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Kinematics

Practice exam-style IB Math AI questions for Kinematics, aligned with the syllabus and grouped by topic.

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Question 1
HL • Paper 1
Easy
Calculator Permitted
HL • Paper 1
Easy
Calculator Permitted

A particle moves along a straight line. Its displacement ss metres from the origin at time tt seconds is given by

s(t)=t3−6t2+9t+2,t≥0s(t)=t^3-6t^2+9t+2, \qquad t\geq 0
A

expressions for the velocity v(t)v(t) and acceleration a(t)a(t) of the particle.

[2]
B

the times at which the particle is instantaneously at rest.

[2]
C

the acceleration of the particle at the first instant when it is at rest.

[1]
Question 2
HL • Paper 1
Easy
Calculator Permitted
HL • Paper 1
Easy
Calculator Permitted

The acceleration of a moving object is a(t)=6t−4 m s−2a(t)=6t-4\ \text{m s}^{-2}. Initially, its velocity is 3 m s−13\ \text{m s}^{-1} and its displacement is −2 m-2\ \text{m}.

A

an expression for the velocity v(t)v(t).

[2]
B

an expression for the displacement s(t)s(t).

[2]
C

the displacement of the object after 22 seconds.

[1]
Question 3
HL • Paper 1
Easy
Calculator Permitted
HL • Paper 1
Easy
Calculator Permitted

The position of an object moving along a straight line is modelled by

x=5cos⁡(0.4t)−2x=5\cos(0.4t)-2

where xx is measured in metres and tt in seconds.

A

x˙\dot{x} and x¨\ddot{x}.

[3]
B

the first time after t=0t=0 when the object is instantaneously at rest.

[2]
Question 4
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The velocity-time graph of a cyclist is formed by straight-line segments joining the points (0,2)(0,2), (2,6)(2,6), (5,0)(5,0), (7,−4)(7,-4) and (9,0)(9,0), where time is measured in seconds and velocity in m s−1\text{m s}^{-1}.

A

the cyclist's acceleration during the first 22 seconds.

[2]
B

the cyclist's displacement during the 99 seconds.

[2]
C

the total distance travelled during the 99 seconds.

[2]

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Question 5
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The velocity of a delivery robot moving along a straight corridor is given by

v(t)={4−t,0≤t≤6,t−8,6<t≤10,v(t)= \begin{cases} 4-t, & 0\leq t\leq 6,\\ t-8, & 6<t\leq 10, \end{cases}

where vv is measured in m s−1\text{m s}^{-1}.

A

the times at which the robot changes direction.

[2]
B

the displacement of the robot during the first 1010 seconds.

[2]
C

the total distance travelled during the first 1010 seconds.

[2]
Question 6
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

An object has initial velocity −2 m s−1-2\ \text{m s}^{-1}. Its acceleration is

a(t)={2,0≤t≤3,−1,3<t≤7.a(t)= \begin{cases} 2, & 0\leq t\leq 3,\\ -1, & 3<t\leq 7. \end{cases}
A

the velocity of the object at t=3t=3 and at t=7t=7.

[2]
B

the displacement of the object from t=0t=0 to t=7t=7.

[2]
C

when the object changes direction for 0<t<70<t<7.

[2]
Question 7
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The velocity of a particle moving along a straight line is

v(t)=t2−5t+4,0≤t≤5v(t)=t^2-5t+4, \qquad 0\leq t\leq 5

where vv is measured in m s−1\text{m s}^{-1}.

A

the times at which the particle changes direction.

[2]
B

the displacement of the particle during the first 55 seconds.

[2]
C

the total distance travelled by the particle during the first 55 seconds.

[3]
Question 8
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A particle moves in the positive direction along a straight line. Its velocity v m s−1v\ \text{m s}^{-1} is related to its displacement s ms\ \text{m} by

v=16−s,0≤s<16v=\sqrt{16-s}, \qquad 0\leq s<16

At t=0t=0, the particle is at s=0s=0.

A

the acceleration of the particle.

[2]
B

the speed of the particle when s=7s=7.

[1]
C

the time taken for the particle to travel from s=0s=0 to s=7s=7.

[2]

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Question 9
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A particle moves along a straight line with velocity

v(t)=12te−t,t≥0v(t)=12te^{-t}, \qquad t\geq 0

where vv is measured in m s−1\text{m s}^{-1}.

A

an expression for the acceleration of the particle.

[2]
B

the time at which the velocity is greatest and this greatest velocity.

[2]
C

the exact distance travelled during the first 22 seconds.

[2]
Question 10
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A particle starts at the origin and moves along a straight line with velocity

v(t)=3t2−12t+9,t≥0v(t)=3t^2-12t+9, \qquad t\geq 0
A

the first time after t=0t=0 when the particle returns to the origin.

[2]
B

the total distance travelled before the particle first returns to the origin.

[3]
Question 11
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The velocity of a particle is

v(t)=t(t−3)2,0≤t≤5v(t)=t(t-3)^2, \qquad 0\leq t\leq 5
A

an expression for the acceleration of the particle.

[1]
B

the intervals during which the particle is speeding up.

[3]
C

why the particle does not change direction at t=3t=3.

[1]
Question 12
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The displacement of a particle from an origin is

s(t)=t3−6t2+9t,0≤t≤3.5s(t)=t^3-6t^2+9t, \qquad 0\leq t\leq 3.5

where ss is measured in metres.

A

the times in the interval when the particle is instantaneously at rest.

[2]
B

the greatest displacement of the particle from the origin during the interval.

[2]
C

the total distance travelled during the interval.

[2]

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Question 13
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
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A particle moves along a straight line with velocity

v(t)=t2−kt+3,t≥0v(t)=t^2-kt+3, \qquad t\geq0

where tt is measured in seconds and v(t)v(t) is measured in m s−1\text{m s}^{-1}; the polynomial uses the numerical value of tt, and kk is a dimensionless constant. The particle changes direction at t=1t=1.

A

the value of kk.

[2]
B

the other time at which the particle changes direction.

[1]
C

the total distance travelled during the first 44 seconds.

[2]
Question 14
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The displacement of a buoy moving vertically is modelled by

s(t)=20sin⁡(0.15t),0≤t≤30s(t)=20\sin(0.15t), \qquad 0\leq t\leq30

where ss is measured in metres and tt in seconds.

A

expressions for the velocity and acceleration of the buoy.

[2]
B

the first time after t=0t=0 when the buoy is instantaneously at rest.

[2]
C

the total distance travelled by the buoy during the 3030 seconds.

[2]
Question 15
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A train moves along a straight track. Its acceleration a m s−2a\ \text{m s}^{-2} depends on its displacement s ms\ \text{m} according to

a=−0.4sa=-0.4s

When s=0s=0, its speed is 6 m s−16\ \text{m s}^{-1} and it is moving in the positive direction.

A

the speed vv satisfies v2=36−0.4s2v^2=36-0.4s^2.

[3]
B

the displacement at which the train first comes to rest.

[2]
Question 16
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A particle moves along a straight line with velocity

v(t)=(t−2)2(t−5),0≤t≤6v(t)=(t-2)^2(t-5), \qquad 0\leq t\leq6

Here, tt is measured in seconds and v(t)v(t) is measured in m s−1\text{m s}^{-1}.

A

all the times at which the particle is instantaneously at rest.

[2]
B

which of these times corresponds to a change of direction. your answer.

[2]
C

the total distance travelled by the particle during the 66 seconds.

[3]

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Question 17
HL • Paper 2
Medium
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HL • Paper 2
Medium
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A shuttle moves along a straight track. Its velocity at time tt seconds is given by

v(t)=4cos⁡(πt4),0≤t≤8v(t)=4\cos\left(\frac{\pi t}{4}\right),\qquad 0\leq t\leq 8

where velocity is measured in m s−1\text{m s}^{-1}. Initially, the shuttle is 33 metres from an origin OO. Take the positive direction to be away from OO, and let s(t)s(t) be the signed position relative to OO, so that s(0)=3 ms(0)=3\ \text{m}.

A
I.

an expression for the acceleration of the shuttle.

[2]
II.

the times at which the shuttle changes direction.

[2]
B

the total distance travelled and the greatest displacement of the shuttle from OO during the interval.

[4]
Question 18
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A cart moves along a straight test track with acceleration

a(t)=6t−6a(t)=6t-6

where t≥0t\geq0. At t=0t=0, its velocity is 4 m s−14\ \text{m s}^{-1} and its displacement is −3 m-3\ \text{m}.

A
I.

the velocity of the cart at time tt.

[2]
II.

the displacement of the cart at time tt.

[2]
B

the minimum speed of the cart during the first 33 seconds and the total distance it travels during this interval.

[4]
Question 19
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Two automated carts, AA and BB, move along the same straight rail. Their displacements from an origin at time tt seconds are

sA=t2+2t,sB=30−4ts_A=t^2+2t,\qquad s_B=30-4t

where displacement is measured in metres.

A
I.

the velocity and acceleration of each cart.

[2]
II.

the first time at which the carts meet.

[2]
B

the total distance travelled by the two carts before they meet and their relative speed at the meeting time.

[4]
Question 20
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A drone moves horizontally in the positive direction. Its velocity v m s−1v\ \text{m s}^{-1} is related to its displacement s ms\ \text{m} by

v=10s+2,s≥0v=\frac{10}{s+2},\qquad s\geq0

At t=0t=0, the drone is at s=0s=0.

A
I.

Using a=vdvdsa=v\dfrac{dv}{ds}, the acceleration in terms of ss.

[2]
II.

the acceleration when s=3s=3.

[2]
B

the time taken for the drone to reach s=8s=8 and its average speed during this journey.

[4]

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Question 21
HL • Paper 2
Hard
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HL • Paper 2
Hard
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The horizontal position of a sensor attached to a machine is modelled by

x(t)=6e−0.2tcos⁡t,t≥0x(t)=6e^{-0.2t}\cos t,\qquad t\geq0

where xx is measured in metres and tt in seconds.

A
I.

x˙\dot{x}.

[2]
II.

x¨\ddot{x}.

[2]
B

the first time after t=0t=0 when the sensor is instantaneously at rest. its position and whether this position is a local maximum or a local minimum.

[4]
Question 22
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

An elevator moves vertically, with upward taken as positive. Its velocity is

v(t)=(t−1)(t−3),0≤t≤5v(t)=(t-1)(t-3),\qquad 0\leq t\leq5

where velocity is measured in m s−1\text{m s}^{-1}. At t=0t=0, the elevator is 44 metres above a reference level.

A
I.

the acceleration and the times when the elevator changes direction.

[2]
II.

the elevator's displacement from the reference level at t=5t=5.

[2]
B

the total distance travelled by the elevator and its greatest height above the reference level during the interval.

[4]
Question 23
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A particle has acceleration

a(t)=3sin⁡(πt3),0≤t≤6a(t)=3\sin\left(\frac{\pi t}{3}\right),\qquad 0\leq t\leq6

Initially, its velocity is −9π m s−1-\dfrac{9}{\pi}\ \text{m s}^{-1} and its displacement is 1010 metres.

A
I.

an expression for the velocity.

[2]
II.

the times when the particle changes direction.

[2]
B

the total distance travelled and the greatest displacement from the origin during the interval.

[4]
Question 24
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Two runners, AA and BB, move in the positive direction along a straight path. At t=0t=0, runner AA is at the origin and runner BB is 99 metres ahead. Runner AA has velocity vA=2t m s−1v_A=2t\ \text{m s}^{-1}, while runner BB runs at a constant velocity of 6 m s−16\ \text{m s}^{-1}.

A
I.

expressions for the displacement of each runner from the origin.

[2]
II.

the time when runner AA first catches runner BB.

[2]
B

After runner AA passes runner BB, when the runners are first 2020 metres apart. their relative speed at this time.

[4]

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Question 25
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A camera drone moves along a straight tunnel. Its displacement from the tunnel entrance is modelled by

s(t)=0.05t3−0.9t2+4ts(t)=0.05t^3-0.9t^2+4t

where ss is measured in metres, tt is measured in seconds and 0≤t≤120\leq t\leq12.

A

Consider the velocity and direction of motion of the drone.

I.

an expression for the velocity of the drone.

[2]
II.

the times at which the drone changes direction.

[2]
B

the displacement of the drone from its starting point after 1212 seconds.

[2]
C

the total distance travelled by the drone during the 1212 seconds.

[2]
Question 26
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The velocity of a ferry moving towards and away from a terminal is modelled by

v(t)=3sin⁡(πt6)−1v(t)=3\sin\left(\frac{\pi t}{6}\right)-1

where vv is measured in m s−1\text{m s}^{-1} and 0≤t≤120\leq t\leq12. Positive velocity represents motion towards the terminal.

A

Consider the intersections of the velocity graph with the time axis.

I.

the two times at which the ferry is instantaneously at rest.

[2]
II.

the interval during which the ferry moves towards the terminal.

[2]
B

the net displacement of the ferry over the 1212 seconds and its sign.

[2]
C

the total distance travelled by the ferry during the 1212 seconds.

[2]
Question 27
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The velocity of an experimental racing vehicle is modelled by

v(t)=t(12−t)e−0.1t,0≤t≤12v(t)=t(12-t)e^{-0.1t},\qquad 0\leq t\leq12

where vv is measured in m s−1\text{m s}^{-1}.

A

Consider the acceleration of the vehicle.

I.

a(t)=e−0.1t(12−3.2t+0.1t2)a(t)=e^{-0.1t}(12-3.2t+0.1t^2)
[2]
II.

the time at which the vehicle reaches its greatest velocity.

[2]
B

the greatest velocity of the vehicle.

[2]
C

the total distance travelled during the 1212 seconds.

[2]
Question 28
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A ball is projected vertically upwards from a platform. Taking upwards as positive, its acceleration is modelled by

a(t)=−9.8−0.4ta(t)=-9.8-0.4t

Initially, the ball has velocity 18 m s−118\ \text{m s}^{-1} and displacement s=0s=0.

A

the velocity and displacement models.

I.

v(t)v(t).

[2]
II.

s(t)s(t).

[2]
B

the greatest height reached above the platform.

[2]
C

the speed of the ball when it first returns to the level of the platform.

[2]

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Question 29
HL • Paper 3
Hard
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HL • Paper 3
Hard
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The velocity of a solar-powered cart was recorded during a test. A quadratic regression model is to be used for the data.

t [s]

v [m s^-1]

0

1

2

5

4

7.4

6

8.2

8

7.4

10

5

A

Use the velocity data to and analyze a quadratic model.

I.

the quadratic regression equation in the form v=at2+bt+cv=at^2+bt+c.

[2]
II.

Use the model to the maximum velocity and the time at which it occurs.

[2]
B

Assuming the model continues to apply, the first time after t=10t=10 at which the cart comes to rest and the total distance travelled by then.

[2]
C

Give one reason why the stopping prediction in part (b) should be treated with caution.

[2]
Question 30
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A particle moves along a straight line with velocity

v(t)=t3−6t2+8tv(t)=t^3-6t^2+8t

for 0≤t≤50\leq t\leq5. Initially, its displacement from an origin is 22 metres.

A
I.

the acceleration and the times when the particle is instantaneously at rest.

[2]
II.

the intervals during which the particle is speeding up.

[2]
B

the total distance travelled and the greatest displacement from the origin during the interval.

[4]
Question 31
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A particle starts at an origin and moves along a straight line. Its velocity is modelled by

v(t)=Ae−t−1v(t)=Ae^{-t}-1

where A>0A>0. The particle first comes to rest at t=ln⁡4t=\ln4.
Here, tt is the numerical time in seconds, v(t)v(t) and AA are measured in m s−1\text{m s}^{-1}, and displacement is measured in metres.

A
I.

the value of AA.

[2]
II.

the acceleration at t=ln⁡4t=\ln4 and why the particle changes direction at this time.

[2]
B

the first time after t=0t=0 when the particle returns to the origin and the total distance travelled by then.

[4]
Question 32
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
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The velocity of a maintenance vehicle is modelled by

v(t)=2+3sin⁡t,0≤t≤2πv(t)=2+3\sin t,\qquad 0\leq t\leq2\pi

where vv is measured in m s−1\text{m s}^{-1}. The vehicle starts at displacement s=−1 ms=-1\ \text{m}.

A
I.

the two times when the vehicle is instantaneously at rest.

[2]
II.

the direction of motion on each of the three resulting time intervals.

[2]
B

the net displacement and the total distance travelled during the interval.

[4]

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Question 33
HL • Paper 2
Hard
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HL • Paper 2
Hard
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A particle moves along a straight line with velocity

v(t)=te−t/2(6−t),0≤t≤10v(t)=te^{-t/2}(6-t),\qquad 0\leq t\leq10

where tt is measured in seconds and v(t)v(t) is measured in metres per second.

A
I.

the acceleration of the particle.

[2]
II.

when the particle changes direction.

[2]
B

the maximum speed and the total distance travelled during the interval.

[4]
Question 34
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A boat moves along a straight channel. Its acceleration is

a(t)={2t,0≤t≤2,−2,2<t≤6,a(t)=\begin{cases}2t,&0\leq t\leq2,\\-2,&2<t\leq6,\end{cases}

where acceleration is measured in m s−2\text{m s}^{-2}. Initially, the boat's velocity is −1 m s−1-1\ \text{m s}^{-1}.

A
I.

a piecewise expression for the velocity.

[2]
II.

the times when the boat changes direction.

[2]
B

the displacement and the total distance travelled during the first 66 seconds.

[4]
Question 35
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
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The position of a robotic arm along a straight guide is modelled by

x(t)=t+4sin⁡t,0≤t≤2πx(t)=t+4\sin t,\qquad 0\leq t\leq2\pi

where tt is measured in seconds, angles are in radians, and xx is measured in metres.

A
I.

the velocity and acceleration.

[2]
II.

the times at which the position is stationary and classify each stationary position.

[2]
B

the total distance travelled by the robotic arm during the interval.

[4]
Question 36
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
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A capsule moves in the positive direction along a straight tunnel. Its speed v m s−1v\ \text{m s}^{-1} and displacement s ms\ \text{m} satisfy

v2=36+4s−s2v^2=36+4s-s^2

Initially, s=0s=0.

A
I.

the acceleration is a=2−sa=2-s.

[2]
II.

the displacement at which the speed is greatest and this greatest speed.

[2]
B

the displacement at which the capsule first comes to rest and the time taken to reach this displacement.

[4]

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Question 37
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A test vehicle moves in the positive direction along a straight track. Its acceleration depends on its displacement ss according to

a=6−0.5sa=6-0.5s

At s=0s=0, the vehicle has speed 2 m s−12\ \text{m s}^{-1}.

A

Use the relationship between acceleration, velocity and displacement.

I.

the velocity satisfies v2=4+12s−0.5s2v^2=4+12s-0.5s^2.

[2]
II.

the greatest speed attained before the vehicle first comes to rest.

[2]
B

the displacement at which the vehicle first comes to rest.

[2]
C

the time taken for the vehicle to reach this displacement.

[2]
Question 38
HL • Paper 3
Hard
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HL • Paper 3
Hard
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During a braking test, the acceleration aa of a car is related to its velocity vv by

a=−0.04v2a=-0.04v^2

At t=0t=0, the car passes a marker with velocity 20 m s−120\ \text{m s}^{-1} and displacement s=0s=0.

A

Consider the velocity and displacement during braking.

I.

that v(t)=201+0.8tv(t)=\frac{20}{1+0.8t} satisfies the acceleration model and the initial condition.

[2]
II.

an expression for s(t)s(t).

[2]
B

Safety regulations define the car as effectively stopped when its speed first reaches 2 m s−12\ \text{m s}^{-1}. the distance travelled by this time.

[2]
C

why the model does not a finite time at which the car is completely at rest.

[2]
Question 39
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
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A capsule moves along a straight evacuation tube. Its velocity depends on its displacement according to

v=6e−0.03s,s≥0v=6e^{-0.03s},\qquad s\geq0

where ss is measured in metres and vv in m s−1\text{m s}^{-1}.

A

Consider the capsule's acceleration and travel time.

I.

the acceleration is a=−0.03v2a=-0.03v^2.

[2]
II.

the time taken to travel the first 4040 metres.

[2]
B

the displacement at which the speed is half its initial value.

[2]
C

The speed is to fall to a fraction qq, where 0<q<10<q<1, of its initial value. Obtain an expression for the required displacement and how it changes as qq decreases.

[2]
Question 40
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

An elevator undergoes a six-second acceleration phase. Its acceleration is modelled by

a(t)=kt(6−t),0≤t≤6a(t)=kt(6-t),\qquad 0\leq t\leq6

The elevator starts from rest and has velocity 12 m s−112\ \text{m s}^{-1} at t=6t=6.

A

the parameter and the resulting velocity.

I.

the value of kk.

[2]
II.

an expression for v(t)v(t).

[2]
B

the distance travelled during the acceleration phase.

[2]
C

For a general acceleration phase of duration TT, where T>0T>0, suppose a(t)=kt(T−t)a(t)=kt(T-t), the elevator starts from rest and reaches velocity V>0V>0 at time TT. the distance travelled is VT2\frac{VT}{2}.

[2]

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Question 41
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A magnetic transport pod moves in the positive direction. Its acceleration depends on its displacement according to

a=2s+1,s≥0a=\frac{2}{s+1},\qquad s\geq0

At s=0s=0, its speed is 1 m s−11\ \text{m s}^{-1}.

A

Use a=vdvdsa=v\frac{dv}{ds} to the motion.

I.

v2=1+4ln⁡(s+1)v^2=1+4\ln(s+1).

[2]
II.

the speed when s=9s=9.

[2]
B

the displacement at which the pod first reaches a speed of 5 m s−15\ \text{m s}^{-1}.

[2]
C

how the pod can continue to speed up even though its acceleration decreases as ss increases.

[2]
Question 42
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A maintenance shuttle moves along a straight rail. Its acceleration during an eight-second control cycle is

a(t)=3sin⁡(πt8),0≤t≤8a(t)=3\sin\left(\frac{\pi t}{8}\right),\qquad 0\leq t\leq8

Initially, its velocity is −2 m s−1-2\ \text{m s}^{-1} and its displacement is zero.

A

the velocity and the reversal time.

I.

an expression for v(t)v(t).

[2]
II.

the time at which the shuttle changes direction.

[2]
B

the displacement of the shuttle during the control cycle.

[2]
C

the total distance travelled during the control cycle.

[2]
Question 43
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A robotic cart has velocity

v(t)=t2−kt+6v(t)=t^2-kt+6

where t≥0t\geq0 is the numerical time in seconds, v(t)v(t) is measured in m s−1\text{m s}^{-1}, and kk is a positive numerical constant.

A

First consider the case k=5k=5.

I.

the times at which the cart changes direction.

[2]
II.

the total distance travelled during the first 55 seconds.

[2]
B

the value of kk for which the cart is instantaneously at rest once but does not change direction.

[2]
C

the range of values of kk for which the cart changes direction twice for t>0t>0.

[2]
Question 44
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A delivery pod travels along a straight underground route. Its velocity is modelled by

v(t)=1+5cos⁡(πt10),0≤t≤20v(t)=1+5\cos\left(\frac{\pi t}{10}\right),\qquad 0\leq t\leq20

where vv is measured in m s−1\text{m s}^{-1}.

A

Consider the direction of travel of the pod.

I.

the times at which the pod changes direction.

[2]
II.

the proportion of the 2020 seconds during which the pod moves in the positive direction.

[2]
B

the displacement of the pod during the 2020 seconds.

[2]
C

the total distance travelled during the 2020 seconds.

[2]

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Question 45
HL • Paper 2
Hard
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HL • Paper 2
Hard
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A vehicle moves along a straight road in the positive direction. During braking, its acceleration depends on its displacement ss from the point where braking begins according to

a=−1−0.05sa=-1-0.05s

At s=0s=0, the vehicle's speed is 12 m s−112\ \text{m s}^{-1}.

A
I.

the speed satisfies v2=144−2s−0.05s2v^2=144-2s-0.05s^2.

[2]
II.

the stopping distance and the magnitude of the acceleration when the vehicle stops.

[2]
B

the time taken for the vehicle to stop.

[4]
Question 46
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Two autonomous vehicles, AA and BB, move along the same straight test track. At t=0t=0, vehicle AA is at the origin and vehicle BB is DD metres ahead. Their velocities are

vA(t)=6+2sin⁡(0.5t),vB(t)=4+0.3tv_A(t)=6+2\sin(0.5t),\qquad v_B(t)=4+0.3t

Initially, both vehicles move in the positive direction.

A

For part (a)(i), leave the initial separation as DD. For part (a)(ii), take D=12D=12 m.

I.

expressions for the positions sA(t)s_A(t) and sB(t)s_B(t).

[2]
II.

the first time at which AA catches BB.

[2]
B

For this part, take D=12D=12 m. the velocity of each vehicle at the first meeting and which vehicle is moving faster.

[2]
C

The initial separation is changed from 1212 metres to DD metres. the greatest value of DD for which AA catches BB at least once.

[2]
Question 47
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A carriage moves in the positive direction along a straight track. Its velocity is related to its displacement by

v=(s+4)(20−s),0≤s≤20v=\sqrt{(s+4)(20-s)},\qquad 0\leq s\leq20

where ss is measured in metres.

A

the acceleration and speed of the carriage.

I.

the acceleration is a=8−sa=8-s.

[2]
II.

the maximum speed of the carriage.

[2]
B

the time taken for the carriage to travel from s=0s=0 to s=20s=20.

[2]
C

whether the carriage spends more time accelerating or decelerating before reaching s=20s=20. your answer.

[2]
Question 48
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A moving platform has velocity

v(t)=2+4cos⁡tv(t)=2+4\cos t

where vv is measured in m s−1\text{m s}^{-1} and tt is measured in seconds. The argument of the cosine is in radians, and over one period 0≤t≤2π s0\leq t\leq2\pi\ \text{s}.

A

Consider the platform's motion over one period.

I.

the times at which the platform changes direction.

[2]
II.

the displacement over one period.

[2]
B

the exact total distance travelled during the period.

[2]
C

A general platform has velocity v(t)=c+Bcos⁡tv(t)=c+B\cos t, where 0<c<B0<c<B. Let θ=cos⁡−1(−cB)\theta=\cos^{-1}\left(-\frac{c}{B}\right). its total distance over one period is

4cθ−2πc+4Bsin⁡θ4c\theta-2\pi c+4B\sin\theta
[2]

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Integration