3.02d Constant acceleration: SUVAT formulae

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.
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   2. Hence find the shorter of the two possible times for the frigate to intercept the ship.
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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.
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      \includegraphics[max width=\textwidth, alt={}]{bcd20c69-cace-408c-8961-169c19ff0231-24_2488_1728_219_141}

2 One end of a light elastic string, of natural length 0.6 m and modulus of elasticity 30 N , is attached to a fixed point \(O\). A particle \(P\) of weight 48 N is attached to the other end of the string. \(P\) is released from rest at a point \(d \mathrm {~m}\) vertically below \(O\). Subsequently \(P\) just reaches \(O\).

1. Find \(d\).
2. Find the magnitude and direction of the acceleration of \(P\) when it has travelled 1.3 m from its point of release.

3 A light elastic rope has natural length 15 m . One end of the rope is attached to a fixed point O and the other end is attached to a small rock of mass 12 kg . When the rock is hanging in equilibrium vertically below O , the length of the rope is 15.8 m .

1. Show that the modulus of elasticity of the rope is 2205 N . The rock is pulled down to the point 20 m vertically below O , and is released from rest in this position. It moves upwards, and comes to rest instantaneously, with the rope slack, at the point A .
2. Find the acceleration of the rock immediately after it is released.
3. Use an energy method to find the distance OA. At time \(t\) seconds after release, the rope is still taut and the displacement of the rock below the equilibrium position is \(x\) metres.
4. Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = - 12.25 x\).
5. Write down an expression for \(x\) in terms of \(t\), and hence find the time between releasing the rock and the rope becoming slack.

3. A car starts from rest at the point \(O\) and moves along a straight line. The car accelerates to a maximum velocity, \(V \mathrm {~ms} ^ { - 1 }\), before decelerating and coming to rest again at the point \(A\). The acceleration of the car during this journey, \(a \mathrm {~ms} ^ { - 2 }\), is modelled by the formula $$a = \frac { 500 - k x } { 150 }$$ where \(x\) is the distance in metres of the car from \(O\).
Using this model and given that the car is travelling at \(16 \mathrm {~ms} ^ { - 1 }\) when it is 40 m from \(O\),

1. find \(k\),
2. show that \(V = 41\), correct to 2 significant figures,
3. find the distance \(O A\).

1 A camshaft inside an engine is rotating with angular speed \(42 \mathrm { rads } ^ { - 1 }\). When the throttle is opened the camshaft speeds up with constant angular acceleration, and 8 seconds after the throttle was opened the angular speed is \(76 \mathrm { rad } \mathrm { s } ^ { - 1 }\).

1. Find the angular acceleration of the camshaft.
2. Find the time taken for the camshaft to turn through 810 radians from the moment that the throttle was opened.

4 \includegraphics[max width=\textwidth, alt={}, center]{6e3d5f5e-7ffa-4111-903d-468fb4d20192-2_364_1313_1224_376} An unidentified aircraft \(U\) is flying horizontally with constant velocity \(250 \mathrm {~ms} ^ { - 1 }\) in the direction with bearing \(040 ^ { \circ }\). Two spotter planes \(P\) and \(Q\) are flying horizontally at the same height as \(U\), and at one instant \(P\) is 15000 m due west of \(U\), and \(Q\) is 15000 m due east of \(U\) (see diagram).

1. Plane \(P\) is flying with constant velocity \(210 \mathrm {~ms} ^ { - 1 }\) in the direction with bearing \(070 ^ { \circ }\).

1 Alan is running in a straight line on a bearing of \(090 ^ { \circ }\) at a constant speed of \(4 \mathrm {~ms} ^ { - 1 }\). Ben sees Alan when they are 50 m apart and Alan is on a bearing of \(060 ^ { \circ }\) from Ben. Ben sets off immediately to intercept Alan by running at a constant speed of \(6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Calculate the bearing on which Ben should run to intercept Alan.
2. Calculate the magnitude of the velocity of Ben relative to Alan and find the time it takes, from the moment Ben sees Alan, for Ben to intercept Alan.

1 A turntable is rotating at \(3 \mathrm { rad } \mathrm { s } ^ { - 1 }\). The turntable is then accelerated so that after 4 revolutions it is rotating at \(12.4 \mathrm { rad } \mathrm { s } ^ { - 1 }\). Assuming that the angular acceleration of the turntable is constant,

1. find the angular acceleration,
2. find the time taken to increase its angular speed from \(3 \mathrm { rad } \mathrm { s } ^ { - 1 }\) to \(12.4 \mathrm { rad } \mathrm { s } ^ { - 1 }\).

3 Two planes, \(A\) and \(B\), flying at the same altitude, are participating in an air show. Initially the planes are 400 m apart and plane \(B\) is on a bearing of \(130 ^ { \circ }\) from plane \(A\). Plane \(A\) is moving due south with a constant speed of \(75 \mathrm {~ms} ^ { - 1 }\). Plane \(B\) is moving at a constant speed of \(40 \mathrm {~ms} ^ { - 1 }\) and has set a course to get as close as possible to \(A\).

1. Find the bearing of the course set by \(B\) and the shortest distance between the two planes in the subsequent motion.
2. Find the total distance travelled by \(A\) and \(B\) from the instant when they are initially 400 m apart to the point of their closest approach.

2 A ship \(S\) is travelling with constant speed \(5 \mathrm {~ms} ^ { - 1 }\) on a course with bearing \(325 ^ { \circ }\). A second ship \(T\) observes \(S\) when \(S\) is 9500 m from \(T\) on a bearing of \(060 ^ { \circ }\) from \(T\). Ship \(T\) sets off in pursuit, travelling with constant speed \(8.5 \mathrm {~ms} ^ { - 1 }\) in a straight line.

1. Find the bearing of the course which \(T\) should take in order to intercept \(S\).
2. Find the distance travelled by \(S\) from the moment that \(T\) sets off in pursuit until the point of interception.

2. Two constant forces \(\mathbf { F } _ { 1 }\) and \(\mathbf { F } _ { 2 }\) are the only forces acting on a particle \(P\) of mass 2 kg . The particle is initially at rest at the point \(A\) with position vector \(( - 2 \mathbf { i } - \mathbf { j } - 4 \mathbf { k } ) \mathrm { m }\). Four seconds later, \(P\) is at the point \(B\) with position vector \(( 6 \mathbf { i } - 5 \mathbf { j } + 8 \mathbf { k } ) \mathrm { m }\). Given that \(\mathbf { F } _ { 1 } = ( 12 \mathbf { i } - 4 \mathbf { j } + 6 \mathbf { k } ) \mathrm { N }\), find

1. \(\mathbf { F } _ { 2 }\),
2. the work done on \(P\) as it moves from \(A\) to \(B\).

1. At time \(t = 0\), a particle \(P\) of mass 3 kg is at rest at the point \(A\) with position vector \(( \mathbf { j } - 3 \mathbf { k } ) \mathrm { m }\). Two constant forces \(\mathbf { F } _ { 1 }\) and \(\mathbf { F } _ { 2 }\) then act on the particle \(P\) and it passes through the point \(B\) with position vector \(( 8 \mathbf { i } - 3 \mathbf { j } + 5 \mathbf { k } ) \mathrm { m }\).

Given that \(\mathbf { F } _ { 1 } = ( 4 \mathbf { i } - 2 \mathbf { j } + 5 \mathbf { k } ) \mathrm { N }\) and \(\mathbf { F } _ { 2 } = ( 8 \mathbf { i } - 4 \mathbf { j } + 7 \mathbf { k } ) \mathrm { N }\) and that \(\mathbf { F } _ { 1 }\) and \(\mathbf { F } _ { 2 }\) are the only two forces acting on \(P\), find the velocity of \(P\) as it passes through \(B\), giving your answer as a vector.

2 A car of mass 1350 kg travels along a straight horizontal road. Throughout this question the resistance force to the motion of the car is modelled as constant and equal to 920 N .

1. Calculate the power, in kW , developed by the car when the car is travelling at a constant speed of \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The car is now used to tow a caravan of mass 1050 kg along the same road. When the car tows the caravan at a constant speed of \(20 \mathrm {~ms} ^ { - 1 }\) the power developed by the car is 45 kW .
2. Find the additional resistance force due to the caravan. In the remaining parts of this question the power developed by the car is constant and equal to 68 kW and the resistance force due to the caravan is modelled as constant and equal to the value found in part (ii). When the car and caravan pass a point A on the same straight horizontal road the speed of the car and caravan is \(20 \mathrm {~ms} ^ { - 1 }\).
3. Find the acceleration of the car and caravan at point A . The car and caravan later pass a point B on the same straight horizontal road with speed \(28 \mathrm {~ms} ^ { - 1 }\). The distance \(A B\) is \(1024 m\).
4. Find the time taken for the car and caravan to travel from point A to point B .
5. Suggest one way in which any of the modelling assumptions used in this question could have been improved.

2 A car of mass 1400 kg , travels along a straight horizontal road AB , after which it descends a hill BC inclined at a constant angle of \(7 ^ { \circ }\) to the horizontal (see diagram). \(\mathrm { A } , \mathrm { B }\) and C all lie in the same vertical plane. Throughout the entire journey, the total resistance to the car's motion is constant. \includegraphics[max width=\textwidth, alt={}, center]{cad8805d-59f6-4ed2-81f4-9e8c749461f5-3_232_1227_392_251} Between A and B, the car moves at a constant speed of \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), and the power developed by the car is a constant \(P \mathrm {~W}\). When the car reaches B , the engine is switched off and the car travels down a line of greatest slope from \(B\) to \(C\) with an acceleration of \(0.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). The resistance to motion is unchanged.

1. Determine the value of \(P\). When the car reaches C it turns round and travels back up the hill towards B at a constant speed of \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The power developed by the car between C and B is a constant 16 kW . The resistance to motion is unchanged.
2. Determine the value of \(v\).

9 In this question you must show detailed reasoning. A car accelerates from rest along a straight level road. The velocity of the car after 8 s is \(25.6 \mathrm {~ms} ^ { - 1 }\).
In one model for the motion, the velocity \(v \mathrm {~ms} ^ { - 1 }\) at time \(t\) seconds is given by \(v = 1.2 t ^ { 2 } - k t ^ { 3 }\), where \(k\) is a constant and \(0 \leqslant t \leqslant 8\).

1. The model gives the correct velocity of \(25.6 \mathrm {~ms} ^ { - 1 }\) at time 8 s . Show that \(k = 0.1\). A second model for the motion uses constant acceleration.
2. Find the value of the acceleration which gives the correct velocity of \(25.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at time 8 s .
3. Show that these two models give the same value for the displacement in the first 8 s .

6 In this question take \(g\) as \(10 \mathrm {~m \mathrm {~s} ^ { - 2 }\).} A small ball is released from rest. It falls for 2 seconds and is then brought to rest over the next 5 seconds. This motion is modelled in the speed-time graph Fig. 6. \begin{figure}[h] \end{figure} For this model,

1. calculate the distance fallen from \(t = 0\) to \(t = 7\),
2. find the acceleration of the ball from \(t = 2\) to \(t = 6\), specifying the direction,
3. obtain an expression in terms of \(t\) for the downward speed of the ball from \(t = 2\) to \(t = 6\),
4. state the assumption that has been made about the resistance to motion from \(t = 0\) to \(t = 2\). The part of the motion from \(t = 2\) to \(t = 7\) is now modelled by \(v = - \frac { 3 } { 2 } t ^ { 2 } + \frac { 19 } { 2 } t + 7\).
5. Verify that \(v\) agrees with the values given in Fig. 6 at \(t = 2 , t = 6\) and \(t = 7\).
6. Calculate the distance fallen from \(t = 2\) to \(t = 7\) according to this model.

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.