s = 1.28 \text{ m (3 s.f.)}$

Taking upward as positive, with origin at ground level: $s = s_0 + ut + \frac{1}{2}at^2$ For P (starting at A, 10 m above ground): $s_P = 10 + 5t + \frac{1}{2}(-9.8)t^2$ $s_P = 10 + 5t - 4.9t^2$

Q starts at $t = 1$, so Q has been moving for $(t-1)$ seconds: For Q (starting at B, ground level): $s_Q = 0 + 7(t-1) + \frac{1}{2}(-9.8)(t-1)^2$ $s_Q = 7(t-1) - 4.9(t-1)^2$ $s_Q = 7t - 7 - 4.9(t^2 - 2t + 1)$ $s_Q = 7t - 7 - 4.9t^2 + 9.8t - 4.9$ $s_Q = -4.9t^2 + 16.8t - 11.9$

$10 + 5T - 4.9T^2 = -4.9T^2 + 16.8T - 11.9$ $10 + 5T = 16.8T - 11.9$ $10 + 11.9 = 16.8T - 5T$ $21.9 = 11.8T$ $T = \frac{21.9}{11.8}$ $T = 1.855...$ $T = 1.86 \text{ s (3 s.f.)}$

$v = u + at$ For particle P: ; $u = 5$ m s$^{-1}$ ; $a = -9.8$ m s$^{-2}$ ; $t = 1.86$ s $v_P = 5 + (-9.8)(1.86)$ $= 5 - 18.228$ $= -13.228...$ $= -13.2 \text{ m s}^{-1}$ The negative sign indicates P is moving downward.

Speed = $|v|$ $\text{Speed} = |-13.2|$ $= 13.2 \text{ m s}^{-1}$

R moves from $-2\mathbf{i}+\mathbf{j}$ to $9\mathbf{i}+23\mathbf{j}$ in $11$ s. Its displacement is $11\mathbf{i}+22\mathbf{j}$, so its velocity is $\mathbf{i}+2\mathbf{j}$.

$\mathbf{r}=(-2\mathbf{i}+\mathbf{j})+t(\mathbf{i}+2\mathbf{j})=(t-2)\mathbf{i}+(2t+1)\mathbf{j}$.

Point C has position vector $25\mathbf{i}+25\mathbf{j}$. Since S moves with velocity $-\mathbf{i}-\mathbf{j}$, $\mathbf{s}=(25-t)\mathbf{i}+(25-t)\mathbf{j}$.

$\overrightarrow{SR}=\mathbf{r}-\mathbf{s}=[(t-2)-(25-t)]\mathbf{i}+[(2t+1)-(25-t)]\mathbf{j}$.

Therefore $\overrightarrow{SR}=(2t-27)\mathbf{i}+(3t-24)\mathbf{j}$ m.

The distance squared is $d^2=(2t-27)^2+(3t-24)^2$.

$d^2=4t^2-108t+729+9t^2-144t+576=13t^2-252t+1305$.

The quadratic is minimum at $t=\dfrac{252}{2\cdot13}=\dfrac{126}{13}=9.69\ldots$ s.

Take distances from A. Then C is at $1$ m, D is at $3$ m, the rod's centre of mass is at $2.5$ m, and B is at $5$ m.

Taking moments about D: $R_C(2)+Mg(2)=10g(3)+5g(0.5)$.

$2R_C+2Mg=30g+2.5g=32.5g$, so $R_C=16.25g-Mg=(16.25-M)g$.

Taking moments about C: $10g(1)+R_D(2)=5g(1.5)+Mg(4)$.

$10g+2R_D=7.5g+4Mg$, so $2R_D=(4M-2.5)g$ and $R_D=(2M-1.25)g$.

For the rod to rest on both supports, both reactions must be non-negative.

$R_C\ge0$ gives $16.25-M\ge0$, so $M\le16.25$. Also $R_D\ge0$ gives $2M-1.25\ge0$, so $M\ge0.625$. Therefore $0.625\le M\le16.25$.

From 2:00 to 2:30 is $0.5$ hours. The velocity changes from $3\mathbf{i}+5\mathbf{j}$ to $\mathbf{i}+7\mathbf{j}$.

$\mathbf{a}=\dfrac{(\mathbf{i}+7\mathbf{j})-(3\mathbf{i}+5\mathbf{j})}{0.5}=-4\mathbf{i}+4\mathbf{j}$.

At time $t$ hours after 2:00, $\mathbf{v}_P=(3-4t)\mathbf{i}+(5+4t)\mathbf{j}$.

At time $T$, the velocity is parallel to $-\mathbf{i}+2\mathbf{j}$, so $\dfrac{5+4T}{3-4T}=\dfrac{2}{-1}=-2$.

$5+4T=-2(3-4T)=-6+8T$. Hence $11=4T$ and $T=\dfrac{11}{4}$ hours.

For equal velocities, $3-4t=-4-2t$. This gives $7=2t$, so $t=3.5$.

At $t=3.5$, P's j-component is $5+4(3.5)=19$. Q's j-component is $\mu+3(3.5)=\mu+10.5$.

$\mu+10.5=19$, so $\mu=8.5$.

The initial velocity is $(5\mathbf{i}-8\mathbf{j})$ and the acceleration is $(-\lambda\mathbf{i}+2\lambda\mathbf{j})$. At $t=5$, $\mathbf{v}=(5\mathbf{i}-8\mathbf{j})+5(-\lambda\mathbf{i}+2\lambda\mathbf{j})$.

So $\mathbf{v}=(5-5\lambda)\mathbf{i}+(-8+10\lambda)\mathbf{j}$.

The speed is $13$, so $(5-5\lambda)^2+(-8+10\lambda)^2=13^2$.

Expanding gives $25-50\lambda+25\lambda^2+64-160\lambda+100\lambda^2=169$. Therefore $125\lambda^2-210\lambda-80=0$, and dividing by $5$ gives $25\lambda^2-42\lambda-16=0$.

Solving $25\lambda^2-42\lambda-16=0$ gives $\lambda=2$ or $\lambda=-\dfrac{8}{25}$. Since $\lambda$ is positive, $\lambda=2$.

$\mathbf{v}(4)=(5\mathbf{i}-8\mathbf{j})+4(-2\mathbf{i}+4\mathbf{j})=-3\mathbf{i}+8\mathbf{j}$.

The motion is north-west. The angle west of north satisfies $\tan\alpha=\dfrac{3}{8}$, so $\alpha=20.6^\circ$. The bearing is $360^\circ-20.6^\circ=339.4^\circ$, so $339^\circ$ to the nearest degree.

$\mathbf{v}=(14\mathbf{i}-5\mathbf{j})+2(-4\mathbf{i}+\mathbf{j})=6\mathbf{i}-3\mathbf{j}$.

The speed is $\sqrt{6^2+(-3)^2}=\sqrt{45}=3\sqrt5\ \mathrm{m\,s^{-1}}$.

The velocity $6\mathbf{i}-3\mathbf{j}$ is below the positive $\mathbf{i}$ direction. The acute angle satisfies $\tan\theta=\dfrac{3}{6}$.

$\theta=26.6^\circ$, so the size of the angle is $27^\circ$ to the nearest degree.

$\mathbf{v}(T)=(14-4T)\mathbf{i}+(-5+T)\mathbf{j}$.

The direction is parallel to $2\mathbf{i}-3\mathbf{j}$, so the component ratio must match: $\dfrac{14-4T}{2}=\dfrac{-5+T}{-3}$.

$-3(14-4T)=2(-5+T)$, so $-42+12T=-10+2T$. Hence $10T=32$ and $T=3.2$.

From 1 pm to 3 pm is 2 hours. The displacement of A is $(55\mathbf{i}+34\mathbf{j})-(25\mathbf{i}+10\mathbf{j})=30\mathbf{i}+24\mathbf{j}$ km, so $\mathbf{v}_A=15\mathbf{i}+12\mathbf{j}\ \mathrm{km\,h^{-1}}$.

At time $t$ hours after 1 pm, $\mathbf{r}_A=(25\mathbf{i}+10\mathbf{j})+t(15\mathbf{i}+12\mathbf{j})=(25+15t)\mathbf{i}+(10+12t)\mathbf{j}$.

The speed of A is $\sqrt{15^2+12^2}=\sqrt{369}\ \mathrm{km\,h^{-1}}$. Converting to $\mathrm{m\,s^{-1}}$, $V=\sqrt{369}\times\dfrac{1000}{3600}=\dfrac{5\sqrt{41}}{6}\ \mathrm{m\,s^{-1}}$.

At 2:30 pm, $t=1.5$ hours after 1 pm. For B, $\mathbf{r}_B=(35\mathbf{i}+51\mathbf{j})+1.5(20\mathbf{i}-6\mathbf{j})=65\mathbf{i}+42\mathbf{j}$.

Set $\mathbf{r}_A=65\mathbf{i}+42\mathbf{j}$. From the i-component, $25+15t=65$, so $t=\dfrac{8}{3}$.

When $t=\dfrac{8}{3}$, $10+12t=10+32=42$. Both components match, so A also passes through P.

The velocity is $12\mathbf{i}-16\mathbf{j}$, so the speed is $\sqrt{12^2+(-16)^2}=20\ \mathrm{km\,h^{-1}}$.

At time $t$ hours after midnight, $\mathbf{s}=(19\mathbf{i}+22\mathbf{j})+t(12\mathbf{i}-16\mathbf{j})=(19+12t)\mathbf{i}+(22-16t)\mathbf{j}$.

The lighthouse is at $26\mathbf{i}+15\mathbf{j}$. Therefore $\overrightarrow{LS}=\mathbf{s}-\mathbf{l}=(12t-7)\mathbf{i}+(7-16t)\mathbf{j}$.

The square of the distance is $d^2=(12t-7)^2+(7-16t)^2=400t^2-392t+98$.

This quadratic is minimum when $t=\dfrac{392}{2\cdot400}=0.49$.

At $t=0.49$, $d^2=1.96$, so $d=1.4\ \mathrm{km}$.

The unsafe distance is $1.3$ km. Since $1.4>1.3$, it is safe for the ship to continue its course.

The total force is $(p\mathbf{i}+q\mathbf{j})+(2q\mathbf{i}+p\mathbf{j})=(p+2q)\mathbf{i}+(p+q)\mathbf{j}$. Since the mass is $2$ kg and acceleration is $\mathbf{i}-\mathbf{j}$, the total force is $2\mathbf{i}-2\mathbf{j}$.

$p+2q=2$ and $p+q=-2$.

Subtract the equations to get $q=4$. Then $p+4=-2$, so $p=-6$.

The acceleration is $\mathbf{i}-\mathbf{j}$. Its dot product with $\mathbf{j}$ is $-1$, so the angle with the positive $\mathbf{j}$ direction is obtuse. Since the components are equal in magnitude, the angle is $135^\circ$.

$\mathbf{v}=(3\mathbf{i}-4\mathbf{j})+T(\mathbf{i}-\mathbf{j})=(3+T)\mathbf{i}+(-4-T)\mathbf{j}$.

The velocity is parallel to $11\mathbf{i}-13\mathbf{j}$, so $\dfrac{3+T}{11}=\dfrac{-4-T}{-13}$.

$13(3+T)=11(4+T)$, so $39+13T=44+11T$. Hence $2T=5$ and $T=2.5$.

From 7:00 to 7:36 is $36$ minutes, which is $0.6$ hours.

Displacement $=(4\mathbf{i}+6\mathbf{j})-(-2\mathbf{i}+9\mathbf{j})=6\mathbf{i}-3\mathbf{j}$. Divide by $0.6$: $\mathbf{v}=10\mathbf{i}-5\mathbf{j}$.

$\mathbf{r}=(-2\mathbf{i}+9\mathbf{j})+t(10\mathbf{i}-5\mathbf{j})=(-2+10t)\mathbf{i}+(9-5t)\mathbf{j}$.

At 8:48, $t=1.8$. The ship position is $(-2+18)\mathbf{i}+(9-9)\mathbf{j}=16\mathbf{i}$. The lighthouse is due east, so it has the same j-component, $0$.

At 9:00, $t=2$. The ship position is $18\mathbf{i}-\mathbf{j}$. The lighthouse is north-east of the ship, so its east and north displacements from the ship are equal.

Since the lighthouse has j-component $0$, from $18\mathbf{i}-\mathbf{j}$ it is $1$ km north, so it is also $1$ km east. Hence its position is $19\mathbf{i}$.

Due south of the lighthouse means the ship has i-component $19$. Set $-2+10t=19$, so $t=2.1$.

At $t=2.1$, $\mathbf{r}=(-2+21)\mathbf{i}+(9-10.5)\mathbf{j}=19\mathbf{i}-1.5\mathbf{j}$.

$\mathbf{r}_A=(11\mathbf{i}+11\mathbf{j})+t(3\mathbf{i}-\mathbf{j})=(11+3t)\mathbf{i}+(11-t)\mathbf{j}$.

At $t=6$, Agatha is at $(11+18)\mathbf{i}+(11-6)\mathbf{j}=29\mathbf{i}+5\mathbf{j}$.

$\mathbf{r}_B=(7\mathbf{i}+16\mathbf{j})+t(4\mathbf{i}-2\mathbf{j})=(7+4t)\mathbf{i}+(16-2t)\mathbf{j}$.

Set $7+4t=29$ to get $t=5.5$. Also $16-2t=5$ gives $t=5.5$. So Brionie also passes through P at $t=5.5$ s.

$\overrightarrow{AB}=\mathbf{r}_B-\mathbf{r}_A=[7+4t-(11+3t)]\mathbf{i}+[16-2t-(11-t)]\mathbf{j}$.

So $\overrightarrow{AB}=(t-4)\mathbf{i}+(5-t)\mathbf{j}$ m.

$AB^2=(t-4)^2+(5-t)^2=2(t-4.5)^2+0.5$.

The minimum occurs when $t=4.5$, giving $AB^2=0.5$. Hence the minimum distance is $\sqrt{0.5}=0.707\ldots\ \mathrm{m}$.

Given: $\mathbf{v}_A = 2\mathbf{i} + 3\mathbf{j} + (\mathbf{i} - 4\mathbf{j})t$ This is in form: $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ Therefore: ; Initial velocity: $\mathbf{u}_A = 2\mathbf{i} + 3\mathbf{j}$ ; Acceleration: $\mathbf{a}_A = \mathbf{i} - 4\mathbf{j}$ km h$^{-2}$

$|\mathbf{a}| = \sqrt{a_x^2 + a_y^2}$ $|\mathbf{a}_A| = \sqrt{1^2 + (-4)^2}$ $= \sqrt{1 + 16}$ $= \sqrt{17}$ $= 4.123...$ $= 4.1 \text{ km h}^{-2} \text{ (2 sf)}$

Given: ; At $t = 2$: $\mathbf{v}_B = 4\mathbf{i} + \mathbf{j}$ ; At $t = 5$: $\mathbf{v}_B = \mathbf{i} - 5\mathbf{j}$ $\mathbf{a} = \frac{\Delta \mathbf{v}}{\Delta t}$ $\mathbf{a}_B = \frac{(\mathbf{i} - 5\mathbf{j}) - (4\mathbf{i} + \mathbf{j})}{5 - 2}$ $= \frac{(1-4)\mathbf{i} + (-5-1)\mathbf{j}}{3}$ $= \frac{-3\mathbf{i} - 6\mathbf{j}}{3}$ $= -\mathbf{i} - 2\mathbf{j} \text{ km h}^{-2}$

$\mathbf{v}(t) = \mathbf{v}(t_0) + \mathbf{a}(t - t_0)$ From $t = 2$ back to $t = 0$: $\mathbf{v}_B(0) = \mathbf{v}_B(2) + \mathbf{a}_B(0 - 2)$ $= (4\mathbf{i} + \mathbf{j}) + (-\mathbf{i} - 2\mathbf{j})(-2)$ $= 4\mathbf{i} + \mathbf{j} + 2\mathbf{i} + 4\mathbf{j}$ $= 6\mathbf{i} + 5\mathbf{j} \text{ km h}^{-1}$

For boat A: $\mathbf{v}_A(t) = 2\mathbf{i} + 3\mathbf{j} + (\mathbf{i} - 4\mathbf{j})t = (2+t)\mathbf{i} + (3-4t)\mathbf{j}$ For boat B: $\mathbf{v}_B(t) = (6\mathbf{i} + 5\mathbf{j}) + (-\mathbf{i} - 2\mathbf{j})t$ $= (6-t)\mathbf{i} + (5-2t)\mathbf{j}$

$|\mathbf{v}_A|^2 = (2+t)^2 + (3-4t)^2$ $= 4 + 4t + t^2 + 9 - 24t + 16t^2$ $= 17t^2 - 20t + 13$ $|\mathbf{v}_B|^2 = (6-t)^2 + (5-2t)^2$ $= 36 - 12t + t^2 + 25 - 20t + 4t^2$ $= 5t^2 - 32t + 61$

$17t^2 - 20t + 13 = 5t^2 - 32t + 61$ $12t^2 + 12t - 48 = 0$ $t^2 + t - 4 = 0$

$t = \frac{-1 \pm \sqrt{1 + 16}}{2}$ $= \frac{-1 \pm \sqrt{17}}{2}$ Taking positive value: $T_1 = \frac{-1 + \sqrt{17}}{2}$

Velocities parallel when: $\frac{2 + T_2}{6 - T_2} = \frac{3 - 4T_2}{5 - 2T_2}$

$(2 + T_2)(5 - 2T_2) = (3 - 4T_2)(6 - T_2)$ $10 - 4T_2 + 5T_2 - 2T_2^2 = 18 - 3T_2 - 24T_2 + 4T_2^2$ $10 + T_2 - 2T_2^2 = 18 - 27T_2 + 4T_2^2$ $0 = 4T_2^2 + 2T_2^2 - 27T_2 - T_2 + 18 - 10$ $0 = 6T_2^2 - 28T_2 + 8$

$0 = 3T_2^2 - 14T_2 + 4$ Therefore: $p = -14$ and $q = 4$

For a bearing angle $\alpha$ with $\tan\alpha=\dfrac43$, use $\sin\alpha=\dfrac45$ and $\cos\alpha=\dfrac35$. Therefore $\mathbf{v}_B=20\sin\alpha\mathbf{i}+20\cos\alpha\mathbf{j}=16\mathbf{i}+12\mathbf{j}$.

At noon, B is at $10\mathbf{i}+5\mathbf{j}$. With constant velocity $16\mathbf{i}+12\mathbf{j}$, $\mathbf{s}=(10+16t)\mathbf{i}+(5+12t)\mathbf{j}$.

A has position vector $\mathbf{r}=40t\mathbf{i}+20\mathbf{j}$. Hence $\overrightarrow{AB}=\mathbf{s}-\mathbf{r}=(10+16t-40t)\mathbf{i}+(5+12t-20)\mathbf{j}$.

This simplifies to $\overrightarrow{AB}=(10-24t)\mathbf{i}+(12t-15)\mathbf{j}$ km.

For collision, both components must be zero. From $10-24t=0$, $t=\dfrac{5}{12}$. From $12t-15=0$, $t=\dfrac54$. These times are different, so the boats never collide.

A bearing of $225^\circ$ means B is south-west of A, so the i and j components of $\overrightarrow{AB}$ are equal and negative. Set $10-24t=12t-15$.

The equation gives $t=\dfrac{25}{36}$. Then each component is $-\dfrac{20}{3}$, so the distance is $\sqrt{\left(\dfrac{20}{3}\right)^2+\left(\dfrac{20}{3}\right)^2}=\dfrac{20\sqrt2}{3}\ \mathrm{km}$.

$\mathbf{r}=(13\mathbf{i}+5\mathbf{j})+t(3\mathbf{i}-10\mathbf{j})=(13+3t)\mathbf{i}+(5-10t)\mathbf{j}$.

$\mathbf{s}=(3\mathbf{i}-5\mathbf{j})+t(15\mathbf{i}+14\mathbf{j})=(3+15t)\mathbf{i}+(-5+14t)\mathbf{j}$.

$\overrightarrow{AB}=\mathbf{s}-\mathbf{r}=[3+15t-(13+3t)]\mathbf{i}+[-5+14t-(5-10t)]\mathbf{j}$.

This simplifies to $\overrightarrow{AB}=(12t-10)\mathbf{i}+(24t-10)\mathbf{j}$ km.

The square of the distance is $d^2=(12t-10)^2+(24t-10)^2$.

$d^2=720t^2-720t+200$. This quadratic is minimum at $t=\dfrac{720}{2\cdot720}=\dfrac12$.

At $t=\dfrac12$, $\overrightarrow{AB}=(-4)\mathbf{i}+2\mathbf{j}$, so the shortest distance is $\sqrt{(-4)^2+2^2}=\sqrt{20}=2\sqrt5\ \mathrm{km}$.

At closest approach, B is west and north of A. The angle west of north satisfies $\tan\theta=\dfrac{4}{2}=2$, so $\theta=63.4^\circ$. Bearing $=360^\circ-63.4^\circ=296.6^\circ$, which is $297^\circ$.

From 12:00 to 12:30, the displacement of P is $(20\mathbf{i}+34\mathbf{j})-(15\mathbf{i}+36\mathbf{j})=5\mathbf{i}-2\mathbf{j}$ km. The time is $0.5$ hours, so $\mathbf{v}_P=\dfrac{5\mathbf{i}-2\mathbf{j}}{0.5}=10\mathbf{i}-4\mathbf{j}$.

At $t$ hours after 12:00, $\mathbf{p}=(15\mathbf{i}+36\mathbf{j})+t(10\mathbf{i}-4\mathbf{j})=(15+10t)\mathbf{i}+(36-4t)\mathbf{j}$.

At 13:30, $t=1.5$. Then $\mathbf{p}=(15+15)\mathbf{i}+(36-6)\mathbf{j}=30\mathbf{i}+30\mathbf{j}$. Also $\mathbf{q}=(42-8(1.5))\mathbf{i}+(9+14(1.5))\mathbf{j}=30\mathbf{i}+30\mathbf{j}$.

Since both position vectors are the same, the collision point has position vector $30\mathbf{i}+30\mathbf{j}$ km.

At 14:30, $t=2.5$, so $\mathbf{p}=(15+25)\mathbf{i}+(36-10)\mathbf{j}=40\mathbf{i}+26\mathbf{j}$.

At 12:30, $t=0.5$, so $\mathbf{q}=38\mathbf{i}+16\mathbf{j}$. From 12:30 to 14:30 is $2$ hours due north at $15\ \mathrm{km\,h^{-1}}$, so the extra displacement is $30\mathbf{j}$. Thus Q is at $38\mathbf{i}+46\mathbf{j}$.

The separation vector is $(40-38)\mathbf{i}+(26-46)\mathbf{j}=2\mathbf{i}-20\mathbf{j}$. The distance is $\sqrt{2^2+20^2}=\sqrt{404}=2\sqrt{101}\ \mathrm{km}$.

The velocity of S is $10\mathbf{i}+15\mathbf{j}$. Its speed is the magnitude: $\sqrt{10^2+15^2}=\sqrt{325}=5\sqrt{13}\ \mathrm{km\,h^{-1}}$.

B has velocity $-8\mathbf{i}+10\mathbf{j}$, so it moves west and north. The angle west of north satisfies $\tan\theta=\dfrac{8}{10}$, giving $\theta=38.66^\circ$. The bearing is $360^\circ-38.66^\circ=321^\circ$ to the nearest degree.

At time $t$ hours after 9 am, $\mathbf{r}_S=(10t)\mathbf{i}+(15t)\mathbf{j}$. Also $\mathbf{r}_B=(50\mathbf{i}+10\mathbf{j})+t(-8\mathbf{i}+10\mathbf{j})=(50-8t)\mathbf{i}+(10+10t)\mathbf{j}$.

$\overrightarrow{SB}=\mathbf{r}_B-\mathbf{r}_S=[(50-8t)-10t]\mathbf{i}+[(10+10t)-15t]\mathbf{j}=(50-18t)\mathbf{i}+(10-5t)\mathbf{j}$.

For a collision, both components of $\overrightarrow{SB}$ must be zero at the same time. From $50-18t=0$, $t=\dfrac{25}{9}$. From $10-5t=0$, $t=2$. These are different, so B and S do not collide.

When the separation is along a $45^\circ$ diagonal, the two components have equal magnitude. Using the given condition, set $50-18t=10-5t$, so $40=13t$ and $t=\dfrac{40}{13}$.

At $t=\dfrac{40}{13}$, both components are $-\dfrac{70}{13}$. Therefore $d=\sqrt{\left(\dfrac{70}{13}\right)^2+\left(\dfrac{70}{13}\right)^2}=\dfrac{70\sqrt2}{13}=7.6\ \mathrm{km}$ to 2 significant figures.

$|\mathbf{v}_A|=\sqrt{3^2+12^2}=\sqrt{153}=3\sqrt{17}\ \mathrm{km\,h^{-1}}$.

At time $t$, $\mathbf{a}=(-9\mathbf{i}+6\mathbf{j})+t(3\mathbf{i}+12\mathbf{j})=(-9+3t)\mathbf{i}+(6+12t)\mathbf{j}$.

$\mathbf{b}=(16\mathbf{i}+6\mathbf{j})+t(p\mathbf{i}+q\mathbf{j})=(16+pt)\mathbf{i}+(6+qt)\mathbf{j}$.

$\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=[25+(p-3)t]\mathbf{i}+[(q-12)t]\mathbf{j}$.

Given $\overrightarrow{AB}=(25-12t)\mathbf{i}-9t\mathbf{j}$, compare coefficients: $p-3=-12$ and $q-12=-9$.

Therefore $p=-9$ and $q=3$.

Use $|\overrightarrow{AB}|=15$: $(25-12t)^2+(-9t)^2=15^2$.

This gives $225t^2-600t+400=0$, so $(3t-4)^2=0$ and $t=\dfrac43$.

At $t=\dfrac43$, $\overrightarrow{AB}=9\mathbf{i}-12\mathbf{j}$. Therefore $\overrightarrow{BA}=-9\mathbf{i}+12\mathbf{j}$.

A is north-west of B. The angle west of north satisfies $\tan\theta=\dfrac{9}{12}$, so $\theta=36.9^\circ$. Bearing $=360^\circ-36.9^\circ=323^\circ$ to the nearest degree.

Q has velocity $20\mathbf{i}-20\mathbf{j}$, so it travels south-east with equal east and south components. Its bearing is $135^\circ$.

P starts at the origin and has velocity $15\mathbf{i}$, so $\mathbf{p}=15t\mathbf{i}$.

Q starts at $200\mathbf{j}$ and has velocity $20\mathbf{i}-20\mathbf{j}$, so $\mathbf{q}=20t\mathbf{i}+(200-20t)\mathbf{j}$.

$\overrightarrow{PQ}=\mathbf{q}-\mathbf{p}=(20t-15t)\mathbf{i}+(200-20t)\mathbf{j}=5t\mathbf{i}+(200-20t)\mathbf{j}$.

At $t=10$, $\overrightarrow{PQ}=50\mathbf{i}$. This means Q is east of P, so P is west of Q. The bearing of P from Q is $270^\circ$.

Q is north-east of P when the i and j components of $\overrightarrow{PQ}$ are equal and positive. So $5t=200-20t$.

The equation gives $t=8$. Then $\overrightarrow{PQ}=40\mathbf{i}+40\mathbf{j}$, so the distance is $\sqrt{40^2+40^2}=40\sqrt2\ \mathrm{m}$.

For distance $200$ m, $(5t)^2+(200-20t)^2=200^2$.

This simplifies to $425t^2-8000t=0$, so $t(425t-8000)=0$. Hence $t=0$ or $t=\dfrac{8000}{425}=\dfrac{320}{17}$ s.