3.02e Two-dimensional constant acceleration: with vectors

2 The acceleration of a particle of mass 4 kg is given by \(\mathbf { a } = ( 9 \mathbf { i } - 4 t \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { 2 }\), where \(\mathbf { i }\) and \(\mathbf { j }\) are unit vectors and \(t\) is the time in seconds.

1. Find the acceleration of the particle when \(t = 0\) and also when \(t = 3\).
2. Calculate the force acting on the particle when \(t = 3\). The particle has velocity \(( 4 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } { } ^ { 1 }\) when \(t = 1\).
3. Find an expression for the velocity of the particle at time \(t\).

2 In this question, positions are given relative to a fixed origin, O. The \(x\)-direction is east and the \(y\)-direction north; distances are measured in kilometres. Two boats, the Rosemary and the Sage, are having a race between two points A and B.
The position vector of the Rosemary at time \(t\) hours after the start is given by $$\mathbf { r } = \binom { 3 } { 2 } + \binom { 6 } { 8 } t , \text { where } 0 \leqslant t \leqslant 2 .$$ The Rosemary is at point A when \(t = 0\), and at point B when \(t = 2\).

1. Find the distance AB .
2. Show that the Rosemary travels at constant velocity. The position vector of the Sage is given by $$\mathbf { r } = \binom { 3 ( 2 t + 1 ) } { 2 \left( 2 t ^ { 2 } + 1 \right) }$$
3. Plot the points A and B . Draw the paths of the two boats for \(0 \leqslant t \leqslant 2\).
4. What can you say about the result of the race?
5. Find the speed of the Sage when \(t = 2\). Find also the direction in which it is travelling, giving your answer as a compass bearing, to the nearest degree.
6. Find the displacement of the Rosemary from the Sage at time \(t\) and hence calculate the greatest distance between the boats during the race.

7 A projectile P travels in a vertical plane over level ground. Its position vector \(\mathbf { r }\) at time \(t\) seconds after projection is modelled by $$\mathbf { r } = \binom { x } { y } = \binom { 0 } { 5 } + \binom { 30 } { 40 } t - \binom { 0 } { 5 } t ^ { 2 }$$ where distances are in metres and the origin is a point on the level ground.

1. Write down
   (A) the height from which P is projected,
   (B) the value of \(g\) in this model.
2. Find the displacement of P from \(t = 3\) to \(t = 5\).
3. Show that the equation of the trajectory is $$y = 5 + \frac { 4 } { 3 } x - \frac { x ^ { 2 } } { 180 }$$

3 A particle, of mass 10 kg , moves on a smooth horizontal plane. At time \(t\) seconds, the acceleration of the particle is given by $$\left\{ \left( 40 t + 3 t ^ { 2 } \right) \mathbf { i } + 20 \mathrm { e } ^ { - 4 t } \mathbf { j } \right\} \mathrm { m } \mathrm {~s} ^ { - 2 }$$ where the vectors \(\mathbf { i }\) and \(\mathbf { j }\) are perpendicular unit vectors.

1. At time \(t = 1\), the velocity of the particle is \(\left( 6 \mathbf { i } - 5 \mathrm { e } ^ { - 4 } \mathbf { j } \right) \mathrm { m } \mathrm { s } ^ { - 1 }\). Find the velocity of the particle at time \(t\).
2. Calculate the initial speed of the particle.

6 A ship and a navy frigate are a distance of 8 km apart, with the frigate on a bearing of \(120 ^ { \circ }\) from the ship, as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{bcd20c69-cace-408c-8961-169c19ff0231-16_451_549_411_760} The ship travels due east at a constant speed of \(50 \mathrm {~km} \mathrm {~h} ^ { - 1 }\). The frigate travels at a constant speed of \(35 \mathrm {~km} \mathrm {~h} ^ { - 1 }\).

1. 1. Find the bearings, to the nearest degree, of the two possible directions in which the frigate can travel to intercept the ship.
      [0pt] [5 marks]
   2. Hence find the shorter of the two possible times for the frigate to intercept the ship.
      [0pt] [5 marks]
2. The captain of the frigate would like the frigate to travel at less than \(35 \mathrm {~km} \mathrm {~h} ^ { - 1 }\). Find the minimum speed at which the frigate can travel to intercept the ship.
   [0pt] [3 marks] \(7 \quad\) A particle is projected from a point \(O\) on a plane which is inclined at an angle \(\theta\) to the horizontal. The particle is projected up the plane with velocity \(u\) at an angle \(\alpha\) above the horizontal. The particle strikes the plane for the first time at a point \(A\). The motion of the particle is in a vertical plane which contains the line \(O A\). \includegraphics[max width=\textwidth, alt={}, center]{bcd20c69-cace-408c-8961-169c19ff0231-20_469_624_502_685}
   1. Find, in terms of \(u , \theta , \alpha\) and \(g\), the time taken by the particle to travel from \(O\) to \(A\).
   2. The particle is moving horizontally when it strikes the plane at \(A\). By using the identity \(\sin ( P - Q ) = \sin P \cos Q - \cos P \sin Q\), or otherwise, show that $$\tan \alpha = k \tan \theta$$ where \(k\) is a constant to be determined.
      [0pt] [5 marks]
      \includegraphics[max width=\textwidth, alt={}]{bcd20c69-cace-408c-8961-169c19ff0231-24_2488_1728_219_141}

1. \hspace{0pt} [In this question the unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are due east and due north respectively.]

A coastguard patrol boat \(C\) is moving with constant velocity \(( 8 \mathbf { i } + u \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }\). Another ship \(S\) is moving with constant velocity \(( 12 \mathbf { i } + 16 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }\).

1. Find, in terms of \(u\), the velocity of \(C\) relative to \(S\). At noon, \(S\) is 10 km due west of \(C\).
   If \(C\) is to intercept \(S\),
2. 1. find the value of \(u\).
   2. Using this value of \(u\), find the time at which \(C\) would intercept \(S\). If instead, at noon, \(C\) is moving with velocity \(( 8 \mathbf { i } + 8 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }\) and continues at this constant velocity,
3. find the distance of closest approach of \(C\) to \(S\).

7 At midnight, ship \(A\) is 70 km due north of ship \(B\). Ship \(A\) travels with constant velocity \(20 \mathrm {~km} \mathrm {~h} ^ { - 1 }\) in the direction with bearing \(140 ^ { \circ }\). Ship \(B\) travels with constant velocity \(15 \mathrm {~km} \mathrm {~h} ^ { - 1 }\) in the direction with bearing \(025 ^ { \circ }\).

1. Find the magnitude and direction of the velocity of \(A\) relative to \(B\).
2. Find the distance between the ships when they are at their closest, and find the time when this occurs.

3 From a speedboat, a ship is sighted on a bearing of \(045 ^ { \circ }\). The ship has constant velocity \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in the direction with bearing \(120 ^ { \circ }\). The speedboat travels in a straight line with constant speed \(15 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and intercepts the ship.

1. Find the bearing of the course of the speedboat.
2. Find the magnitude of the velocity of the ship relative to the speedboat.

4 A cruise ship \(C\) is sailing due north at a constant speed of \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A boat \(B\), initially 2000 m due west of \(C\), sails with constant speed \(11 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) on a straight line course which takes it as close as possible to \(C\).

1. Find the bearing of the direction in which \(B\) sails.
2. Find the shortest distance between \(B\) and \(C\) in the subsequent motion.

4 \includegraphics[max width=\textwidth, alt={}, center]{a9e010ce-c3a8-4f95-a154-fd16ef3e5e5b-2_823_650_1318_751} A boat \(A\) is travelling with constant speed \(6.3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) on a course with bearing \(075 ^ { \circ }\). Boat \(B\) is travelling with constant speed \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) on a course with bearing \(025 ^ { \circ }\). At one instant, \(A\) is 2500 m due north of \(B\) (see diagram).

1. Find the magnitude and bearing of the velocity of \(A\) relative to \(B\).
2. Find the shortest distance between \(A\) and \(B\) in the subsequent motion.

4 From a helicopter, a small plane is spotted 3750 m away on a bearing of \(075 ^ { \circ }\). The plane is at the same altitude as the helicopter, and is flying with constant speed \(62 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in a horizontal straight line on a bearing of \(295 ^ { \circ }\). The helicopter flies with constant speed \(48 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in a straight line, and intercepts the plane.

1. Find the bearings of the two possible directions in which the helicopter could fly.
2. Given that interception occurs in the shorter of the two possible times, find the time taken to make the interception. \includegraphics[max width=\textwidth, alt={}, center]{afecdb38-c372-480a-9d6d-fafe6a371dc2-3_668_298_260_922} A uniform lamina of mass 63 kg occupies the region bounded by the \(x\)-axis, the \(y\)-axis, and the curve \(y = 8 - x ^ { 3 }\) for \(0 \leqslant x \leqslant 2\). The unit of length is the metre. The vertices of the lamina are \(O ( 0,0 )\), \(A ( 2,0 )\) and \(B ( 0,8 )\) (see diagram).
3. Show that the moment of inertia of this lamina about \(O B\) is \(56 \mathrm {~kg} \mathrm {~m} ^ { 2 }\). It is given that the moment of inertia of the lamina about \(O A\) is \(1036.8 \mathrm {~kg} \mathrm {~m} ^ { 2 }\), and the centre of mass of the lamina has coordinates \(\left( \frac { 4 } { 5 } , \frac { 24 } { 7 } \right)\). The lamina is free to rotate in a vertical plane about a fixed horizontal axis passing through \(O\) and perpendicular to the lamina. Starting with the lamina at rest with \(B\) vertically above \(O\), a couple of constant anticlockwise moment 800 Nm is applied to the lamina.
4. Show that the lamina begins to rotate anticlockwise.
5. Find the angular speed of the lamina at the instant when \(O B\) first becomes horizontal. \includegraphics[max width=\textwidth, alt={}, center]{afecdb38-c372-480a-9d6d-fafe6a371dc2-4_709_752_267_699} A smooth circular wire, with centre \(O\) and radius \(a\), is fixed in a vertical plane, and the point \(A\) is on the wire at the same horizontal level as \(O\). A small bead \(B\) of mass \(m\) can move freely on the wire. A light elastic string, with natural length \(a\) and modulus of elasticity \(\sqrt { 3 } m g\), passes through a fixed ring at \(A\), and has one end fixed at \(O\) and the other end attached to \(B\). The section \(A B\) of the string is at an angle \(\theta\) above the horizontal, where \(- \frac { 1 } { 2 } \pi < \theta < \frac { 1 } { 2 } \pi\), so that \(O B\) is at an angle \(2 \theta\) to the horizontal (see diagram).
6. Taking \(O\) as the reference level for gravitational potential energy, show that the total potential energy of the system is $$m g a ( \sqrt { 3 } + \sqrt { 3 } \cos 2 \theta + \sin 2 \theta ) .$$
7. Find the two values of \(\theta\) for which the system is in equilibrium.
8. For each position of equilibrium, determine whether it is stable or unstable. \includegraphics[max width=\textwidth, alt={}, center]{afecdb38-c372-480a-9d6d-fafe6a371dc2-5_478_1403_267_372} A thin horizontal rail is fixed at a height of 0.6 m above horizontal ground. A non-uniform straight \(\operatorname { rod } A B\) has mass 6 kg and length 3 m ; its centre of mass \(G\) is 2 m from \(A\) and 1 m from \(B\), and its moment of inertia about a perpendicular axis through its mid-point \(M\) is \(4.9 \mathrm {~kg} \mathrm {~m} ^ { 2 }\). The rod is placed in a vertical plane perpendicular to the rail, with \(A\) on the ground and \(M\) in contact with the rail. It is released from rest in this position, and begins to rotate about \(M\), without slipping on the rail. When the angle between \(A B\) and the upward vertical is \(\theta\) radians, the rod has angular speed \(\omega \mathrm { rad } \mathrm { s } ^ { - 1 }\), the frictional force in the direction \(A B\) is \(F \mathrm {~N}\), and the normal reaction is \(R \mathrm {~N}\) (see diagram).
9. Show that \(\omega ^ { 2 } = 4.8 - 12 \cos \theta\).
10. Find the angular acceleration of the rod in terms of \(\theta\).
11. Show that \(F = 94.8 \cos \theta - 14.4\), and find \(R\) in terms of \(\theta\).
12. Given that the coefficient of friction between the rod and the rail is 0.9 , show that the rod will slip on the rail before \(B\) hits the ground.

4 \includegraphics[max width=\textwidth, alt={}, center]{ea62d6d9-ac2f-44e7-8bfb-ae9aeea7109b-2_688_777_1382_683} From a boat \(B\), a cruiser \(C\) is observed 3500 m away on a bearing of \(040 ^ { \circ }\). The cruiser \(C\) is travelling with constant speed \(15 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) along a straight line course with bearing \(110 ^ { \circ }\) (see diagram). The boat \(B\) travels with constant speed \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) on a straight line course which takes it as close as possible to the cruiser \(C\).

1. Show that the bearing of the course of \(B\) is \(073 ^ { \circ }\), correct to the nearest degree.
2. Find the magnitude and the bearing of the velocity of \(C\) relative to \(B\).
3. Find the shortest distance between \(B\) and \(C\) in the subsequent motion.

6 Two ships \(P\) and \(Q\) are moving on straight courses with constant speeds. At one instant \(Q\) is 80 km from \(P\) on a bearing of \(220 ^ { \circ }\). Three hours later, \(Q\) is 36 km due south of \(P\).

1. Show that the velocity of \(Q\) relative to \(P\) is \(19.1 \mathrm {~km} \mathrm {~h} ^ { - 1 }\) in the direction with bearing \(063.8 ^ { \circ }\) (both correct to 3 significant figures).
2. Find the shortest distance between the two ships in the subsequent motion. Given that the speed of \(P\) is \(28 \mathrm {~km} \mathrm {~h} ^ { - 1 }\) and \(Q\) is travelling in the direction with bearing \(105 ^ { \circ }\), find
3. the bearing of the direction in which \(P\) is travelling,
4. the speed of \(Q\).

3 \includegraphics[max width=\textwidth, alt={}, center]{ab760a4b-e0ec-4256-838f-ed6c762ff18b-2_460_388_1160_826} A ship \(S\) is travelling with constant velocity \(15 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) on a course with bearing \(120 ^ { \circ }\). A patrol boat \(B\) observes the ship when \(S\) is due north of \(B\). The patrol boat \(B\) then moves with constant speed \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in a straight line (see diagram).

1. Given that \(V = 18\), find the bearing of the course of \(B\) such that \(B\) intercepts \(S\).
2. Given instead that \(V = 9\), find the bearing of the course of \(B\) such that \(B\) passes as close as possible to \(S\).
3. Find the smallest value of \(V\) for which it is possible for \(B\) to intercept \(S\).

3 A particle Q of mass \(m\) moves in a horizontal plane under the action of a single force \(\mathbf { F }\). At time \(t , \mathrm { Q }\) has velocity \(\binom { 2 } { 3 t - 2 }\).

1. Find an expression for \(\mathbf { F }\) in terms of \(m\). At time \(t\), the displacement of Q is given by \(\mathbf { r } = \binom { x } { y }\). When \(t = 1 , \mathrm { Q }\) is at the point with position vector \(\binom { 4 } { - 4 }\).
2. Find the equation of the path of Q , giving your answer in the form \(y = a x ^ { 2 } + b x + c\), where \(a\), \(b\) and \(c\) are constants to be determined.
3. What can you deduce about the path of Q from the value of the constant \(c\) you found in part (b)?

1. \begin{figure}[h] \end{figure} Figure 1 represents the plan of part of a smooth horizontal floor, where \(W _ { 1 }\) and \(W _ { 2 }\) are two fixed parallel vertical walls. The walls are 3 metres apart. A particle lies at rest at a point \(O\) on the floor between the two walls, where the point \(O\) is \(d\) metres, \(0 < d \leqslant 3\), from \(W _ { 1 }\) At time \(t = 0\), the particle is projected from \(O\) towards \(W _ { 1 }\) with speed \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in a direction perpendicular to the walls. The coefficient of restitution between the particle and each wall is \(\frac { 2 } { 3 }\) The particle returns to \(O\) at time \(t = T\) seconds, having bounced off each wall once.

1. Show that \(T = \frac { 45 - 5 d } { 4 u }\) The value of \(u\) is fixed, the particle still hits each wall once but the value of \(d\) can now vary.
2. Find the least possible value of \(T\), giving your answer in terms of \(u\). You must give a reason for your answer.

8 In this question the value of \(\boldsymbol { g \) should be taken as \(\mathbf { 1 0 } \mathbf { m ~ s } ^ { \mathbf { - 2 } }\).} As shown in Fig. 8, particles A and B are projected towards one another. Each particle has an initial speed of \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) vertically and \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) horizontally. Initially A and B are 70 m apart horizontally and B is 15 m higher than A . Both particles are projected over horizontal ground. \begin{figure}[h] \end{figure}

1. Show that, \(t\) seconds after projection, the height in metres of each particle above its point of projection is \(10 t - 5 t ^ { 2 }\).
2. Calculate the horizontal range of A . Deduce that A hits the horizontal ground between the initial positions of A and B .
3. Calculate the horizontal distance travelled by B before reaching the ground.
4. Show that the paths of the particles cross but that the particles do not collide if they are projected at the same time. In fact, particle A is projected 2 seconds after particle B .
5. Verify that the particles collide 0.75 seconds after A is projected.

2 In this question the value of \(g\) should be taken as \(10 \mathrm {~m \mathrm {~s} ^ { 2 }\).} As shown in Fig. 8, particles A and B are projected towards one another. Each particle has an initial speed of \(10 \mathrm {~m} \mathrm {~s} ^ { 1 }\) vertically and \(20 \mathrm {~m} \mathrm {~s} { } ^ { 1 }\) horizontally. Initially A and B are 70 m apart horizontally and B is 15 m higher than A . Both particles are projected over horizontal ground. \begin{figure}[h] \end{figure}

1. Show that, \(t\) seconds after projection, the height in metres of each particle above its point of projection is \(10 t - 5 t ^ { 2 }\).
2. Calculate the horizontal range of A . Deduce that A hits the horizontal ground between the initial positions of A and B .
3. Calculate the horizontal distance travelled by B before reaching the ground.
4. Show that the paths of the particles cross but that the particles do not collide if they are projected at the same time. In fact, particle A is projected 2 seconds after particle B .
5. Verify that the particles collide 0.75 seconds after A is projected.

4 Water flows in a constant direction at a constant speed of \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A boat travels in the water at a speed of \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) relative to the water.

1. The direction in which the boat travels relative to the water is perpendicular to the direction of motion of the water. The resultant velocity of the boat is \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(74 ^ { \circ }\) to the direction of motion of the water, as shown in the diagram. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{6151e6ab-30af-4d1c-ab4a-e7dbad170cbf-004_120_164_662_488} \captionsetup{labelformat=empty} \caption{Velocity of the water}
   \end{figure} \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{6151e6ab-30af-4d1c-ab4a-e7dbad170cbf-004_126_186_667_890} \captionsetup{labelformat=empty} \caption{Velocity of the boat relative to the water}
   \end{figure}
   1. Find \(V\).
   2. Show that \(u = 3.44\), correct to three significant figures.
2. The boat changes course so that it travels relative to the water at an angle of \(45 ^ { \circ }\) to the direction of motion of the water. The resultant velocity of the boat is now of magnitude \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The velocity of the water is unchanged, as shown in the diagram below. $$\xrightarrow { 3.44 \mathrm {~m} \mathrm {~s} ^ { - 1 } }$$
   \includegraphics[max width=\textwidth, alt={}]{6151e6ab-30af-4d1c-ab4a-e7dbad170cbf-004_132_273_1493_895}
   Velocity of the boat relative to the water \includegraphics[max width=\textwidth, alt={}, center]{6151e6ab-30af-4d1c-ab4a-e7dbad170cbf-004_232_355_1498_1384} Find the value of \(v\).
   (4 marks)

4 Water flows in a constant direction at a constant speed of \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A boat travels in the water at a speed of \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) relative to the water.

1. The direction in which the boat travels relative to the water is perpendicular to the direction of motion of the water. The resultant velocity of the boat is \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(74 ^ { \circ }\) to the direction of motion of the water, as shown in the diagram. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{c220e6c4-2676-4022-8301-7d720dc082b2-3_120_164_662_488} \captionsetup{labelformat=empty} \caption{Velocity of the water}
   \end{figure} \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{c220e6c4-2676-4022-8301-7d720dc082b2-3_126_186_667_890} \captionsetup{labelformat=empty} \caption{Velocity of the boat relative to the water}
   \end{figure}
   1. Find \(V\).
   2. Show that \(u = 3.44\), correct to three significant figures.
2. The boat changes course so that it travels relative to the water at an angle of \(45 ^ { \circ }\) to the direction of motion of the water. The resultant velocity of the boat is now of magnitude \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The velocity of the water is unchanged, as shown in the diagram below. $$\xrightarrow { 3.44 \mathrm {~m} \mathrm {~s} ^ { - 1 } }$$
   \includegraphics[max width=\textwidth, alt={}]{c220e6c4-2676-4022-8301-7d720dc082b2-3_132_273_1493_895}
   Velocity of the boat relative to the water \includegraphics[max width=\textwidth, alt={}, center]{c220e6c4-2676-4022-8301-7d720dc082b2-3_232_355_1498_1384} Find the value of \(v\).
   (4 marks)

8 A boat is initially at the origin, heading due east at \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). It then experiences a constant acceleration of \(( - 0.2 \mathbf { i } + 0.25 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 2 }\). The unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are directed east and north respectively.

1. State the initial velocity of the boat as a vector.
2. Find an expression for the velocity of the boat \(t\) seconds after it has started to accelerate.
3. Find the value of \(t\) when the boat is travelling due north.
4. Find the bearing of the boat from the origin when the boat is travelling due north.

11 In this question the unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are in the directions east and north respectively. A particle of mass 0.12 kg is moving so that its position vector \(\mathbf { r }\) metres at time \(t\) seconds is given by \(\mathbf { r } = 2 t ^ { 3 } \mathbf { i } + \left( 5 t ^ { 2 } - 4 t \right) \mathbf { j }\).

1. Show that when \(t = 0.7\) the bearing on which the particle is moving is approximately \(044 ^ { \circ }\).
2. Find the magnitude of the resultant force acting on the particle at the instant when \(t = 0.7\).
3. Determine the times at which the particle is moving on a bearing of \(045 ^ { \circ }\).

1. \hspace{0pt} [In this question \(\mathbf { i }\) and \(\mathbf { j }\) are horizontal unit vectors due east and due north respectively and position vectors are given relative to the fixed point \(O\).]

A particle \(P\) moves with constant acceleration.
At time \(t = 0\), the particle is at \(O\) and is moving with velocity ( \(2 \mathbf { i } - 3 \mathbf { j }\) ) \(\mathrm { ms } ^ { - 1 }\) At time \(t = 2\) seconds, \(P\) is at the point \(A\) with position vector ( \(7 \mathbf { i } - 10 \mathbf { j }\) ) m.

1. Show that the magnitude of the acceleration of \(P\) is \(2.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) At the instant when \(P\) leaves the point \(A\), the acceleration of \(P\) changes so that \(P\) now moves with constant acceleration ( \(4 \mathbf { i } + 8.8 \mathbf { j }\) ) \(\mathrm { m } \mathrm { s } ^ { - 2 }\) At the instant when \(P\) reaches the point \(B\), the direction of motion of \(P\) is north east.
2. Find the time it takes for \(P\) to travel from \(A\) to \(B\).

1. \hspace{0pt} [In this question \(\mathbf { i }\) and \(\mathbf { j }\) are horizontal unit vectors due east and due north respectively]

A radio controlled model boat is placed on the surface of a large pond.
The boat is modelled as a particle.
At time \(t = 0\), the boat is at the fixed point \(O\) and is moving due north with speed \(0.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
Relative to \(O\), the position vector of the boat at time \(t\) seconds is \(\mathbf { r }\) metres.
At time \(t = 15\), the velocity of the boat is \(( 10.5 \mathbf { i } - 0.9 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\).
The acceleration of the boat is constant.

1. Show that the acceleration of the boat is \(( 0.7 \mathbf { i } - 0.1 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 2 }\).
2. Find \(\mathbf { r }\) in terms of \(t\).
3. Find the value of \(t\) when the boat is north-east of \(O\).
4. Find the value of \(t\) when the boat is moving in a north-east direction.