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Motion graphs (displacement–time & velocity–time)

20 questions

LessonNot started
Two cyclists, PP and QQ, are riding in the same direction along a straight horizontal road. The diagram shows the velocity–time graphs for their motion.96540QPt (s)v (m s⁻¹)At time t=0t = 0 seconds, cyclist QQ, who is riding with constant speed 6 m s−16\text{ m s}^{-1}, is overtaken by cyclist PP, whose speed is then 9 m s−19\text{ m s}^{-1}. From t=0t = 0 cyclist PP slows down with constant deceleration for 40 seconds. When t=40t = 40 the speed of PP is 5 m s−15\text{ m s}^{-1}, and PP keeps this speed in the subsequent motion.
(a) Calculate the deceleration of PP.

(2 marks)
(b) Determine the value of tt when QQ overtakes PP.

(4 marks)
●●●●●Level 46 marksStart
In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

A particle PP moves along a straight line. At time t=0t = 0, PP is at rest at the fixed point OO on the line.

At time tt seconds, t⩾0t \geqslant 0, the acceleration of PP, in the direction of increasing displacement from OO, is k(5−2t)k(5 - 2t) m s−2^{-2}, where kk is a positive constant.

Given that the speed of PP is 1818 m s−1^{-1} when t=2t = 2
(a) show that k=3k = 3

(3 marks)

At time tt seconds, the displacement of PP from OO is ss metres.
(b) Show that s=12t2(15−2t)s = \frac{1}{2}t^2(15 - 2t)

(2 marks)
(c) Find the total distance travelled by PP before it returns to OO.

(4 marks)
●●●●●Level 49 marksStart
Three points AA, BB and CC lie in that order on a straight horizontal road.

A bus moves along the road from AA to CC.
  • The bus passes AA with speed 3 m s−13\ \text{m s}^{-1}
  • The bus moves with constant acceleration from AA to BB, reaching BB with speed V m s−1V\ \text{m s}^{-1}
  • The bus then moves with constant deceleration from BB until it comes to rest at CC
The bus is modelled as a particle.
(a) Sketch a speed-time graph for the motion of the bus from AA to CC.

(3 marks)

The distance ABAB is 165165 m and the distance BCBC is 120120 m.
(b) Find, in terms of VV,

(i) the time taken by the bus to travel from AA to BB

(ii) the time taken by the bus to travel from BB to CC

(4 marks)

The bus takes 4242 s to travel from AA to CC.
(c) Find the value of VV.

(2 marks)
(d) Find the deceleration of the bus as it moves from BB to CC.

(2 marks)
●●●●●Level 411 marksStart
A car moves along a straight horizontal road. At time t=0t = 0, the car passes the point OO with speed 5 m s−15\ \text{m s}^{-1}1025350.8−0.2−1t (s)a (m s⁻²)Figure 1 (not accurately drawn)

Figure 1 shows the acceleration-time graph for the motion of the car in the interval 0⩽t⩽350 \leqslant t \leqslant 35, where tt seconds is the time after the car passes OO. The acceleration of the car is
  • 0.8 m s−20.8\ \text{m s}^{-2} for 0<t<100 < t < 10
  • −0.2 m s−2-0.2\ \text{m s}^{-2} for 10<t<2510 < t < 25
  • −1 m s−2-1\ \text{m s}^{-2} for 25<t<3525 < t < 35
The car is modelled as a particle.
(a) Show that the speed of the car when t=25t = 25 is 10 m s−110\ \text{m s}^{-1}

(3 marks)
(b) Sketch a speed-time graph for the motion of the car in the interval 0⩽t⩽350 \leqslant t \leqslant 35

(3 marks)
(c) Find the total distance travelled by the car in the interval 0⩽t⩽350 \leqslant t \leqslant 35

(4 marks)
●●●●●Level 410 marksStart
A small stone is projected vertically upwards with speed 16.8 m s−116.8\ \text{m s}^{-1} from a point AA at the top of a vertical cliff. The point AA is 38.538.5 m vertically above the point BB. The point BB lies on horizontal ground at the foot of the cliff.

The stone moves freely under gravity until it hits the ground at BB with speed V m s−1V\ \text{m s}^{-1}. After hitting the ground the stone does not rebound.
(a) Find the value of VV.

(3 marks)
(b) Find the time taken for the stone to reach BB.

(3 marks)

The point CC is 9.89.8 m vertically above AA.
(c) Find the length of time for which the stone is above CC.

(4 marks)
(d) Sketch a speed-time graph for the motion of the stone from the instant it is projected to the instant it reaches BB.

(No further calculations are required.)

(2 marks)
●●●●●Level 412 marksStart
A particle PP moves along a straight line. At time tt seconds, where t⩾0t \geqslant 0, the velocity of PP is vv m s⁻¹, where

v=3t2−12t+9v = 3t^2 - 12t + 9

At t=0t = 0, PP is at the point OO on the line.
(a) Find the values of tt at which PP is instantaneously at rest. (2 marks)
(b) Find the total distance travelled by PP in the first 44 seconds. (6 marks)
(c) Find the magnitude of the acceleration of PP when t=4t = 4. (2 marks)
●●●●●Level 410 marksStart
10557515UTime, t (s)Speed (m s⁻¹)Figure 1 — not accurately drawn

Figure 1 shows a speed-time graph for a model of the motion of a cyclist riding a 10001000 m time trial in 7575 s.

The cyclist
  • starts from rest at time t=0t = 0 and accelerates at a constant rate, reaching a speed of 15 m s−115\ \text{m s}^{-1} when t=10t = 10
  • then rides at a constant speed of 15 m s−115\ \text{m s}^{-1} from t=10t = 10 until t=55t = 55
  • then decelerates at a constant rate from t=55t = 55 to t=75t = 75, crossing the finish line with speed U m s−1U\ \text{m s}^{-1}
Using the model,
(a) find the acceleration of the cyclist during the first 1010 s of the ride, stating the units of your answer,

(2 marks)
(b) find the distance covered by the cyclist during the first 5555 s of the ride,

(3 marks)
(c) find the value of UU.

(3 marks)
●●●●●Level 48 marksStart

Constant acceleration (suvat)

54 questions

LessonNot started
[In this question, i\mathbf{i} and j\mathbf{j} are horizontal unit vectors and position vectors are given relative to a fixed origin OO.]

A particle PP is moving on a smooth horizontal plane.

The particle has constant acceleration (2i+3j) m s−2(2\mathbf{i} + 3\mathbf{j})\ \text{m s}^{-2}

At time t=0t = 0, PP passes through the point AA.

At time t=3t = 3 s, PP passes through the point BB.

The velocity of PP as it passes through AA is (5i−11j) m s−1(5\mathbf{i} - 11\mathbf{j})\ \text{m s}^{-1}
(a) Find the speed of PP as it passes through BB.

(4 marks)

The position vector of AA is (−9i−20j)(-9\mathbf{i} - 20\mathbf{j}) m.

At time t=Tt = T seconds, where T>3T > 3, PP passes through the point CC.

The position vector of CC is (41i+cj)(41\mathbf{i} + c\mathbf{j}) m.
(b) Find the value of TT.

(3 marks)
(c) Find the value of cc.

(3 marks)
●●●●●Level 410 marksStart
[In this question, i\mathbf{i} and j\mathbf{j} are horizontal unit vectors.]

A particle PP of mass 6 kg is at rest at the point AA on a smooth horizontal plane.

At time t=0t = 0, two forces, F1=(7i−4j)\mathbf{F}_1 = (7\mathbf{i} - 4\mathbf{j}) N and F2=(λi+μj)\mathbf{F}_2 = (\lambda\mathbf{i} + \mu\mathbf{j}) N, where λ\lambda and μ\mu are constants, are applied to PP

Given that PP moves in the direction of the vector (4i+3j)(4\mathbf{i} + 3\mathbf{j})
(a) show that 3λ−4μ+37=03\lambda - 4\mu + 37 = 0

(4 marks)

At time t=6t = 6 seconds, PP passes through the point BB.

Given that λ=5\lambda = 5
(b) find the length of ABAB.

(5 marks)
●●●●●Level 49 marksStart
The diagram shows a small block BB, of mass 0.60.6 kg, on a rough horizontal surface, and a particle PP, of mass 0.90.9 kg, on a smooth plane inclined at 40°40° to the horizontal. The horizontal surface meets the inclined plane along an edge, at which a small smooth pulley is fixed. BB and PP are attached to the ends of a light inextensible string which passes over the pulley; the string is taut, the part of the string from BB to the pulley is horizontal and the part from PP to the pulley is parallel to a line of greatest slope of the plane.BP40°The system is released from rest. In the first 0.50.5 seconds of the motion PP moves 0.350.35 m down the plane and BB does not reach the pulley.
(a) Find the tension in the string during the first 0.50.5 seconds of the motion.

(4 marks)
(b) Determine the coefficient of friction between BB and the horizontal surface.

(5 marks)
●●●●●Level 49 marksStart
The points AA and BB are the lower and upper ends, respectively, of a line of greatest slope on a rough plane inclined at an angle θ\theta to the horizontal, where sin⁡θ=513\sin\theta = \frac{5}{13}. The distance ABAB is 0.750.75 m (see diagram).0.75 mθ5.5 m s⁻¹ABThe coefficient of friction between the plane and a particle PP is 0.450.45. The particle PP is projected from AA towards BB, up the line of greatest slope, with speed 5.55.5 m s−1^{-1}.

Show that the speed of PP at BB is 4.34.3 m s−1^{-1}.

(6 marks)
●●●●●Level 46 marksStart
Two cyclists, PP and QQ, are riding in the same direction along a straight horizontal road. The diagram shows the velocity–time graphs for their motion.96540QPt (s)v (m s⁻¹)At time t=0t = 0 seconds, cyclist QQ, who is riding with constant speed 6 m s−16\text{ m s}^{-1}, is overtaken by cyclist PP, whose speed is then 9 m s−19\text{ m s}^{-1}. From t=0t = 0 cyclist PP slows down with constant deceleration for 40 seconds. When t=40t = 40 the speed of PP is 5 m s−15\text{ m s}^{-1}, and PP keeps this speed in the subsequent motion.
(a) Calculate the deceleration of PP.

(2 marks)
(b) Determine the value of tt when QQ overtakes PP.

(4 marks)
●●●●●Level 46 marksStart
Figure 1250 kg400 kgαDiagram not accurately drawn.

A quad bike of mass 400400 kg is towing a cart of mass 250250 kg up a straight road, as shown in Figure 1. The road is inclined at an angle α\alpha to the horizontal, where sin⁡α=114\sin\alpha = \frac{1}{14}

The driving force produced by the engine of the quad bike is 10001000 N and the quad bike and the cart move with acceleration 0.4 m s−20.4\ \text{m s}^{-2}

The non-gravitational resistance to the motion of
  • the quad bike is modelled as a constant force of magnitude 3R3R newtons
  • the cart is modelled as a constant force of magnitude 2R2R newtons
The quad bike and the cart are modelled as particles. The tow bar between the quad bike and the cart is modelled as a light rod that is parallel to the direction of motion.

Using the model,
(a) show that R=57R = 57

(4 marks)
(b) find the tension in the tow bar.

(3 marks)

When the quad bike and the cart are moving at 9 m s−19\ \text{m s}^{-1}, the tow bar breaks.

Given that the non-gravitational resistance to the motion of the cart is unchanged,
(c) use the model to find the further distance moved by the cart before it first comes to rest.

(4 marks)
●●●●●Level 411 marksStart
Figure 1PQαDiagram not accurately drawn.

A fixed rough plane is inclined at an angle α\alpha to the horizontal, where tan⁡α=34\tan\alpha = \frac{3}{4}. A small smooth pulley is fixed at the top of the plane.

A particle PP of mass 55 kg is at rest on the plane. One end of a light inextensible string is attached to PP. The string passes over the pulley and its other end is attached to a particle QQ of mass 44 kg, which hangs freely, vertically below the pulley, as shown in Figure 1.

The part of the string from PP to the pulley lies along a line of greatest slope of the plane. The coefficient of friction between PP and the plane is μ\mu.

The system is in limiting equilibrium, with the string taut and PP on the point of slipping up the plane.
(a) Find the value of μ\mu.

(8 marks)

The string breaks and PP begins to slide down the plane.
(b) Find the time taken for PP to slide 2.52.5 m down the plane.

(4 marks)
●●●●●Level 412 marksStart
Figure 13.2 mP (2m kg)Q (5m kg)Diagram not accurately drawn.

In a table game, two discs, PP and QQ, rest on a rough horizontal board. Disc PP has mass 2m2m kg and disc QQ has mass 5m5m kg. The distance between PP and QQ is 3.23.2 m, as shown in Figure 1. The discs are modelled as particles.

Disc PP is struck so that it receives an impulse of magnitude λ\lambda N s in the direction PQPQ. Immediately after the impulse is applied, the speed of PP is 8 m s−18\text{ m s}^{-1}
(a) Find λ\lambda in terms of mm.

(2 marks)

Immediately before PP collides with QQ, the speed of PP is 6 m s−16\text{ m s}^{-1}

The coefficient of friction between PP and the board is μ\mu
(b) Find the value of μ\mu

(7 marks)

Immediately after PP collides with QQ, the speed of QQ is 2.2 m s−12.2\text{ m s}^{-1}
(c) Find the speed of PP immediately after the collision.

(3 marks)
●●●●●Level 412 marksStart
Three points AA, BB and CC lie in that order on a straight horizontal road.

A bus moves along the road from AA to CC.
  • The bus passes AA with speed 3 m s−13\ \text{m s}^{-1}
  • The bus moves with constant acceleration from AA to BB, reaching BB with speed V m s−1V\ \text{m s}^{-1}
  • The bus then moves with constant deceleration from BB until it comes to rest at CC
The bus is modelled as a particle.
(a) Sketch a speed-time graph for the motion of the bus from AA to CC.

(3 marks)

The distance ABAB is 165165 m and the distance BCBC is 120120 m.
(b) Find, in terms of VV,

(i) the time taken by the bus to travel from AA to BB

(ii) the time taken by the bus to travel from BB to CC

(4 marks)

The bus takes 4242 s to travel from AA to CC.
(c) Find the value of VV.

(2 marks)
(d) Find the deceleration of the bus as it moves from BB to CC.

(2 marks)
●●●●●Level 411 marksStart
A car moves along a straight horizontal road. At time t=0t = 0, the car passes the point OO with speed 5 m s−15\ \text{m s}^{-1}1025350.8−0.2−1t (s)a (m s⁻²)Figure 1 (not accurately drawn)

Figure 1 shows the acceleration-time graph for the motion of the car in the interval 0⩽t⩽350 \leqslant t \leqslant 35, where tt seconds is the time after the car passes OO. The acceleration of the car is
  • 0.8 m s−20.8\ \text{m s}^{-2} for 0<t<100 < t < 10
  • −0.2 m s−2-0.2\ \text{m s}^{-2} for 10<t<2510 < t < 25
  • −1 m s−2-1\ \text{m s}^{-2} for 25<t<3525 < t < 35
The car is modelled as a particle.
(a) Show that the speed of the car when t=25t = 25 is 10 m s−110\ \text{m s}^{-1}

(3 marks)
(b) Sketch a speed-time graph for the motion of the car in the interval 0⩽t⩽350 \leqslant t \leqslant 35

(3 marks)
(c) Find the total distance travelled by the car in the interval 0⩽t⩽350 \leqslant t \leqslant 35

(4 marks)
●●●●●Level 410 marksStart
A lorry moves along a straight horizontal road with constant acceleration.

The lorry passes the point AA with speed u m s−1u\ \text{m s}^{-1} and, 1616 s later, passes the point BB with speed 22 m s−122\ \text{m s}^{-1}

The distance ABAB is 256256 m.

The lorry is modelled as a particle.
(a) Show that u=10u = 10

(2 marks)
(b) Find the time taken for the lorry to travel from AA to the midpoint of ABAB.

(5 marks)

The mass of the lorry is 24002400 kg. As the lorry moves from AA to BB, the resistance to the motion of the lorry is modelled as a constant force of magnitude 900900 N.
(c) Find the magnitude of the driving force produced by the engine of the lorry as it moves from AA to BB.

(3 marks)
●●●●●Level 410 marksStart
P (3 kg)αH NFigure 1 (not accurately drawn)

A small package PP of mass 33 kg is placed on a rough ramp. The ramp is modelled as a rough plane inclined at an angle α\alpha to the horizontal, where tan⁡α=512\tan\alpha = \frac{5}{12}, and the package is modelled as a particle.

The package is held in equilibrium by a horizontal force of magnitude HH newtons, as shown in Figure 1. The line of action of the horizontal force lies in a vertical plane containing a line of greatest slope of the ramp.

The coefficient of friction between the package and the ramp is 14\frac{1}{4}
(a) Find the smallest possible value of HH.

(6 marks)

The horizontal force is now removed and the package starts to slide down the ramp from rest.

The package slides 22 m down the ramp in the first TT seconds after the force is removed.
(b) Find the value of TT.

(6 marks)
●●●●●Level 412 marksStart
A particle PP of mass 3m3m and a particle QQ of mass kmkm, where kk is a constant, are moving in opposite directions along the same straight line on a smooth horizontal surface. Particle PP collides directly with QQ.

Immediately before the collision the speed of PP is 7u7u and the speed of QQ is 2u2u. Immediately after the collision the speed of PP is vv and the speed of QQ is 4v4v. As a result of the collision
  • the direction of motion of PP is unchanged
  • the direction of motion of QQ is reversed
The coefficient of restitution between PP and QQ is 23\frac23
(a) Show that v=2uv = 2u

(3 marks)
(b) Find the value of kk.

(3 marks)

After the collision, QQ hits a smooth fixed vertical wall that is perpendicular to the direction of motion of QQ, and rebounds. When PP and QQ first collide, they are at a distance dd from the wall. They collide for a second time at a distance 13d\frac13 d from the wall. The coefficient of restitution between QQ and the wall is ff.
(c) Find the value of ff.

(5 marks)
●●●●●Level 411 marksStart
A particle AA of mass 22 kg lies on a rough plane inclined at an angle α\alpha to the horizontal, where sin⁡α=35\sin\alpha = \dfrac35. The particle is attached to one end of a light inextensible string. The string lies along a line of greatest slope of the plane and passes over a small smooth pulley PP fixed at the top of the plane. The other end of the string is attached to a particle BB of mass 33 kg, which hangs freely with the string vertical. BB is 11 m above horizontal ground, and the plane is long enough that AA does not reach PP in the motion that follows.

The coefficient of friction between AA and the plane is 12\dfrac12. The system is released from rest with the string taut. Take g=9.8g = 9.8 m s⁻².
(a) Show that the acceleration of AA is 1.961.96 m s⁻² and find the tension in the string. (5 marks)

When BB reaches the ground it does not rebound.
(b) Find the further distance that AA travels up the plane after BB reaches the ground, before AA first comes to instantaneous rest. (3 marks)
(c) Determine whether AA remains at rest at this point. (2 marks)
●●●●●Level 410 marksStart
A small stone is released from rest at the top of a deep well and falls freely towards the water at the bottom of the well.

The stone passes through two different fixed points, AA and BB, before reaching the water.

At AA the stone has velocity u m s−1u\ \text{m s}^{-1}

At BB the stone has velocity 4u m s−14u\ \text{m s}^{-1}

It can be assumed that:

* the motion is due to gravitational force only * the acceleration due to gravity remains constant throughout.
(a) Show that the time taken for the stone to travel from AA to BB is 3ug\dfrac{3u}{g} seconds.

(2 marks)
(b) Point AA is hh metres above the water.

Show that

h>15u22gh > \frac{15u^2}{2g}

Fully justify your answer.

(3 marks)
●●●●●Level 45 marksStart
In this question use g=9.8 m s−2g = 9.8\ \text{m s}^{-2}

A machine for cricket catching practice fires cricket balls upwards. Each ball leaves the machine at ground level with a speed of 11.2 m s−111.2\ \text{m s}^{-1}

The manufacturer claims that the balls can reach a maximum height of 6.46.4 metres above the ground.
(a) Suppose that the machine fires the balls vertically upwards.
(a) (i) Verify the manufacturer's claim.

(ii) State two modelling assumptions you have made in verifying this claim.

(4 marks)
(b) In fact the machine fires the balls in a direction anywhere between 00 and 3030 degrees from the vertical.

The range of maximum heights, hh metres, above the ground which can be reached by the balls may be expressed as

k<h≤6.4k < h \le 6.4

Find the value of kk

(4 marks)
●●●●●Level 48 marksStart
In this question use g=9.8 m s−2g = 9.8\ \text{m s}^{-2}

Two sledges, AA and BB, are connected to each other by a light, rigid pole. Sledge AA has a mass of 1212 kilograms. Sledge BB has a mass of 88 kilograms.

A child pulls sledge AA using a rope inclined at an angle of 20∘20^\circ above the horizontal. The tension in the rope is 4040 newtons.

As a result the sledges move forward, from rest, in a straight line across horizontal snow with acceleration 0.75 m s−20.75\ \text{m s}^{-2} as shown in the diagram below.BA20°As it moves sledge AA experiences a total resistance force of 1414 N
(a) Show that the total resistance force experienced by sledge BB is approximately 8.68.6 N

(4 marks)
(b) At the instant that the sledges reach a speed of 3 m s−13\ \text{m s}^{-1} the child lets go of the rope.

The sledges decelerate at a constant rate until they come to rest. It can be assumed that the resistance forces remain unchanged.

Find the distance the sledges travel after the child lets go of the rope.

(3 marks)
(c) State one modelling assumption that you have used about the pole when answering part (a).

(1 mark)
●●●●●Level 48 marksStart
Two particles, PP and QQ, move in the same horizontal plane.

Particle PP is initially at rest at the point with position vector (2i+j)(2\mathbf{i} + \mathbf{j}) metres and moves with constant acceleration (3i−2j) m s−2(3\mathbf{i} - 2\mathbf{j})\ \text{m s}^{-2}

Particle QQ moves in a straight line, passing through the points with position vectors (10i−3j)(10\mathbf{i} - 3\mathbf{j}) metres and (i+cj)(\mathbf{i} + c\mathbf{j}) metres, where cc is a constant.

PP and QQ are moving along parallel paths.
(a) Show that c=3c = 3

(4 marks)
(b) (i) Find an expression for the position vector of PP at time tt seconds.

(ii) Hence, prove that the paths of PP and QQ are not collinear.

(4 marks)
●●●●●Level 48 marksStart
In this question use g=9.8 m s−2g = 9.8\ \text{m s}^{-2}

Aisha is running a goalkeeping drill with Jonah on a large playing field. The playing field can be assumed to be level and horizontal.

Aisha and Jonah are initially standing next to each other.

Aisha throws a ball forward, from a height of 1.61.6 metres above the playing field, at an angle of 65∘65^\circ above the horizontal with a speed of 10 m s−110\ \text{m s}^{-1}

Exactly 0.250.25 seconds after the ball is thrown, Jonah sets off from Aisha and runs across the playing field, in a straight line with a constant acceleration a m s−2a\ \text{m s}^{-2}

Jonah catches the ball when it is 0.40.4 metres above the playing field, as shown in the diagram below.1.6 m0.4 m65°10 m s⁻¹Find aa

(7 marks)
●●●●●Level 47 marksStart
In this question use g=9.8 m s−2g = 9.8\ \text{m s}^{-2}

Mei and Tomasz are practising catching on a sports field. The sports field can be assumed to be level and horizontal.

Mei stands at the point MM and Tomasz stands at the point TT, which is 55 metres in front of Mei.

Mei throws a ball forward, towards Tomasz, from a height of 1.91.9 metres above the sports field, at an angle of 55∘55^\circ above the horizontal with a speed of 9 m s−19\ \text{m s}^{-1}

Exactly 0.40.4 seconds after the ball is thrown, Tomasz sets off from rest at TT and runs directly away from Mei, in a straight line with a constant acceleration a m s−2a\ \text{m s}^{-2}

Tomasz catches the ball when it is 1.21.2 metres above the sports field, as shown in the diagram below.1.9 m5 m1.2 m55°9 m s⁻¹MTFind aa

(7 marks)
●●●●●Level 47 marksStart
10557515UTime, t (s)Speed (m s⁻¹)Figure 1 — not accurately drawn

Figure 1 shows a speed-time graph for a model of the motion of a cyclist riding a 10001000 m time trial in 7575 s.

The cyclist
  • starts from rest at time t=0t = 0 and accelerates at a constant rate, reaching a speed of 15 m s−115\ \text{m s}^{-1} when t=10t = 10
  • then rides at a constant speed of 15 m s−115\ \text{m s}^{-1} from t=10t = 10 until t=55t = 55
  • then decelerates at a constant rate from t=55t = 55 to t=75t = 75, crossing the finish line with speed U m s−1U\ \text{m s}^{-1}
Using the model,
(a) find the acceleration of the cyclist during the first 1010 s of the ride, stating the units of your answer,

(2 marks)
(b) find the distance covered by the cyclist during the first 5555 s of the ride,

(3 marks)
(c) find the value of UU.

(3 marks)
●●●●●Level 48 marksStart

Vertical motion under gravity

14 questions

LessonNot started
A small stone is projected vertically upwards with speed 16.8 m s−116.8\ \text{m s}^{-1} from a point AA at the top of a vertical cliff. The point AA is 38.538.5 m vertically above the point BB. The point BB lies on horizontal ground at the foot of the cliff.

The stone moves freely under gravity until it hits the ground at BB with speed V m s−1V\ \text{m s}^{-1}. After hitting the ground the stone does not rebound.
(a) Find the value of VV.

(3 marks)
(b) Find the time taken for the stone to reach BB.

(3 marks)

The point CC is 9.89.8 m vertically above AA.
(c) Find the length of time for which the stone is above CC.

(4 marks)
(d) Sketch a speed-time graph for the motion of the stone from the instant it is projected to the instant it reaches BB.

(No further calculations are required.)

(2 marks)
●●●●●Level 412 marksStart
A small stone is released from rest at the top of a deep well and falls freely towards the water at the bottom of the well.

The stone passes through two different fixed points, AA and BB, before reaching the water.

At AA the stone has velocity u m s−1u\ \text{m s}^{-1}

At BB the stone has velocity 4u m s−14u\ \text{m s}^{-1}

It can be assumed that:

* the motion is due to gravitational force only * the acceleration due to gravity remains constant throughout.
(a) Show that the time taken for the stone to travel from AA to BB is 3ug\dfrac{3u}{g} seconds.

(2 marks)
(b) Point AA is hh metres above the water.

Show that

h>15u22gh > \frac{15u^2}{2g}

Fully justify your answer.

(3 marks)
●●●●●Level 45 marksStart
In this question use g=9.8 m s−2g = 9.8\ \text{m s}^{-2}

A machine for cricket catching practice fires cricket balls upwards. Each ball leaves the machine at ground level with a speed of 11.2 m s−111.2\ \text{m s}^{-1}

The manufacturer claims that the balls can reach a maximum height of 6.46.4 metres above the ground.
(a) Suppose that the machine fires the balls vertically upwards.
(a) (i) Verify the manufacturer's claim.

(ii) State two modelling assumptions you have made in verifying this claim.

(4 marks)
(b) In fact the machine fires the balls in a direction anywhere between 00 and 3030 degrees from the vertical.

The range of maximum heights, hh metres, above the ground which can be reached by the balls may be expressed as

k<h≤6.4k < h \le 6.4

Find the value of kk

(4 marks)
●●●●●Level 48 marksStart

Kinematics with calculus (a → v → s)

34 questions

LessonNot started
In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

[In this question i\mathbf{i} and j\mathbf{j} are horizontal perpendicular unit vectors.]

A particle PP of mass 0.50.5 kg moves on a smooth horizontal plane. For 0≤t≤20 \le t \le 2, its position vector relative to a fixed origin OO at time tt seconds is r\mathbf{r} metres, where

r=(t3−8t)i+(t−t2)j\mathbf{r} = (t^3 - 8t)\mathbf{i} + (t - t^2)\mathbf{j}
(a) Show that the speed of PP at the instant when t=2t = 2 is 5 m s−15\text{ m s}^{-1}

(3 marks)

At the instant when t=2t = 2, PP receives an impulse of (−8i−j)(-8\mathbf{i} - \mathbf{j}) N s.
(b) Find the angle through which the direction of motion of PP is turned by the impulse.

(4 marks)
(c) Find the increase in the kinetic energy of PP caused by the impulse.

(2 marks)
●●●●●Level 49 marksStart
In this question the unit vectors i\mathbf{i} and j\mathbf{j} are in the directions east and north respectively.

A single force F\mathbf{F} N acts on a particle PP, which moves on a smooth horizontal surface. The velocity of PP at time tt seconds, where t≥0t \geq 0, is v m s−1\mathbf{v}\ \text{m s}^{-1}, where

v=(t2−4t+7)i+(3t−5)j.\mathbf{v} = (t^2 - 4t + 7)\mathbf{i} + (3t - 5)\mathbf{j}.
(a) Show that PP is never stationary.

(2 marks)
(b) Find, in terms of i\mathbf{i} and j\mathbf{j}, the acceleration of PP at time tt.

(1 mark)

The mass of PP is 0.60.6 kg.
(c) Determine the magnitude of F\mathbf{F} when PP is moving in the direction of the vector 3i−2j3\mathbf{i} - 2\mathbf{j}. Give your answer correct to 3 significant figures.

(5 marks)
●●●●●Level 48 marksStart
A train travels along a straight horizontal track, passing three signals, AA, BB and CC, in that order. At time tt seconds after passing AA, the velocity of the train is v m s−1v\text{ m s}^{-1}, where vv is modelled by the equation

v=at2+bt+c,v = at^2 + bt + c,

where aa, bb and cc are constants.

The train passes AA with velocity 12 m s−112\text{ m s}^{-1}.
(a) State the value of cc.

(1 mark)

The train passes BB when t=8t = 8, and at that instant its deceleration is 0.5 m s−20.5\text{ m s}^{-2}. The train passes CC when t=20t = 20, with velocity 4.8 m s−14.8\text{ m s}^{-1}.
(b) Determine the values of aa and bb.

(4 marks)
(c) Find, to the nearest metre, the distance between signals AA and CC.

(2 marks)
●●●●●Level 47 marksStart
In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

A particle PP moves along a straight line. At time t=0t = 0, PP is at rest at the fixed point OO on the line.

At time tt seconds, t⩾0t \geqslant 0, the acceleration of PP, in the direction of increasing displacement from OO, is k(5−2t)k(5 - 2t) m s−2^{-2}, where kk is a positive constant.

Given that the speed of PP is 1818 m s−1^{-1} when t=2t = 2
(a) show that k=3k = 3

(3 marks)

At time tt seconds, the displacement of PP from OO is ss metres.
(b) Show that s=12t2(15−2t)s = \frac{1}{2}t^2(15 - 2t)

(2 marks)
(c) Find the total distance travelled by PP before it returns to OO.

(4 marks)
●●●●●Level 49 marksStart
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

[In this question, i\mathbf{i} and j\mathbf{j} are horizontal perpendicular unit vectors.]

A particle PP moves on a smooth horizontal plane. At time tt seconds, t⩾0t \geqslant 0, the position vector of PP relative to a fixed origin OO is r\mathbf{r} metres, where

r=(2t2−5t)i+(t3−6t2+9t)j\mathbf{r} = (2t^2 - 5t)\mathbf{i} + (t^3 - 6t^2 + 9t)\mathbf{j}
(a) Find the speed of PP at the instant when PP is moving in the direction of the vector i\mathbf{i}

(5 marks)
(b) Find the exact magnitude of the acceleration of PP when t=3.5t = 3.5

(4 marks)
(c) Show that PP never returns to OO

(2 marks)
●●●●●Level 411 marksStart
A remote-controlled car moves on a horizontal surface. The unit vectors i\mathbf{i} and j\mathbf{j} are directed east and north respectively.

At time tt seconds, where 0≤t≤50 \le t \le 5, the velocity of the car is

v=(t2−4t+1)i+(2t−5)j\mathbf{v} = \left(t^2 - 4t + 1\right)\mathbf{i} + (2t - 5)\mathbf{j} m s⁻¹.
(a) Find the acceleration of the car when t=2t = 2, and state the direction in which it acts. (2 marks)
(b) Find the value of tt at which the car is moving in the direction of the vector i+j\mathbf{i} + \mathbf{j}. Fully justify your answer. (4 marks)
(c) Find the distance of the car from its starting position when t=5t = 5. (3 marks)
●●●●●Level 49 marksStart
A particle PP of mass 0.50.5 kg moves in a horizontal plane under the action of a single force F\mathbf{F} newtons.

At time tt seconds the position vector of PP is

r=(t3−3t)i+2t2 jmetres\mathbf{r} = \left(t^3 - 3t\right)\mathbf{i} + 2t^2\,\mathbf{j} \quad \text{metres}
(a) Find the velocity of PP when t=2t = 2. (3 marks)
(b) Find the speed of PP when t=2t = 2, giving your answer to 33 significant figures. (2 marks)
(c) Find the magnitude of F\mathbf{F} when t=2t = 2, giving your answer to 33 significant figures. (4 marks)
●●●●●Level 49 marksStart
A particle PP moves in a horizontal plane. At time tt seconds (t≥0t \ge 0), its position vector relative to a fixed origin OO is

r=(2t2−3t)i+(4t−t2)j\mathbf{r} = (2t^2 - 3t)\mathbf{i} + (4t - t^2)\mathbf{j} metres.
(a) Find the velocity v\mathbf{v} of PP at time tt, and show that the acceleration of PP is constant. (3 marks)
(b) Find the speed of PP when t=3t = 3. (2 marks)
(c) Find the value of tt at which PP is moving parallel to i\mathbf{i}. (2 marks)
(d) Find the distance of PP from OO at this time, giving your answer in exact form. (3 marks)
●●●●●Level 410 marksStart
A particle PP moves along a straight line. At time tt seconds, where t⩾0t \geqslant 0, the velocity of PP is vv m s⁻¹, where

v=3t2−12t+9v = 3t^2 - 12t + 9

At t=0t = 0, PP is at the point OO on the line.
(a) Find the values of tt at which PP is instantaneously at rest. (2 marks)
(b) Find the total distance travelled by PP in the first 44 seconds. (6 marks)
(c) Find the magnitude of the acceleration of PP when t=4t = 4. (2 marks)
●●●●●Level 410 marksStart
A particle PP moves in a horizontal plane. At time tt seconds, where t⩾0t \geqslant 0, the acceleration of PP is (6t i+4j)(6t\,\mathbf{i} + 4\mathbf{j}) m s⁻². When t=0t = 0, PP is at the point AA with position vector (7i−5j)(7\mathbf{i} - 5\mathbf{j}) m relative to a fixed origin OO, and the velocity of PP is (−5i−j)(-5\mathbf{i} - \mathbf{j}) m s⁻¹.
(a) Show that PP is never instantaneously at rest. (3 marks)
(b) Find the value of tt at which the velocity of PP is perpendicular to its acceleration, giving a reason for rejecting any other solutions of your equation. (4 marks)
(c) Find the distance of PP from OO at this time. (3 marks)
●●●●●Level 410 marksStart
A particle PP moves along a straight line. At time tt seconds, where t⩾0t \geqslant 0, the acceleration of PP is aa m s⁻², where

a=6t−4a = 6t - 4

When t=0t = 0, PP is at the origin OO and is moving with velocity 55 m s⁻¹.
(a) Find an expression for the velocity of PP at time tt. (2 marks)
(b) Find the displacement of PP from OO when t=3t = 3. (3 marks)
(c) Show that PP never comes to rest. (2 marks)
●●●●●Level 47 marksStart
A particle is moving in a straight line through the origin OO

The displacement of the particle, rr metres, from OO, at time tt seconds is given by

r=p+1.5t−qe−0.4tr = p + 1.5t - q\text{e}^{-0.4t}

where pp and qq are constants.

When t=3t = 3, the acceleration of the particle is −0.5 m s−2-0.5\ \text{m s}^{-2}
(a) Show that q≈10.4q \approx 10.4

(5 marks)
(b) The particle has an initial displacement of −6-6 metres.

Find the value of pp

Give your answer to two significant figures.

(2 marks)
●●●●●Level 47 marksStart
A particle moves in a plane with velocity v=(6t−4) i+3t2 j m s−1\mathbf{v} = (6t - 4)\,\mathbf{i} + 3t^2\,\mathbf{j}\ \text{m s}^{-1}. When t=0t = 0 its position vector relative to a fixed origin OO is (2i+j)(2\mathbf{i} + \mathbf{j}) m. Find
(a) its position vector at time tt,
(b) its position vector when t=2t = 2, and (c) its distance from OO when t=2t = 2.
●●●●●Level 45 marksStart
A particle moves in a plane with acceleration a=2t i+(6t2−4) j m s−2\mathbf{a} = 2t\,\mathbf{i} + (6t^2 - 4)\,\mathbf{j}\ \text{m s}^{-2}. When t=0t = 0 its velocity is (3i+j) m s−1(3\mathbf{i} + \mathbf{j})\ \text{m s}^{-1}. Find
(a) its velocity at time tt,
(b) its velocity when t=2t = 2, and (c) its speed when t=2t = 2.
●●●●●Level 45 marksStart

Motion in two dimensions with vectors

48 questions

LessonNot started
[In this question, i\mathbf{i} and j\mathbf{j} are horizontal unit vectors and position vectors are given relative to a fixed origin OO.]

A particle PP is moving on a smooth horizontal plane.

The particle has constant acceleration (2i+3j) m s−2(2\mathbf{i} + 3\mathbf{j})\ \text{m s}^{-2}

At time t=0t = 0, PP passes through the point AA.

At time t=3t = 3 s, PP passes through the point BB.

The velocity of PP as it passes through AA is (5i−11j) m s−1(5\mathbf{i} - 11\mathbf{j})\ \text{m s}^{-1}
(a) Find the speed of PP as it passes through BB.

(4 marks)

The position vector of AA is (−9i−20j)(-9\mathbf{i} - 20\mathbf{j}) m.

At time t=Tt = T seconds, where T>3T > 3, PP passes through the point CC.

The position vector of CC is (41i+cj)(41\mathbf{i} + c\mathbf{j}) m.
(b) Find the value of TT.

(3 marks)
(c) Find the value of cc.

(3 marks)
●●●●●Level 410 marksStart
[In this question i\mathbf{i} and j\mathbf{j} are horizontal perpendicular unit vectors.]

A particle PP of mass 22 kg moves on a smooth horizontal surface with velocity 5j m s−15\mathbf{j}\text{ m s}^{-1}. It receives an impulse λ(i−j)\lambda(\mathbf{i} - \mathbf{j}) N s, where λ\lambda is a positive constant, and as a result it loses 10.510.5 J of kinetic energy.
(a) Show that λ2−10λ+21=0\lambda^2 - 10\lambda + 21 = 0

(5 marks)

Given also that the magnitude of the impulse is less than 66 N s,
(b) find the velocity of PP immediately after the impulse, giving your answer in terms of i\mathbf{i} and j\mathbf{j}.

(3 marks)
(c) Find the angle through which the impulse turns the direction of motion of PP.

(2 marks)
●●●●●Level 410 marksStart
[In this question i\mathbf{i} and j\mathbf{j} are horizontal perpendicular unit vectors.]

A box of mass 2.52.5 kg slides across a smooth horizontal floor with velocity (2i−3j) m s−1(2\mathbf{i} - 3\mathbf{j})\text{ m s}^{-1}. Someone kicks the box. The kick gives it a horizontal impulse I\mathbf{I} N s, where ∣I∣=102|\mathbf{I}| = 10\sqrt2, and the box then slides with velocity (λi+j) m s−1(\lambda\mathbf{i} + \mathbf{j})\text{ m s}^{-1}, where λ\lambda is a positive constant. The box is modelled as a particle.
(a) Show that (λ−2)2=16(\lambda - 2)^2 = 16

(4 marks)
(b) Hence find I\mathbf{I} in terms of i\mathbf{i} and j\mathbf{j}.

(3 marks)
(c) Find the angle between I\mathbf{I} and the direction in which the box was sliding before the kick.

(2 marks)
●●●●●Level 49 marksStart
In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

[In this question i\mathbf{i} and j\mathbf{j} are horizontal perpendicular unit vectors.]

A particle PP of mass 0.50.5 kg moves on a smooth horizontal plane. For 0≤t≤20 \le t \le 2, its position vector relative to a fixed origin OO at time tt seconds is r\mathbf{r} metres, where

r=(t3−8t)i+(t−t2)j\mathbf{r} = (t^3 - 8t)\mathbf{i} + (t - t^2)\mathbf{j}
(a) Show that the speed of PP at the instant when t=2t = 2 is 5 m s−15\text{ m s}^{-1}

(3 marks)

At the instant when t=2t = 2, PP receives an impulse of (−8i−j)(-8\mathbf{i} - \mathbf{j}) N s.
(b) Find the angle through which the direction of motion of PP is turned by the impulse.

(4 marks)
(c) Find the increase in the kinetic energy of PP caused by the impulse.

(2 marks)
●●●●●Level 49 marksStart
[In this question i\mathbf{i} and j\mathbf{j} are horizontal perpendicular unit vectors.]

A particle PP of mass 0.60.6 kg is moving on a smooth horizontal surface with velocity (2i+j) m s−1(2\mathbf{i} + \mathbf{j})\text{ m s}^{-1} when it receives an impulse λ(−i+j)\lambda(-\mathbf{i} + \mathbf{j}) N s, where λ\lambda is a constant. Immediately after receiving the impulse, PP is moving in a direction perpendicular to its direction of motion immediately before it received the impulse.
(a) Show that λ=3\lambda = 3

(5 marks)
(b) Find the increase in the kinetic energy of PP caused by the impulse.

(3 marks)
●●●●●Level 48 marksStart
A particle PP of mass mm kg is moving on a smooth horizontal surface under the action of two constant horizontal forces (3i−5j)(3\mathbf{i} - 5\mathbf{j}) N and (ai+bj)(a\mathbf{i} + b\mathbf{j}) N. The resultant of these two forces is R\mathbf{R} N. It is given that R\mathbf{R} acts in a direction which is parallel to the vector 2i+j2\mathbf{i} + \mathbf{j}.
(a) Show that a−2b+13=0a - 2b + 13 = 0.

(3 marks)

It is given that a=5a = 5 and that PP moves with an acceleration of magnitude 1.25 m s−21.25\ \text{m s}^{-2}.
(b) Determine the value of mm.

(4 marks)
●●●●●Level 47 marksStart
In this question the unit vectors i\mathbf{i} and j\mathbf{j} are in the directions east and north respectively.

A single force F\mathbf{F} N acts on a particle PP, which moves on a smooth horizontal surface. The velocity of PP at time tt seconds, where t≥0t \geq 0, is v m s−1\mathbf{v}\ \text{m s}^{-1}, where

v=(t2−4t+7)i+(3t−5)j.\mathbf{v} = (t^2 - 4t + 7)\mathbf{i} + (3t - 5)\mathbf{j}.
(a) Show that PP is never stationary.

(2 marks)
(b) Find, in terms of i\mathbf{i} and j\mathbf{j}, the acceleration of PP at time tt.

(1 mark)

The mass of PP is 0.60.6 kg.
(c) Determine the magnitude of F\mathbf{F} when PP is moving in the direction of the vector 3i−2j3\mathbf{i} - 2\mathbf{j}. Give your answer correct to 3 significant figures.

(5 marks)
●●●●●Level 48 marksStart
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

[In this question, i\mathbf{i} and j\mathbf{j} are horizontal perpendicular unit vectors.]

A particle PP moves on a smooth horizontal plane. At time tt seconds, t⩾0t \geqslant 0, the position vector of PP relative to a fixed origin OO is r\mathbf{r} metres, where

r=(2t2−5t)i+(t3−6t2+9t)j\mathbf{r} = (2t^2 - 5t)\mathbf{i} + (t^3 - 6t^2 + 9t)\mathbf{j}
(a) Find the speed of PP at the instant when PP is moving in the direction of the vector i\mathbf{i}

(5 marks)
(b) Find the exact magnitude of the acceleration of PP when t=3.5t = 3.5

(4 marks)
(c) Show that PP never returns to OO

(2 marks)
●●●●●Level 411 marksStart
[In this question i\mathbf{i} and j\mathbf{j} are unit vectors, with i\mathbf{i} horizontal and j\mathbf{j} vertical.](6i + 14j) m s⁻¹ABFigure 1 (not accurately drawn)

Two points, AA and BB, are on horizontal ground. A particle PP is projected from AA at time t=0t = 0 with velocity (6i+14j)(6\mathbf{i} + 14\mathbf{j}) m s−1^{-1} and moves freely under gravity until it hits the ground at BB, as shown in Figure 1.
(a) Find the distance ABAB.

(4 marks)

For an interval of TT seconds during its flight, the speed of PP is less than 7.57.5 m s−1^{-1}.
(b) Find the value of TT.

(3 marks)

When PP is moving perpendicular to the direction in which it was projected, its height above the ground is hh metres.
(c) Find the value of hh.

(4 marks)
●●●●●Level 411 marksStart
A remote-controlled car moves on a horizontal surface. The unit vectors i\mathbf{i} and j\mathbf{j} are directed east and north respectively.

At time tt seconds, where 0≤t≤50 \le t \le 5, the velocity of the car is

v=(t2−4t+1)i+(2t−5)j\mathbf{v} = \left(t^2 - 4t + 1\right)\mathbf{i} + (2t - 5)\mathbf{j} m s⁻¹.
(a) Find the acceleration of the car when t=2t = 2, and state the direction in which it acts. (2 marks)
(b) Find the value of tt at which the car is moving in the direction of the vector i+j\mathbf{i} + \mathbf{j}. Fully justify your answer. (4 marks)
(c) Find the distance of the car from its starting position when t=5t = 5. (3 marks)
●●●●●Level 49 marksStart
A particle PP of mass 0.50.5 kg moves in a horizontal plane under the action of a single force F\mathbf{F} newtons.

At time tt seconds the position vector of PP is

r=(t3−3t)i+2t2 jmetres\mathbf{r} = \left(t^3 - 3t\right)\mathbf{i} + 2t^2\,\mathbf{j} \quad \text{metres}
(a) Find the velocity of PP when t=2t = 2. (3 marks)
(b) Find the speed of PP when t=2t = 2, giving your answer to 33 significant figures. (2 marks)
(c) Find the magnitude of F\mathbf{F} when t=2t = 2, giving your answer to 33 significant figures. (4 marks)
●●●●●Level 49 marksStart
A particle PP moves in a horizontal plane. At time tt seconds (t≥0t \ge 0), its position vector relative to a fixed origin OO is

r=(2t2−3t)i+(4t−t2)j\mathbf{r} = (2t^2 - 3t)\mathbf{i} + (4t - t^2)\mathbf{j} metres.
(a) Find the velocity v\mathbf{v} of PP at time tt, and show that the acceleration of PP is constant. (3 marks)
(b) Find the speed of PP when t=3t = 3. (2 marks)
(c) Find the value of tt at which PP is moving parallel to i\mathbf{i}. (2 marks)
(d) Find the distance of PP from OO at this time, giving your answer in exact form. (3 marks)
●●●●●Level 410 marksStart
A particle PP moves in a horizontal plane. At time tt seconds, where t⩾0t \geqslant 0, the acceleration of PP is (6t i+4j)(6t\,\mathbf{i} + 4\mathbf{j}) m s⁻². When t=0t = 0, PP is at the point AA with position vector (7i−5j)(7\mathbf{i} - 5\mathbf{j}) m relative to a fixed origin OO, and the velocity of PP is (−5i−j)(-5\mathbf{i} - \mathbf{j}) m s⁻¹.
(a) Show that PP is never instantaneously at rest. (3 marks)
(b) Find the value of tt at which the velocity of PP is perpendicular to its acceleration, giving a reason for rejecting any other solutions of your equation. (4 marks)
(c) Find the distance of PP from OO at this time. (3 marks)
●●●●●Level 410 marksStart
Two particles, PP and QQ, are moving in separate straight lines across a smooth horizontal surface.

PP moves with constant velocity (8i−6j) m s−1(8\mathbf{i} - 6\mathbf{j})\ \text{m s}^{-1}

QQ moves from the point with position vector (−5i+9j)(-5\mathbf{i} + 9\mathbf{j}) metres to the point with position vector (15i−6j)(15\mathbf{i} - 6\mathbf{j}) metres during a 2.52.5 second period.
(a) Show that PP and QQ move along parallel lines.

(3 marks)
(b) Megan says

“QQ is also moving with a constant velocity of (8i−6j) m s−1(8\mathbf{i} - 6\mathbf{j})\ \text{m s}^{-1}”

Explain why Megan may be incorrect.

(1 mark)
(c) A third particle, RR, is moving with a constant speed of 6 m s−16\ \text{m s}^{-1}, in a straight line, across the same surface.

PP and RR move along lines that intersect at a fixed point XX.

It is given that:
  • PP passes through XX exactly 11 second after RR passes through XX
  • PP and RR are exactly 5050 metres apart 55 seconds after RR passes through XX
Show that PP and RR move along perpendicular lines.

(5 marks)
●●●●●Level 49 marksStart
Two particles, PP and QQ, move in the same horizontal plane.

Particle PP is initially at rest at the point with position vector (2i+j)(2\mathbf{i} + \mathbf{j}) metres and moves with constant acceleration (3i−2j) m s−2(3\mathbf{i} - 2\mathbf{j})\ \text{m s}^{-2}

Particle QQ moves in a straight line, passing through the points with position vectors (10i−3j)(10\mathbf{i} - 3\mathbf{j}) metres and (i+cj)(\mathbf{i} + c\mathbf{j}) metres, where cc is a constant.

PP and QQ are moving along parallel paths.
(a) Show that c=3c = 3

(4 marks)
(b) (i) Find an expression for the position vector of PP at time tt seconds.

(ii) Hence, prove that the paths of PP and QQ are not collinear.

(4 marks)
●●●●●Level 48 marksStart
A particle, TT, moves around the edges of a triangle ABCABC, in an anticlockwise direction, at a constant speed, as shown in the diagram below.ABCWhen TT moves from AA to BB its velocity is (23 i+2j) m s−1\left(2\sqrt{3}\,\mathbf{i} + 2\mathbf{j}\right)\ \text{m s}^{-1}
(a) Find the speed of TT.

(1 mark)
(b) When TT moves from BB to CC its velocity is (−23 i+2j) m s−1\left(-2\sqrt{3}\,\mathbf{i} + 2\mathbf{j}\right)\ \text{m s}^{-1}

Show that the acute angle ABC=60∘ABC = 60^\circ

(2 marks)
(c) It is given that ABCABC is an equilateral triangle.

TT returns to its initial position after 1818 seconds.

Vertex BB lies at the point with position vector [51]\begin{bmatrix} 5 \\ 1 \end{bmatrix} metres.

Find the position vector of CC.

(3 marks)
●●●●●Level 46 marksStart
A particle is moving such that its position vector, r\mathbf{r} metres, at time tt seconds, is given by

r=e2tcos⁡t i+e2tsin⁡t j\mathbf{r} = \mathrm{e}^{2t}\cos t\,\mathbf{i} + \mathrm{e}^{2t}\sin t\,\mathbf{j}

Show that the magnitude of the acceleration of the particle, a m s−2a\ \text{m s}^{-2}, is given by

a=5e2ta = 5\mathrm{e}^{2t}

Fully justify your answer.

(7 marks)
●●●●●Level 47 marksStart
[In this question i\mathbf{i} and j\mathbf{j} are horizontal unit vectors.]

A box of mass 1.51.5 kg is at rest on a smooth horizontal floor. The box is modelled as a particle.

At time t=0t = 0, two horizontal forces, F1=(5i−4j)\mathbf{F}_1 = (5\mathbf{i} - 4\mathbf{j}) N and F2=(pi+qj)\mathbf{F}_2 = (p\mathbf{i} + q\mathbf{j}) N, where pp and qq are constants, are applied to the box. No other horizontal forces act on the box.

Given that the box moves in the direction of the vector (3i+j)(3\mathbf{i} + \mathbf{j}),
(a) show that p−3q+17=0p - 3q + 17 = 0

(4 marks)

Given that q=7q = 7
(b) find the speed of the box at time t=3t = 3 seconds.

(5 marks)
●●●●●Level 49 marksStart
[In this question i\mathbf{i} and j\mathbf{j} are horizontal unit vectors directed due east and due north respectively and position vectors are given relative to a fixed origin OO.]

A light aircraft is flying at a constant height with constant velocity (80i+150j) km h−1(80\mathbf{i} + 150\mathbf{j})\ \text{km h}^{-1}

At 14:00, the aircraft is at the point with position vector (−20i−30j)(-20\mathbf{i} - 30\mathbf{j}) km.

The aircraft is modelled as a particle moving in a horizontal plane.
(a) Find the speed of the aircraft.

(2 marks)

At time tt hours after 14:00, the position vector of the aircraft is a\mathbf{a} km.
(b) Find an expression for a\mathbf{a} in terms of i\mathbf{i}, j\mathbf{j} and tt.

(2 marks)

A tall radio mast is modelled as being at the point with position vector (10i+19j)(10\mathbf{i} + 19\mathbf{j}) km in the same horizontal plane.

Aircraft are not permitted to fly within 33 km of the mast.
(c) (i) Find the value of tt when the aircraft is closest to the mast.

(ii) Hence determine whether the aircraft can continue on its course without breaking this rule.

(7 marks)
●●●●●Level 411 marksStart
[In this question i\mathbf{i} and j\mathbf{j} are horizontal unit vectors directed due east and due north respectively and position vectors are given relative to a fixed origin OO.]

Two snowmobiles, AA and BB, are crossing a large frozen lake. The snowmobiles are modelled as particles.

Snowmobile AA is moving with constant velocity.

At 10:00, the position vector of AA is (−6i+11j)(-6\mathbf{i} + 11\mathbf{j}) km.

At 10:40, the position vector of AA is (10i−j)(10\mathbf{i} - \mathbf{j}) km.

At time tt hours after 10:00, the position vector of AA is rA\mathbf{r}_A km.
(a) Show that rA=(−6+24t)i+(11−18t)j\mathbf{r}_A = (-6 + 24t)\mathbf{i} + (11 - 18t)\mathbf{j}

(4 marks)

The speed of AA is V m s−1V\ \text{m s}^{-1}
(b) Find the value of VV.

(2 marks)

Snowmobile BB is moving with constant velocity (16i−20j) km h−1(16\mathbf{i} - 20\mathbf{j})\ \text{km h}^{-1}

At 10:00, the position vector of BB is (10i+9j)(10\mathbf{i} + 9\mathbf{j}) km.

At 11:15, BB passes through the point QQ.
(c) Show that AA also passes through QQ.

(5 marks)
●●●●●Level 411 marksStart
At time tt seconds, where t≥0t \ge 0, a particle PP has velocity v m s−1\mathbf{v}\ \text{m s}^{-1} where v=(2t2−5t+1) i+(t2−3t+9) j.\mathbf{v} = (2t^2 - 5t + 1)\,\mathbf{i} + (t^2 - 3t + 9)\,\mathbf{j}.

Find
(a) the speed of PP at time t=0t = 0

(3 marks)
(b) the value of tt when PP is moving parallel to (i+j)(\mathbf{i} + \mathbf{j})

(2 marks)
(c) the acceleration of PP at time tt seconds

(2 marks)
(d) the value of tt when the direction of the acceleration of PP is perpendicular to i\mathbf{i}

(2 marks)
●●●●●Level 49 marksStart
A particle moves in a plane with velocity v=(6t−4) i+3t2 j m s−1\mathbf{v} = (6t - 4)\,\mathbf{i} + 3t^2\,\mathbf{j}\ \text{m s}^{-1}. When t=0t = 0 its position vector relative to a fixed origin OO is (2i+j)(2\mathbf{i} + \mathbf{j}) m. Find
(a) its position vector at time tt,
(b) its position vector when t=2t = 2, and (c) its distance from OO when t=2t = 2.
●●●●●Level 45 marksStart
A particle moves in a plane with acceleration a=2t i+(6t2−4) j m s−2\mathbf{a} = 2t\,\mathbf{i} + (6t^2 - 4)\,\mathbf{j}\ \text{m s}^{-2}. When t=0t = 0 its velocity is (3i+j) m s−1(3\mathbf{i} + \mathbf{j})\ \text{m s}^{-1}. Find
(a) its velocity at time tt,
(b) its velocity when t=2t = 2, and (c) its speed when t=2t = 2.
●●●●●Level 45 marksStart
A particle moves in a plane with velocity v=(t2−2t) i+(3t−4) j m s−1\mathbf{v} = (t^2 - 2t)\,\mathbf{i} + (3t - 4)\,\mathbf{j}\ \text{m s}^{-1}, for t≥0t \ge 0. Find
(a) the value of tt at which it is moving parallel to i\mathbf{i},
(b) the value of tt at which it is moving parallel to j\mathbf{j}, and (c) explain why t=0t = 0 is not a valid answer to part (b).
●●●●●Level 44 marksStart
At time tt seconds a particle has velocity v=(t2−4t+3) i+(2t−4) j m s−1\mathbf{v} = (t^2 - 4t + 3)\,\mathbf{i} + (2t - 4)\,\mathbf{j}\ \text{m s}^{-1}, for t≥0t \ge 0. Find
(a) the speed of the particle when t=0t = 0,
(b) the values of tt at which it is moving parallel to j\mathbf{j},
(c) the acceleration at time tt, and (d) the value of tt at which the acceleration is perpendicular to i\mathbf{i}.
●●●●●Level 46 marksStart
A particle starts from the origin with velocity (3i+j)(3\mathbf{i} + \mathbf{j}) m s−1^{-1} and moves with constant acceleration (2i−4j)(2\mathbf{i} - 4\mathbf{j}) m s−2^{-2}.

Find the distance of the particle from the origin after 22 seconds, giving your answer in the form k34k\sqrt{34} m. (4 marks)
●●●●●Level 44 marksStart
A particle moves with velocity v=(2t−6) i+(4t−3) j\mathbf{v} = (2t - 6)\,\mathbf{i} + (4t - 3)\,\mathbf{j} m s−1^{-1} at time tt seconds.

Find the time at which the particle is moving parallel to j\mathbf{j}, and its speed at that instant. (4 marks)
●●●●●Level 44 marksStart
A particle has acceleration a=6t i−4 j\mathbf{a} = 6t\,\mathbf{i} - 4\,\mathbf{j} m s−2^{-2} at time tt seconds, and initial velocity (i+2j)(\mathbf{i} + 2\mathbf{j}) m s−1^{-1}.

Find the speed of the particle when t=1t = 1, giving your answer in the form k5k\sqrt{5} m s−1^{-1}. (5 marks)
●●●●●Level 45 marksStart
A particle has velocity v=(3t2−4) i+2t j\mathbf{v} = (3t^{2} - 4)\,\mathbf{i} + 2t\,\mathbf{j} m s−1^{-1} at time tt seconds, and is initially at the point with position vector 2i2\mathbf{i} m.

Find the position vector of the particle when t=2t = 2. (5 marks)
●●●●●Level 45 marksStart