Working with Vectors: past paper questions & worked solutions

Take east as $\mathbf{i}$ and north as $\mathbf{j}$. The force due west is $-10\mathbf{i}$. A bearing of $330^\circ$ is $30^\circ$ west of north, so the 8 N force has components $-8\sin30^\circ\mathbf{i}+8\cos30^\circ\mathbf{j}$.

$\mathbf{F}=(-10-4)\mathbf{i}+4\sqrt3\mathbf{j}=-14\mathbf{i}+4\sqrt3\mathbf{j}$.

$|\mathbf{F}|=\sqrt{(-14)^2+(4\sqrt3)^2}=\sqrt{196+48}=\sqrt{244}=2\sqrt{61}\ \mathrm{N}$.

$2\sqrt{61}=15.620\ldots$, so the magnitude is about $16\ \mathrm{N}$ to 2 significant figures.

Take the initial direction of P as positive. Then before impact, $v_P=3u$ and $v_Q=-4u$. The impulse on P acts opposite to P\'s initial motion, so its signed impulse is $-\dfrac{33}{2}mu$.

For P, impulse equals change in momentum: $-\dfrac{33}{2}mu=3m(v_P-3u)$. Dividing by $3m$ gives $v_P-3u=-\dfrac{11u}{2}$, so $v_P=-\dfrac{5u}{2}$.

The negative sign means P has reversed direction. Its speed is therefore $\left|-\dfrac{5u}{2}\right|=\dfrac{5u}{2}$.

Q receives impulse $+\dfrac{33}{2}mu$ in the positive direction. Hence $\dfrac{33}{2}mu=5m(v_Q-(-4u))=5m(v_Q+4u)$.

Dividing by $5m$ gives $v_Q+4u=\dfrac{33u}{10}$, so $v_Q=-\dfrac{7u}{10}$. Q is still moving in its original direction, and its speed is $\dfrac{7u}{10}$.

$\text{Impulse} = \text{Final momentum} - \text{Initial momentum}$ For particle Q: ; Mass = $3m$ ; Initial velocity = $2u$ ; Final velocity = $v$ ; Impulse received = $4.5mu$

$4.5mu = 3m \times v - 3m \times 2u$ $4.5mu = 3mv - 6mu$ $4.5u = 3v - 6u$ $4.5u + 6u = 3v$ $10.5u = 3v$ $v = \frac{10.5u}{3}$ $v = 3.5u$

$\text{Total momentum before} = \text{Total momentum after}$ Before collision: $\text{P momentum} = km \times 5u = 5kmu$ $\text{Q momentum} = 3m \times 2u = 6mu$ $\text{Total} = 5kmu + 6mu$ After collision: $\text{P momentum} = km \times 2u = 2kmu$ $\text{Q momentum} = 3m \times 3.5u = 10.5mu$ $\text{Total} = 2kmu + 10.5mu$

$5kmu + 6mu = 2kmu + 10.5mu$ $5km + 6m = 2km + 10.5m$ $5k + 6 = 2k + 10.5$ $5k - 2k = 10.5 - 6$ $3k = 4.5$ $k = \frac{4.5}{3}$ $k = 1.5$

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 $7.2\ \mathrm{m}$.

The tractor has mass $6000$ kg. Its forward driving force is $7400$ N, and the backward forces are $200$ N resistance and $6000$ N tension.

$7400-200-6000=6000a$, so $1200=6000a$ and $a=0.2\ \mathrm{m\,s^{-2}}$.

The block has mass $2000$ kg. The forward force is the tension $6000$ N and the resistance is $R$ N.

$6000-R=2000(0.2)=400$. Therefore $R=5600\ \mathrm{N}$.

The cable is inextensible, so the tractor and block move together with the same acceleration.

Take the original direction of P as positive. Before the collision, P has velocity $3u$ and Q has velocity $-u$.

After collision, each direction is reversed and each speed is halved. So P has velocity $-\dfrac32u$ and Q has velocity $\dfrac12u$.

$km(3u)+m(-u)=km\left(-\dfrac32u\right)+m\left(\dfrac12u\right)$.

Divide by $mu$: $3k-1=-\dfrac32k+\dfrac12$. Hence $\dfrac92k=\dfrac32$, so $k=\dfrac13$.

Impulse on Q equals change in momentum of Q: $I=m\left(\dfrac12u-(-u)\right)=\dfrac32mu$.

The magnitude of the impulse is $\dfrac32mu$.

Take P's original direction as positive. Q initially has velocity $-2u$. Since Q reverses direction and then has speed $3u$, its final velocity is $3u$.

Impulse equals change in momentum: $I=5m(3u-(-2u))=25mu$.

Before collision, total momentum is $3m(ku)+5m(-2u)=3mku-10mu$.

If P has final velocity $u$, final momentum is $3m(u)+5m(3u)=18mu$. So $3mku-10mu=18mu$, giving $3k=28$ and $k=\dfrac{28}{3}$.

If P has final velocity $-u$, final momentum is $3m(-u)+5m(3u)=12mu$. So $3mku-10mu=12mu$, giving $3k=22$ and $k=\dfrac{22}{3}$.

A bearing of $240^\circ$ is $60^\circ$ west of south. Hence $\mathbf{F}=(-8\sin60^\circ)\mathbf{i}+(-8\cos60^\circ)\mathbf{j}=-4\sqrt3\mathbf{i}-4\mathbf{j}$.

The force $\mathbf{G}$ is due south with magnitude $10$, so $\mathbf{G}=-10\mathbf{j}$. Therefore $\mathbf{R}=\mathbf{F}+\mathbf{G}=-4\sqrt3\mathbf{i}-14\mathbf{j}$.

$|\mathbf{R}|=\sqrt{(-4\sqrt3)^2+(-14)^2}=\sqrt{48+196}=\sqrt{244}\ \mathrm{N}$.

The resultant points south-west. The angle west of south satisfies $\tan\theta=\dfrac{4\sqrt3}{14}=\dfrac{2\sqrt3}{7}$.

$\theta=26.3^\circ$. Bearing $=180^\circ+26.3^\circ=206.3^\circ$, so the bearing is $206^\circ$ to the nearest degree.

Since $\mathbf F$ has magnitude $4$ N on a bearing of $120^\circ$, $2\mathbf F$ has magnitude $8$ N on the same bearing.

$2\mathbf F=8\sin120^\circ\mathbf i+8\cos120^\circ\mathbf j=4\sqrt3\mathbf i-4\mathbf j$. Also $\mathbf G=6\mathbf j$.

$\mathbf P=2\mathbf F+\mathbf G=4\sqrt3\mathbf i+2\mathbf j$.

$|\mathbf P|=\sqrt{(4\sqrt3)^2+2^2}=\sqrt{48+4}=\sqrt{52}$ N.

The vector is in the north-east quadrant. The angle east of north satisfies $\tan\theta=\dfrac{4\sqrt3}{2}=2\sqrt3$.

$\theta=73.9^\circ$ approximately, so the bearing is $074^\circ$ to the nearest degree.

The resultant force is $(2p\mathbf{i}-3p\mathbf{j})+(q\mathbf{i}+\mathbf{j})+(3\mathbf{i}-q\mathbf{j})=(2p+q+3)\mathbf{i}+(-3p+1-q)\mathbf{j}$.

The resultant acts in the direction $4\mathbf{i}-5\mathbf{j}$, so the ratio of the components is the same: $\dfrac{-3p+1-q}{2p+q+3}=\dfrac{-5}{4}$.

Cross-multiplying gives $4(-3p+1-q)=-5(2p+q+3)$. Therefore $-12p+4-4q=-10p-5q-15$, so $2p-q=19$.

With $p=7$, the relation gives $14-q=19$, hence $q=-5$.

Substitute $p=7$ and $q=-5$ into the resultant: $\mathbf{F}=12\mathbf{i}-15\mathbf{j}$. Since the mass is $0.5$ kg, $\mathbf{F}=m\mathbf{a}$ gives $12\mathbf{i}-15\mathbf{j}=0.5\mathbf{a}$.

Dividing by $0.5$ gives $\mathbf{a}=24\mathbf{i}-30\mathbf{j}\ \mathrm{m\,s^{-2}}$.

$\text{Impulse} = \text{Final momentum} - \text{Initial momentum}$ For particle Q (mass = 3m): ; Initial velocity: $-u$ (moving left, negative) ; Final velocity: $+2w$ (moving right, positive) $\text{Impulse} = 3m(2w) - 3m(-u)$ $\frac{7mu}{2} = 6mw + 3mu$ $\frac{7mu}{2} = 3m(2w + u)$ $\frac{7u}{2} = 6w + 3u$ $6w = \frac{7u}{2} - 3u$ $6w = \frac{7u - 6u}{2}$ $6w = \frac{u}{2}$ $w = \frac{u}{12}$

$\text{Total momentum before} = \text{Total momentum after}$ Before collision: $\text{P momentum} = 7m \times ku = 7mku \text{ (right)}$ $\text{Q momentum} = 3m \times (-u) = -3mu \text{ (left)}$ $\text{Total} = 7mku - 3mu$ After collision: $\text{P momentum} = 7m \times w = 7mw \text{ (right)}$ $\text{Q momentum} = 3m \times 2w = 6mw \text{ (right)}$ $\text{Total} = 7mw + 6mw = 13mw$

$7mku - 3mu = 13mw$ $7ku - 3u = 13w$

$7ku - 3u = 13 \times \frac{u}{12}$ $7ku - 3u = \frac{13u}{12}$ $7ku = \frac{13u}{12} + 3u$ $7ku = \frac{13u + 36u}{12}$ $7ku = \frac{49u}{12}$ $7k = \frac{49}{12}$ $k = \frac{49}{84} = \frac{7}{12}$

P could move either right (velocity = $+w$) or left (velocity = $-w$) Case 1: P moves right with velocity $+w$ Already solved: $k = \frac{7}{12}$ Case 2: P moves left with velocity $-w$ Total momentum after = $-7mw + 6mw = -mw$ $7mku - 3mu = -mw$ $7ku - 3u = -w$ $7ku - 3u = -\frac{u}{12}$ $7ku = 3u - \frac{u}{12}$ $7ku = \frac{36u - u}{12}$ $7ku = \frac{35u}{12}$ $k = \frac{35}{84} = \frac{5}{12}$

$\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3=(6\mathbf{i}+8\mathbf{j})+(-16\mathbf{i}+2\mathbf{j})+(-2\mathbf{i}+8\mathbf{j})=-12\mathbf{i}+18\mathbf{j}$.

The mass is $2.5$ kg, so $-12\mathbf{i}+18\mathbf{j}=2.5\mathbf{a}$. Hence $\mathbf{a}=-4.8\mathbf{i}+7.2\mathbf{j}\ \mathrm{m\,s^{-2}}$.

$|\mathbf{a}|=\sqrt{(-4.8)^2+7.2^2}=8.653\ldots$, so the magnitude is $8.7\ \mathrm{m\,s^{-2}}$ to 2 significant figures. Exactly, $|\mathbf{a}|=\dfrac{12\sqrt{13}}{5}$.

With $\mathbf{F}_4=p\mathbf{i}+p\mathbf{j}$, the new resultant is $(p-12)\mathbf{i}+(p+18)\mathbf{j}$.

The resultant acts in the direction $7\mathbf{i}+2\mathbf{j}$, so $\dfrac{p-12}{p+18}=\dfrac{7}{2}$.

$2(p-12)=7(p+18)$, so $2p-24=7p+126$. Therefore $-150=5p$ and $p=-30$.

The two given forces have angle $150^\circ$ between them, so the triangle used for adding them has the supplementary angle $30^\circ$ between the side of length $X$ and the side of length $5\sqrt3$.

The resultant has magnitude $\sqrt{129}$. Therefore $129=X^2+(5\sqrt3)^2-2X(5\sqrt3)\cos30^\circ$.

Since $(5\sqrt3)^2=75$ and $2(5\sqrt3)\cos30^\circ=15$, the equation becomes $129=X^2+75-15X$. Thus $X^2-15X-54=0$, giving $X=18$.

Let $\beta$ be the angle between Q and the resultant. Using the cosine rule in the same triangle, $\dfrac{18}{\sin\beta}=\dfrac{\sqrt{129}}{\sin30^\circ}$.

This gives an obtuse angle for $\beta$. Equivalently, $\cos\beta=\dfrac{(5\sqrt3)^2+129-18^2}{2(5\sqrt3)\sqrt{129}}$, so $\beta=127.6\ldots^\circ$.

To the nearest degree, the angle between Q and the resultant is $128^\circ$.

Take the original direction of A as positive. Before the collision, A has velocity $1.5$ and B has velocity $-1.5$.

After the collision, A keeps the same direction. Let A's speed be $v$. The valid motion has B moving in A's direction with speed $v+1$.

$3m(1.5)+m(-1.5)=3mv+m(v+1)$.

$4.5m-1.5m=3mv+mv+m$, so $3=4v+1$. Hence $v=0.5$, and B's speed is $v+1=1.5$.

For B, impulse equals change in momentum: $I=m(1.5-(-1.5))=3m$.

The magnitude of the impulse is $3m\ \mathrm{N\,s}$.

The boat travels straight, so the sideways components balance: $500\sin40^\circ=P\sin\alpha$.

The speed is constant, so the forward components balance the resistance: $500\cos40^\circ+P\cos\alpha=900$.

From the two equations, $P\sin\alpha=500\sin40^\circ$ and $P\cos\alpha=900-500\cos40^\circ$.

Divide the sine equation by the cosine equation: $\tan\alpha=\dfrac{500\sin40^\circ}{900-500\cos40^\circ}$. This gives $\alpha=31.9^\circ$, so $\alpha=32^\circ$ approximately.

Use $P\sin\alpha=500\sin40^\circ$: $P=\dfrac{500\sin40^\circ}{\sin31.9^\circ}=608.7\ldots\ \mathrm{N}$. Hence $P\approx610\ \mathrm{N}$.

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

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 $t=\dfrac32$.

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.

$\mathbf{F}=(-3\mathbf{i}+2\mathbf{j})+(p\mathbf{i}+q\mathbf{j})=(p-3)\mathbf{i}+(q+2)\mathbf{j}$.

Since the particle starts from rest, its motion is in the direction of the resultant force. This direction is $\mathbf{i}-2\mathbf{j}$, so $\dfrac{q+2}{p-3}=-2$.

$q+2=-2(p-3)=-2p+6$, hence $2p+q-4=0$.

Substitute $p=5$ into $2p+q-4=0$: $10+q-4=0$, so $q=-6$.

The resultant force is $(5-3)\mathbf{i}+(-6+2)\mathbf{j}=2\mathbf{i}-4\mathbf{j}$ N.

The mass is $0.5$ kg, so $\mathbf{a}=\dfrac{2\mathbf{i}-4\mathbf{j}}{0.5}=4\mathbf{i}-8\mathbf{j}$.

Starting from rest, $\mathbf{v}=\mathbf{a}t=(4\mathbf{i}-8\mathbf{j})4=16\mathbf{i}-32\mathbf{j}$.

Speed $=\sqrt{16^2+(-32)^2}=\sqrt{1280}=16\sqrt5\ \mathrm{m\,s^{-1}}$, about $36\ \mathrm{m\,s^{-1}}$.

Equilibrium gives $\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3=\mathbf{0}$. Now $\mathbf{F}_1+\mathbf{F}_2=(5\mathbf{i}+2\mathbf{j})+(-3\mathbf{i}+\mathbf{j})=2\mathbf{i}+3\mathbf{j}$.

So $\mathbf{F}_3=-2\mathbf{i}-3\mathbf{j}$.

The vector $-\mathbf{j}$ points vertically downward. For $\mathbf{F}_3=-2\mathbf{i}-3\mathbf{j}$, the angle from $-\mathbf{j}$ satisfies $\tan\theta=\dfrac{2}{3}$.

$\theta=33.7^\circ$, so the angle is about $34^\circ$.

With $\mathbf{F}_4=\lambda(\mathbf{i}+3\mathbf{j})$, the resultant is $(2+\lambda)\mathbf{i}+(3+3\lambda)\mathbf{j}$.

The mass is $4$ kg and acceleration magnitude is $3.25$, so the resultant force magnitude is $4(3.25)=13$ N.

$(2+\lambda)^2+(3+3\lambda)^2=13^2$.

Expanding gives $10\lambda^2+22\lambda-156=0$. The roots are $3$ and $-5.2$, and since $\lambda$ is positive, $\lambda=3$.

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.

For equilibrium, $\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3=\mathbf{0}$. Hence $\mathbf{F}_3=-(\mathbf{F}_1+\mathbf{F}_2)$.

$\mathbf{F}_1+\mathbf{F}_2=(3c-14)\mathbf{i}+(4c+7)\mathbf{j}$, so $\mathbf{F}_3=(14-3c)\mathbf{i}+(-7-4c)\mathbf{j}$.

When $c=2$, $\mathbf{F}_1=6\mathbf{i}+8\mathbf{j}$ and $\mathbf{F}_2=-14\mathbf{i}+7\mathbf{j}$. Therefore the resultant is $-8\mathbf{i}+15\mathbf{j}$.

The resultant is in the second quadrant. Its angle from the positive $\mathbf{i}$ direction is $180^\circ-\tan^{-1}\left(\dfrac{15}{8}\right)=118.1^\circ$, so the angle is about $118^\circ$.

$|\mathbf{R}|=\sqrt{(-8)^2+15^2}=17\ \mathrm{N}$.

The acceleration magnitude is $8.5\ \mathrm{m\,s^{-2}}$, so $17=8.5m$. Hence $m=2$.

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

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 $T=3.2$.

The car's graph is a horizontal line at $40\ \mathrm{m\,s^{-1}}$. The motorcyclist's graph starts at zero, rises with gradient $4$ until $t=15$, then rises with smaller gradient $1$ for a further $T$ seconds.

Before $t=15$, the motorcyclist's speed is $4t$. Set this equal to the car's speed: $4t_1=40$, so $t_1=10$ s.

After the first $15$ seconds, the motorcyclist's speed is $4(15)=60\ \mathrm{m\,s^{-1}}$.

During the next $T$ seconds the acceleration is $1\ \mathrm{m\,s^{-2}}$, so the speed at catch-up is $60+T$.

Motorcyclist distance $=\dfrac12(15)(60)+\dfrac12[60+(60+T)]T$. Car distance $=40(15+T)$.

$450+60T+\dfrac12T^2=600+40T$. Hence $\dfrac12T^2+20T-150=0$, so $T^2+40T-300=0$ and $k=40$.

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.

For $\mathbf F=10\mathbf i+\mathbf j$, $|\mathbf F|=\sqrt{10^2+1^2}=\sqrt{101}$ N.

The angle that $\mathbf F$ makes with the positive $\mathbf i$ direction is $\alpha$, where $\tan\alpha=\dfrac{1}{10}$. Thus $\alpha=5.71^\circ$ approximately.

The vector $\mathbf i+\mathbf j$ makes an angle of $45^\circ$ with the positive $\mathbf i$ direction.

So the required angle is $45^\circ-5.71^\circ=39.29^\circ$, approximately $39.3^\circ$.

The resultant of $10\mathbf i+\mathbf j$ and $-15\mathbf i+a\mathbf j$ is $-5\mathbf i+(a+1)\mathbf j$.

Parallel but opposite to $2\mathbf i-3\mathbf j$ means the resultant is a positive multiple of $-2\mathbf i+3\mathbf j$.

Let $-5\mathbf i+(a+1)\mathbf j=\lambda(-2\mathbf i+3\mathbf j)$. From the $\mathbf i$ component, $-5=-2\lambda$, so $\lambda=2.5$. Then $a+1=3(2.5)=7.5$, hence $a=6.5$.

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 $\sqrt{0.5}=0.707\ldots\ \mathrm{m}$.

The graph starts at $(0,0)$, rises in a straight line with gradient $0.8$ until time $T$ and speed $V$, then stays horizontal at speed $V$ until $t=30$.

Uniform acceleration gives $V=0.8T$.

The distance is the area under the speed-time graph: $\dfrac12TV+V(30-T)=200$.

Using $V=0.8T$: $\dfrac12T(0.8T)+(0.8T)(30-T)=200$.

$0.4T^2+24T-0.8T^2=200$, so $-0.4T^2+24T-200=0$. Multiplying by $-2.5$ gives $T^2-60T+500=0$, so $k=60$.

$T^2-60T+500=(T-10)(T-50)=0$, so $T=10$ or $50$. Since the whole race lasts $30$ s, $T=10$.

A more realistic model could include reaction time at the start, and could replace the instantaneous change from acceleration to constant speed with a smoother transition.

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 $\sqrt{\left(\dfrac{20}{3}\right)^2+\left(\dfrac{20}{3}\right)^2}=\dfrac{20\sqrt2}{3}\ \mathrm{km}$.

$\mathbf{F}_{\text{resultant}} = \mathbf{F}_1 + \mathbf{F}_2$ $\mathbf{F}_{\text{resultant}} = (-2\mathbf{i} + 3\mathbf{j}) + (4\mathbf{i} + 2\mathbf{j})$ $= (-2 + 4)\mathbf{i} + (3 + 2)\mathbf{j}$ $= 2\mathbf{i} + 5\mathbf{j} \text{ N}$

$\mathbf{F} = m\mathbf{a}$ $2\mathbf{i} + 5\mathbf{j} = 2 \times \mathbf{a}$ $\mathbf{a} = \frac{2\mathbf{i} + 5\mathbf{j}}{2}$ $\mathbf{a} = \mathbf{i} + 2.5\mathbf{j} \text{ m s}^{-2}$ Or: $\mathbf{a} = \mathbf{i} + \frac{5}{2}\mathbf{j}$ m s$^{-2}$

$\mathbf{v} = \mathbf{u} + \mathbf{a}t$ Given: ; $\mathbf{u} = 3\mathbf{i} - 4\mathbf{j}$ m s$^{-1}$ ; $\mathbf{a} = \mathbf{i} + 2.5\mathbf{j}$ m s$^{-2}$ ; $t = 3$ seconds $\mathbf{v} = (3\mathbf{i} - 4\mathbf{j}) + (\mathbf{i} + 2.5\mathbf{j}) \times 3$ $= 3\mathbf{i} - 4\mathbf{j} + 3\mathbf{i} + 7.5\mathbf{j}$ $= 6\mathbf{i} + 3.5\mathbf{j} \text{ m s}^{-1}$

$|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}$ $\text{Speed} = \sqrt{6^2 + 3.5^2}$ $= \sqrt{36 + 12.25}$ $= \sqrt{48.25}$ $= 6.946...$ $= 6.95 \text{ m s}^{-1} \text{ (3 s.f.)}$

$\mathbf{F}_{\text{resultant}} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3$ $= (-2\mathbf{i} + 3\mathbf{j}) + (4\mathbf{i} + 2\mathbf{j}) + (b\mathbf{i} + c\mathbf{j})$ $= (-2 + 4 + b)\mathbf{i} + (3 + 2 + c)\mathbf{j}$ $= (2 + b)\mathbf{i} + (5 + c)\mathbf{j}$

Given that resultant = $\lambda(\mathbf{i} + \mathbf{j})$, we can write: $(2 + b)\mathbf{i} + (5 + c)\mathbf{j} = \lambda\mathbf{i} + \lambda\mathbf{j}$

For this to be in the form $\lambda(\mathbf{i} + \mathbf{j})$, the coefficients must be equal: $2 + b = \lambda$ $5 + c = \lambda$ Therefore: $2 + b = 5 + c$ $b = 5 + c - 2$ $b = 3 + c$ $b - c = 3$

Resultant = $(2 + b)\mathbf{i} + (5 + c)\mathbf{j}$ Since resultant = $\lambda(\mathbf{i} + \mathbf{j})$, we have $2 + b = 5 + c = \lambda$ $|\mathbf{F}_{\text{resultant}}| = \lambda\sqrt{1^2 + 1^2} = \lambda\sqrt{2}$ Given: $|\mathbf{F}_{\text{resultant}}| = 10\sqrt{2}$ $\lambda\sqrt{2} = 10\sqrt{2}$ $\lambda = 10 \quad \text{or} \quad \lambda = -10$ (The negative value comes from considering the opposite direction)

Case 1: $\lambda = 10$ $2 + b = 10 \quad \Rightarrow \quad b = 8$ $5 + c = 10 \quad \Rightarrow \quad c = 5$ Check: $b - c = 8 - 5 = 3$ ✓ Therefore: $\mathbf{F}_3 = 8\mathbf{i} + 5\mathbf{j}$ N Case 2: $\lambda = -10$ $2 + b = -10 \quad \Rightarrow \quad b = -12$ $5 + c = -10 \quad \Rightarrow \quad c = -15$ Check: $b - c = -12 - (-15) = 3$ ✓ Therefore: $\mathbf{F}_3 = -12\mathbf{i} - 15\mathbf{j}$ N

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

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

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

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 $297^\circ$.

A bearing of $150^\circ$ is $30^\circ$ east of south. Therefore the east component is $40\sin150^\circ=20$ and the north component is $40\cos150^\circ=-20\sqrt3$.

$\mathbf v_B=20\mathbf i-20\sqrt3\mathbf j\ \mathrm{km\,h^{-1}}$.

Boat A starts at $60\mathbf i$ and travels due west at $20\ \mathrm{km\,h^{-1}}$, so $\mathbf r=60\mathbf i-20t\mathbf i=(60-20t)\mathbf i$.

Boat B starts at O, so $\mathbf s=t(20\mathbf i-20\sqrt3\mathbf j)=20t\mathbf i-20\sqrt3t\mathbf j$.

$\overrightarrow{AB}=\mathbf s-\mathbf r=(20t-(60-20t))\mathbf i-20\sqrt3t\mathbf j=(40t-60)\mathbf i-20\sqrt3t\mathbf j$.

At noon the distance between the boats is $60$ km. Set the later distance equal to $60$: $(40t-60)^2+(-20\sqrt3t)^2=60^2$.

Expanding gives $2800t^2-4800t=0$, so $t(2800t-4800)=0$. The later time is $t=\dfrac{4800}{2800}=\dfrac{12}{7}$ hours.

$\dfrac{12}{7}$ hours is $1$ hour and $42.857\ldots$ minutes after noon, which rounds to $1{:}43$ pm, or $13{:}43$.

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.