Clastify logo
Subjects
Features
Review
HOT
Tutoring

Kinematics

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

Verified by Karim
Verified by Karim
Paper
Difficulty
Status
Level

This content can't be printed from the browser. Use Export PDF with Pro for an offline copy.

Question 1
SL • Paper 1
Easy
Non Calculator
SL • Paper 1
Easy
Non Calculator

A particle moves in a straight line. Its acceleration at time tt seconds is

a(t)=6t−6a(t)=6t-6

At t=0t=0, its velocity is 8 m s−18\text{ m s}^{-1} and its displacement from the origin is 0 m0\text{ m}.

A

v(t)v(t), the velocity of the particle at time tt.

[2]
B

the time at which the speed of the particle is least.

[2]
C

the displacement of the particle when t=2t=2.

[1]
Question 2
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

A particle moves in a straight line. Its displacement from a fixed origin at time tt seconds, 0≤t≤50\le t\le 5, is given by

s(t)=t3−6t2+9ts(t)=t^3-6t^2+9t

Here, ss is measured in metres.

A

the velocity of the particle and the times at which it is at rest.

[2]
B

the intervals during which the particle is moving in the positive direction.

[1]
C

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

[3]
Question 3
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

A particle moves along a straight line with acceleration

a(t)=4−2ta(t)=4-2t

where tt is measured in seconds. Initially, the particle has velocity −1 m s−1-1\text{ m s}^{-1} and displacement 5 m5\text{ m} from the origin.

A

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

[2]
B

the times in the interval 0≤t≤40\le t\le 4 when the particle is at rest.

[2]
C

whether the displacement has a local maximum or a local minimum at each of these times.

[1]
Question 4
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Two particles, PP and QQ, move along the same straight line. Their displacements from a fixed origin at time tt seconds are

sP(t)=t2−4t+1,sQ(t)=7−2ts_P(t)=t^2-4t+1,\qquad s_Q(t)=7-2t
A

the time t>0t>0 at which the particles meet.

[2]
B

the velocity of each particle at the time they meet.

[2]
C

whether the particles are moving in the same direction when they meet. Give a reason.

[1]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 5
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A particle moves along a straight line. Its displacement from a fixed origin after tt seconds is

s(t)=2t3−15t2+24t+5s(t)=2t^3-15t^2+24t+5

where 0≤t≤50\leq t\leq 5 and ss is measured in metres.

A

expressions for the velocity and acceleration of the particle at time tt.

[2]
B

the times when the particle is instantaneously at rest.

[2]
C

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

[2]
Question 6
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The acceleration of a particle moving along a straight line is

a(t)=1.2t−4.8a(t)=1.2t-4.8

where 0≤t≤60\leq t\leq 6. Initially, the particle is at the origin and has velocity 3 m s−13\ \text{m s}^{-1}.

A

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

[2]
B

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

[2]
C

the total distance travelled by the particle from t=0t=0 to t=6t=6.

[2]
Question 7
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The velocity-time graph of a particle moving along a straight line consists of straight line segments joining five labelled points, as shown.

A

the acceleration of the particle for 2≤t≤52\leq t\leq 5.

[1]
B

the times after t=0t=0 when the particle is instantaneously at rest.

[2]
C

the total distance travelled by the particle from t=0t=0 to t=9t=9.

[3]
Question 8
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A particle moves along a straight line with velocity

v(t)=0.5t2−3t+2v(t)=0.5t^2-3t+2

where 0≤t≤70\leq t\leq 7 and tt is measured in seconds.

A

the acceleration of the particle when t=4t=4.

[1]
B

the times when the particle is instantaneously at rest.

[2]
C

the total distance travelled by the particle from t=0t=0 to t=7t=7.

[3]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 9
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The velocity of a particle moving along a straight line is given by

v(t)=sin⁡t+cos⁡tv(t)=\sin t+\cos t

for 0≤t≤π0\le t\le \pi, where tt is measured in seconds and v(t)v(t) is measured in m s−1\text{m s}^{-1}. The particle is initially at the origin.

A

the acceleration of the particle at t=π2t=\frac{\pi}{2}.

[2]
B

the displacement of the particle from t=0t=0 to t=πt=\pi.

[2]
C

the total distance travelled by the particle from t=0t=0 to t=πt=\pi.

[2]
Question 10
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The velocity v m s−1v\text{ m s}^{-1} of a particle moving in a straight line is defined by

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

the displacement of the particle from t=0t=0 to t=10t=10.

[2]
B

the total distance travelled by the particle from t=0t=0 to t=10t=10.

[4]
Question 11
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A particle moves in the plane. Its position vector at time tt seconds, 0≤t≤40\le t\le 4, is

r(t)=(t2−4t)i+(2t−t2)j\mathbf r(t)=(t^2-4t)\mathbf i+(2t-t^2)\mathbf j

The position is measured in metres.

A

the velocity vector of the particle at time tt.

[2]
B

the time at which the speed of the particle is least.

[2]
C

the least speed of the particle.

[2]
Question 12
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A particle moves in the plane with acceleration

a=−2j\mathbf a=-2\mathbf j

At t=0t=0, its velocity is 3i+4j3\mathbf i+4\mathbf j and its position vector is j\mathbf j.

A

the position vector r(t)\mathbf r(t) of the particle.

[3]
B

the path of the particle satisfies

y=1+4x3−x29y=1+\frac{4x}{3}-\frac{x^2}{9}
[2]
C

the maximum value of yy.

[1]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 13
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A particle moves in the plane with position vector

r(t)=eti+e−tj\mathbf r(t)=e^t\mathbf i+e^{-t}\mathbf j

where t∈Rt\in\mathbb R.

A

the speed of the particle at time tt.

[2]
B

the time at which the speed is least.

[2]
C

, at this time, the velocity and acceleration vectors are perpendicular.

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

A particle moves in a straight line with velocity

v(t)=4sin⁡(0.8t)−1v(t)=4\sin(0.8t)-1

where 0≤t≤80\leq t\leq 8, tt is measured in seconds and vv in m s−1\text{m s}^{-1}.

A

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

[2]
B

the displacement of the particle from t=0t=0 to t=8t=8.

[2]
C

the total distance travelled by the particle from t=0t=0 to t=8t=8.

[3]
Question 15
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The displacement of a particle from a fixed origin is modelled by

s(t)=8ln⁡(t+1)−t2+3s(t)=8\ln(t+1)-t^2+3

where 0≤t≤60\leq t\leq 6, tt is measured in seconds and ss in metres.

A

the velocity of the particle at time tt.

[1]
B

the time when the particle changes direction.

[2]
C

the total distance travelled by the particle from t=0t=0 to t=6t=6.

[3]
Question 16
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A particle moves along a straight line. Let ss be the numerical value of its displacement in metres and let vv be the numerical value of its velocity in m s−1\text{m s}^{-1}. They are related by

v(s)=ss+1,s≥0v(s)=\frac{s}{s+1}, \qquad s\geq 0

The acceleration aa is expressed by its numerical value in m s−2\text{m s}^{-2}.

A

the acceleration is a(s)=s(s+1)3a(s)=\dfrac{s}{(s+1)^3}.

[3]
B

the displacement at which the acceleration is greatest.

[2]
C

the values of ss for which the acceleration is 0.100 m s−20.100\ \text{m s}^{-2}.

[2]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 17
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The position vector of a particle at time tt seconds is

r(t)=(t3−6t)i+4e−t/2j\mathbf r(t)=(t^3-6t)\mathbf i+4e^{-t/2}\mathbf j

where 0≤t≤30\leq t\leq 3 and distances are measured in metres.

A

the velocity vector and the acceleration vector of the particle at time tt.

[2]
B

the speed of the particle when t=2t=2.

[2]
C

the time at which the velocity vector is perpendicular to the acceleration vector.

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

A particle moves along a straight line with velocity

v(t)=0.2(t−0.5)(t−3)(t−6)v(t)=0.2(t-0.5)(t-3)(t-6)

where 0≤t≤80\leq t\leq 8, tt is measured in seconds and vv in m s−1\text{m s}^{-1}.

A

the times when the particle is instantaneously at rest.

[2]
B

the displacement of the particle from t=0t=0 to t=8t=8.

[2]
C

the total distance travelled by the particle from t=0t=0 to t=8t=8.

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

A particle moves along a straight line. Its position coordinate at time tt seconds is

x(t)=t2−4t+3x(t)=t^2-4t+3

where 0≤t≤30\leq t\leq 3 and distances are measured in metres.

A

the velocity of the particle at time tt.

[2]
B

the speed and the magnitude of the acceleration of the particle when t=1t=1.

[2]
C

the times, with 0<t≤30<t\leq 3, when the particle is at the origin.

[2]
Question 20
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A particle moves in a plane with position vector

r(t)=3sin⁡t i+4cos⁡t j\mathbf r(t)=3\sin t\,\mathbf i+4\cos t\,\mathbf j

where 0≤t≤π0\leq t\leq \pi, tt is measured in seconds and distances are measured in metres. This question intentionally extends one-dimensional kinematics to planar motion; numerical integration may be used where necessary.

A

the velocity vector and acceleration vector of the particle at time tt.

[2]
B

the maximum speed of the particle and the time at which it occurs.

[2]
C

the total distance travelled by the particle from t=0t=0 to t=πt=\pi.

[2]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 21
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

A particle AA moves in a straight line. Its displacement from a fixed origin at time tt seconds is given by

sA(t)=t3−9t2+24ts_A(t)=t^3-9t^2+24t

where 0≤t≤60\le t\le 6 and sAs_A is measured in metres.

A
I.

an expression for the velocity of particle AA at time tt.

[2]
II.

the times at which particle AA is instantaneously at rest, and the direction of motion in each of the intervals determined by these times.

[3]
B
I.

the displacement of particle AA from the origin when t=2t=2, t=4t=4 and t=6t=6.

[2]
II.

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

[3]
C

A second particle BB moves along the same line with velocity vB(t)=10−tv_B(t)=10-t, for 0≤t≤100\le t\le 10. Particle BB starts from the origin. the first time at which the distance travelled by particle BB is equal to the total distance travelled by particle AA during the first 66 seconds. If you did not obtain an answer to part (b)(ii), use 44 m44\text{ m}.

[2]
Question 22
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

A particle moves in a straight line with velocity

v(t)=2cos⁡t−1v(t)=2\cos t-1

where 0≤t≤2π0\le t\le 2\pi. Initially, the particle is 1 m1\text{ m} from the origin in the positive direction.

A

(a)

I.

the acceleration of the particle at time tt.

[1]
II.

the times at which the particle is instantaneously at rest.

[3]
B

(b)

I.

an expression for the displacement s(t)s(t) of the particle from the origin.

[2]
II.

the displacement of the particle from t=0t=0 to t=2πt=2\pi.

[2]
C

the total distance travelled by the particle from t=0t=0 to t=2πt=2\pi. If you did not obtain the rest times in part (a)(ii), use t=π3t=\frac{\pi}{3} and t=5π3t=\frac{5\pi}{3}.

[3]
Question 23
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The velocity-time graph of a particle moving in a straight line consists of straight line segments joining the points shown. The displacement of the particle from the origin at t=0t=0 is −5 m-5\text{ m}.

A
I.

the acceleration of the particle for 0≤t≤20\le t\le 2.

[2]
II.

all times at which the particle is instantaneously at rest.

[2]
B
I.

the displacement of the particle from t=0t=0 to t=10t=10.

[2]
II.

the displacement of the particle from the origin at t=10t=10.

[2]
C
I.

the total distance travelled by the particle from t=0t=0 to t=10t=10.

[2]
II.

the first time at which the particle passes through the origin.

[2]
Question 24
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

A particle moves in a straight line with velocity

v(t)=t2−4t+3v(t)=t^2-4t+3

where 0≤t≤50\le t\le 5. The particle starts at the origin. Here, tt is measured in seconds and v(t)v(t) in metres per second.

A
I.

the times at which the particle is instantaneously at rest.

[2]
II.

an expression for the displacement s(t)s(t) of the particle from the origin.

[2]
III.

the particle returns to the origin when t=3t=3.

[1]
B
I.

the intervals during which the particle moves in the positive direction.

[1]
II.

the total distance travelled by the particle from t=0t=0 to t=5t=5. If you did not obtain the displacement function in part (a)(ii), use s(t)=t33−2t2+3ts(t)=\frac{t^3}{3}-2t^2+3t.

[2]
C

the average velocity and the average speed of the particle over the interval 0≤t≤50\le t\le 5.

[2]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 25
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A particle moves in three-dimensional space with position vector

r(t)=3cos⁡t i+3sin⁡t j+4t k\mathbf r(t)=3\cos t\,\mathbf i+3\sin t\,\mathbf j+4t\,\mathbf k

where tt is measured in seconds and distances are measured in metres. Here i\mathbf i, j\mathbf j, and k\mathbf k are mutually perpendicular unit vectors along the positive xx-, yy-, and zz-axes, respectively.

A
I.

the velocity vector and acceleration vector of the particle at time tt.

[2]
II.

the velocity and acceleration vectors are perpendicular for all tt in this interval.

[2]
B
I.

the speed of the particle.

[2]
II.

the total distance travelled by the particle for 0≤t≤π0\le t\le \pi.

[1]
C

the angle between the velocity vector and the positive zz-axis.

[3]
Question 26
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A particle moves in a straight line. Its velocity v m s−1v\text{ m s}^{-1} at time tt seconds satisfies

dvdt=3−v\frac{dv}{dt}=3-v

Initially, v=1v=1 and the particle is at the origin.

A

By solving the differential equation,

v(t)=3−2e−tv(t)=3-2e^{-t}
[3]
B

the time at which the velocity is 2 m s−12\text{ m s}^{-1}.

[1]
C

the displacement of the particle at the time found in part (b).

[2]
Question 27
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A particle moves in a straight line. Its acceleration a m s−2a\text{ m s}^{-2} is related to its displacement s ms\text{ m} from the origin by

a=4sa=4s

When s=0s=0, the particle has velocity 3 m s−13\text{ m s}^{-1}. The particle continues to move in the positive direction.

A

v2=4s2+9v^2=4s^2+9
[3]
B

the velocity of the particle when s=2s=2.

[1]
C

the acceleration of the particle when s=2s=2.

[1]
Question 28
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Two particles, PP and QQ, move in three-dimensional space. Their position vectors at time tt seconds, t≥0t\ge 0, are

rP=(1,0,2)+t(2,−1,1)\mathbf r_P=(1,0,2)+t(2,-1,1) rQ=(5,−2,0)+t(0,1,2)\mathbf r_Q=(5,-2,0)+t(0,1,2)
A

the position vector of PP relative to QQ at time tt.

[2]
B

the time at which the distance between the particles is least.

[3]
C

the least distance between the particles.

[1]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 29
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A lift moves vertically along a straight shaft. Its velocity, in m s−1\text{m s}^{-1}, at time tt seconds is modelled by

v(t)=1.8sin⁡(πt6)−0.3,0≤t≤12v(t)=1.8\sin\left(\frac{\pi t}{6}\right)-0.3,\qquad 0\le t\le 12

Positive velocity is upwards. At t=0t=0, the lift is 2 m2\text{ m} above the ground floor.

A
I.

the times at which the lift is instantaneously at rest.

[3]
II.

the time interval during which the lift is moving upwards.

[2]
B
I.

the height of the lift above the ground floor when it is at its maximum height.

[3]
II.

the displacement of the lift from t=0t=0 to t=12t=12.

[1]
C

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

[3]
Question 30
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A particle moves along a straight line. Its displacement from a fixed origin at time tt seconds is

s(t)=3+6t−2.5t2+0.2t3,0≤t≤8s(t)=3+6t-2.5t^2+0.2t^3,\qquad 0\le t\le 8

where ss is measured in metres.

A
I.

an expression for the velocity of the particle.

[2]
II.

the times at which the particle changes direction.

[3]
B
I.

the acceleration of the particle at time tt.

[1]
II.

the velocity of the particle when its acceleration is zero.

[2]
C

the total distance travelled by the particle during the interval 0≤t≤80\le t\le 8.

[3]
Question 31
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The velocity-time graph of a particle moving in a straight line consists of straight line segments joining the points shown. The velocity is measured in m s−1\text{m s}^{-1} and time is measured in seconds.

A
I.

The graph passes through (0,2)(0,2), (3,5)(3,5), (6,−1)(6,-1), (9,−1)(9,-1) and (12,4)(12,4). the acceleration for 3<t<63<t<6.

[2]
II.

the times when the particle is instantaneously at rest.

[2]
B
I.

the displacement of the particle from t=0t=0 to t=12t=12.

[2]
II.

the total distance travelled from t=0t=0 to t=12t=12.

[2]
C

Given that the particle starts at the origin, the greatest displacement of the particle from the origin during the interval 0≤t≤120\le t\le 12.

[2]
Question 32
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A projectile is launched from a point 1 m1\ \text{m} above horizontal ground. Its position vector at time tt seconds is

r(t)=8ti+(1+12t−4.9t2)j\mathbf r(t)=8t\mathbf i+(1+12t-4.9t^2)\mathbf j

where distances are measured in metres and j\mathbf j is vertically upwards.

A
I.

the velocity vector of the projectile at time tt.

[2]
II.

the acceleration vector of the projectile.

[2]
B
I.

the maximum height reached by the projectile.

[2]
II.

the horizontal distance travelled when the projectile first reaches the ground.

[1]
C

the speed and the direction of motion of the projectile at the instant it reaches the ground. Give the direction as an angle below the horizontal.

[3]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 33
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A particle moves in three-dimensional space. Use the HL vector methods for three-dimensional kinematics in this question. Its position vector at time tt seconds is

r(t)=t2i+sin⁡t j+e−tk\mathbf r(t)=t^2\mathbf i+\sin t\,\mathbf j+e^{-t}\mathbf k

where 0≤t≤40\leq t\leq 4 and distances are measured in metres.

A

the speed of the particle when t=1t=1.

[2]
B

the minimum distance of the particle from the origin during the interval.

[3]
C

whether the particle is moving towards or away from the origin when t=1t=1. your answer.

[2]
Question 34
HL • Paper 3
Medium
Calculator Permitted
HL • Paper 3
Medium
Calculator Permitted

Two delivery carts PP and QQ move along the same straight track. At time tt seconds, PP has velocity vP(t)=2+3e−t/2v_P(t)=2+3e^{-t/2} and starts from the origin. Cart QQ starts 18 m18\ \text{m} ahead of PP and moves with constant velocity 1 m s−11\ \text{m s}^{-1}.

A
I.

the displacement of PP from the origin at time tt.

[2]
II.

the displacement of QQ from the origin at time tt.

[2]
B

the time at which PP catches QQ.

[3]
C

that the carts meet only once for t>0t>0.

[3]
Question 35
SL • Paper 1
Hard
Non Calculator
SL • Paper 1
Hard
Non Calculator

A particle moves along a straight line. Its acceleration at time tt seconds is

a(t)=4cos⁡(2t)a(t)=4\cos(2t)

where 0≤t≤π0\le t\le \pi. Initially its velocity is −1 m s−1-1\text{ m s}^{-1} and it is at the origin.

A
I.

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

[2]
II.

the times at which the particle is instantaneously at rest.

[3]
B
I.

an expression for the displacement s(t)s(t) of the particle from the origin.

[2]
II.

the displacement of the particle from t=0t=0 to t=πt=\pi.

[2]
C

the total distance travelled by the particle from t=0t=0 to t=πt=\pi. If you did not obtain the rest times in part (a)(ii), use t=π12t=\frac{\pi}{12} and t=5π12t=\frac{5\pi}{12}.

[4]
Question 36
SL • Paper 1
Hard
Non Calculator
SL • Paper 1
Hard
Non Calculator

A particle moves along a straight line with velocity

v(t)=e−t−14v(t)=e^{-t}-\frac14

where 0≤t≤ln⁡160\le t\le \ln 16. The particle starts at the origin.

A

Part (a)

I.

the acceleration of the particle at time tt.

[1]
II.

the time at which the particle is instantaneously at rest.

[2]
III.

whether this time gives a maximum or a minimum displacement. Give a reason.

[1]
B

Part (b)

I.

an expression for the displacement s(t)s(t) of the particle from the origin.

[2]
II.

the displacement of the particle from t=0t=0 to t=ln⁡16t=\ln 16.

[1]
C

the total distance travelled by the particle from t=0t=0 to t=ln⁡16t=\ln16. If you did not obtain the time in part (a)(ii), use t=ln⁡4t=\ln4.

[3]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 37
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Two particles PP and QQ move along a straight line. Taking a fixed positive direction along the line, the signed relative displacement of PP from QQ at time tt is

s(t)=((t−1)2+1) ms(t)=\big((t-1)^2+1\big)\ \text{m}

where tt is the numerical value of the time in seconds and 0≤t≤30\le t\le 3.

A
I.

the relative velocity v(t)v(t) and the relative acceleration a(t)a(t).

[2]
II.

the square of the distance between the particles is u4+2u2+1 m2u^4+2u^2+1\ \text{m}^2, where u=t−1u=t-1.

[2]
B
I.

the time at which the distance between the particles is least.

[2]
II.

the least distance between the particles.

[2]
C

whether the particles are moving towards each other or away from each other when t=2t=2. your answer.

[3]
Question 38
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A particle moves in a straight line. Its velocity v m s−1v\text{ m s}^{-1} satisfies the differential equation

dvdt=2−2v\frac{dv}{dt}=2-2v

At t=0t=0, the particle is at the origin and has velocity 5 m s−15\text{ m s}^{-1}.

A
I.

v(t)=1+4e−2tv(t)=1+4e^{-2t}.

[3]
II.

the acceleration of the particle at time tt.

[1]
B
I.

the time at which the acceleration is −1 m s−2-1\text{ m s}^{-2}.

[2]
II.

why the particle never changes direction.

[1]
C

the displacement of the particle from the origin at the time found in part (b)(i). If you did not obtain this time, use t=32ln⁡2t=\frac32\ln2.

[3]
Question 39
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

Two small robots, PP and QQ, move along the same straight track. At t=0t=0, robot PP is at the origin and robot QQ is 18 m18\ \text{m} ahead of PP. Their velocities, in m s−1\text{m s}^{-1}, are

vP(t)=4+2sin⁡(0.5t),vQ(t)=7−0.4tv_P(t)=4+2\sin(0.5t),\qquad v_Q(t)=7-0.4t

for 0≤t≤200\le t\le 20.

A
AI.

expressions for the displacement of each robot from the origin at time tt.

[3]
AII.

the time, after t=0t=0, when the robots meet.

[2]
B

At the instant when the robots meet, whether they are moving in the same direction. Give a reason.

[3]
C
CI.

the position of the robots when they meet.

[2]
CII.

the total distance travelled by robot QQ before the robots meet.

[2]
Question 40
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

A particle moves along a straight line. Its acceleration at time tt seconds is

a(t)=t−5,0≤t≤10a(t)=t-5,\qquad 0\le t\le 10

Initially, the particle is 2 m2\ \text{m} to the negative side of the origin and has velocity 8 m s−18\ \text{m s}^{-1}.

A
I.

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

[2]
II.

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

[3]
B
I.

the times at which the particle is instantaneously at rest.

[2]
II.

the minimum velocity of the particle during the interval.

[1]
C
I.

the times in 0≤t≤100\le t\le 10 when the particle is at the origin.

[2]
II.

the total distance travelled by the particle from t=0t=0 to t=10t=10.

[2]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 41
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

The displacement, in metres, of a sliding door from its closed position is modelled by

s(t)=12(1−e−0.3t)−0.8t,0≤t≤15s(t)=12(1-e^{-0.3t})-0.8t,\qquad 0\le t\le 15

where tt is measured in seconds. Positive displacement means that the door is opening.

A
I.

the velocity of the door at time tt.

[2]
II.

the acceleration of the door at time tt.

[2]
B
I.

the time at which the door is furthest from the closed position.

[2]
II.

the maximum displacement of the door from the closed position.

[2]
C
I.

the times when the speed of the door is 0.5 m s−10.5\ \text{m s}^{-1}.

[2]
II.

the total distance travelled by the door from t=0t=0 to t=15t=15.

[1]
Question 42
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A particle moves along a straight line. Its velocity v m s−1v\ \text{m s}^{-1} satisfies the differential equation

dvdt=0.6(5−v)\frac{dv}{dt}=0.6(5-v)

At t=0t=0, v=−2v=-2 and the displacement from the origin is 0 m0\ \text{m}.

A
I.

v(t)=5−7e−0.6tv(t)=5-7e^{-0.6t}.

[3]
II.

the time when the particle first comes to rest.

[2]
B
I.

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

[2]
II.

the displacement of the particle at t=8t=8.

[1]
C
I.

the total distance travelled by the particle during the first 88 seconds.

[2]
II.

the time at which the particle is 20 m20\ \text{m} from the origin in the positive direction.

[1]
Question 43
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Two drones, AA and BB, move in three-dimensional space. Their position vectors, in metres, at time tt seconds are

rA=(1,2,−1)+t(4,−1,2),rB=(13,−4,5)+t(−1,2,0)\mathbf r_A=(1,2,-1)+t(4,-1,2),\qquad \mathbf r_B=(13,-4,5)+t(-1,2,0)
A
I.

the position vector of drone AA relative to drone BB at time tt.

[2]
II.

the relative velocity of drone AA with respect to drone BB.

[2]
B
I.

the time at which the drones are closest together.

[2]
II.

the minimum distance between the drones.

[2]
C

the times when the drones are exactly 10 m10\ \text{m} apart.

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

A particle moves along a straight line with velocity v(t)=(t−1)(t−4)v(t)=(t-1)(t-4), for 0≤t≤60\le t\le 6, where tt is measured in seconds and vv in m s−1\text{m s}^{-1}.

A
I.

the times at which the particle is instantaneously at rest.

[2]
II.

the intervals on which the particle is moving in the positive direction.

[2]
B

the displacement of the particle from t=0t=0 to t=6t=6.

[2]
C

the total distance travelled by the particle from t=0t=0 to t=6t=6.

[3]
D

The model is generalized by defining the numerical value of the velocity (in m s−1\text{m s}^{-1}) as vp(t)=(t−p)(t−4p)v_p(t)=(t-p)(t-4p) for 0≤t≤6p0\le t\le 6p, where tt denotes the numerical value of time in seconds and p>0p>0 is a dimensionless parameter. the total distance travelled is 15p3 m15p^3\ \text{m}.

[3]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 45
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A particle moves along a straight line. Its acceleration is a(t)=6(t+1)3−1a(t)=\dfrac{6}{(t+1)^3}-1, for 0≤t≤50\le t\le 5. Initially the particle has velocity 0 m s−10\ \text{m s}^{-1} and displacement 2 m2\ \text{m}.

A
I.

v(t)v(t), the velocity at time tt.

[3]
II.

s(t)s(t), the displacement at time tt.

[2]
B

the time after t=0t=0 at which the particle changes direction.

[3]
C

the total distance travelled from t=0t=0 to t=5t=5.

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

Two drones move in three-dimensional space. Their position vectors, in metres, are rP=(3,1,5)+t(2,−1,3)\mathbf r_P=(3,1,5)+t(2,-1,3) and rQ=(13,−3,7)+t(−1,1,4)\mathbf r_Q=(13,-3,7)+t(-1,1,4), where t≥0t\ge 0 is measured in seconds.

A
I.

the position vector of QQ relative to PP at time tt.

[2]
II.

the relative velocity of QQ with respect to PP.

[2]
B

the time at which the distance between the drones is least.

[3]
C

the least distance between the drones.

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

A train starts from rest at a station and travels for 88 seconds before stopping at the next signal. Its velocity is modelled by v(t)=kt(8−t)v(t)=kt(8-t), 0≤t≤80\le t\le 8, where k>0k>0. The distance between the station and the signal is 48 m48\ \text{m}.

A
I.

the value of kk.

[3]
II.

Using the velocity-time graph, the time at which the maximum velocity occurs and the maximum velocity of the train.

[2]
B

the distance travelled in the first 33 seconds.

[2]
C

A similar journey of length DD metres over time TT seconds is modelled by v(t)=kt(T−t)v(t)=kt(T-t), 0≤t≤T0\le t\le T. the maximum velocity is 3D2T\dfrac{3D}{2T}.

[3]
D

For a journey of length 60 m60\ \text{m}, the least possible value of TT if the maximum velocity must not exceed 8 m s−18\ \text{m s}^{-1}.

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

For 0≤t≤40\le t\le 4, the velocity of a particle moving along a straight line is v(t)=t2−4t+3v(t)=t^2-4t+3.

A
I.

the acceleration of the particle at time tt.

[2]
II.

the times at which the particle is at rest.

[2]
B

the displacement from t=0t=0 to t=4t=4.

[2]
C

the total distance travelled from t=0t=0 to t=4t=4.

[2]
D

The model is changed to v(t)=t2−4t+cv(t)=t^2-4t+c on the same interval. the value of cc for which the displacement from t=0t=0 to t=4t=4 is zero.

[2]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 49
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A particle moves in the plane with position vector r(t)=ti+ln⁡(1+t)j\mathbf r(t)=t\mathbf i+\ln(1+t)\mathbf j, for 0≤t≤50\le t\le 5, where tt is measured in seconds and distances are measured in metres. This question assesses the planar extension of one-dimensional motion; use vector differentiation for velocity and acceleration, and distance travelled by integrating speed.

A
I.

the velocity vector at time tt.

[2]
II.

the speed at time tt.

[2]
B

the speed is decreasing for 0≤t≤50\le t\le 5.

[3]
C

the total distance travelled by the particle from t=0t=0 to t=5t=5.

[2]
D

the magnitude of the acceleration when t=2t=2.

[2]
Question 50
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A particle moves along a straight line with velocity v(t)=e−t(4−t)v(t)=e^{-t}(4-t) for all t≥0t\ge 0. Initially, its displacement from the origin is 00.

A
I.

the time at which the particle changes direction.

[2]
II.

the acceleration at time tt.

[2]
B

the displacement of the particle from t=0t=0 to t=6t=6.

[2]
C

the total distance travelled from t=0t=0 to t=6t=6.

[3]
D

the limiting displacement as t→∞t\to\infty.

[1]
Question 51
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A particle moves in the plane with position vector

r(t)=(t2−2t)i+(t33−2t)j\mathbf r(t)=(t^2-2t)\mathbf i+\left(\frac{t^3}{3}-2t\right)\mathbf j

where 0≤t≤30\le t\le 3 and distances are measured in metres.

A
I.

the velocity vector and acceleration vector of the particle at time tt.

[2]
II.

the time at which the velocity is parallel to the xx-axis.

[2]
B
I.

the square of the speed is t4−8t+8t^4-8t+8.

[2]
II.

the time at which the speed is least.

[2]
C
I.

the least speed of the particle.

[2]
II.

, at the time when the speed is least, the velocity and acceleration vectors are perpendicular. If you did not obtain the time in part (b)(ii), use t=23t=\sqrt[3]{2}.

[2]
Question 52
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A particle moves in a straight line. Its acceleration a m s−2a\text{ m s}^{-2} is related to its displacement s ms\text{ m} from the origin by

a=−sa=-s

When t=0t=0, the particle is at the origin and has velocity 2 m s−12\text{ m s}^{-1}. Initially the particle moves in the positive direction.

A
I.

Using a=vdvdsa=v\frac{dv}{ds}, v2=4−s2v^2=4-s^2.

[3]
II.

the maximum displacement of the particle from the origin.

[1]
B
I.

For the initial motion in the positive direction, s=2sin⁡ts=2\sin t.

[3]
II.

the time at which the particle first reaches its maximum displacement.

[2]
C

The motion continues according to s=2sin⁡ts=2\sin t for 0≤t≤π0\le t\le \pi. the total distance travelled by the particle during this interval.

[3]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 53
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A particle moves in a straight line. Its displacement from a fixed origin at time tt seconds is

s(t)=t2−4t+∣t−2∣+3s(t)=t^2-4t+|t-2|+3

where 0≤t≤50\le t\le 5 and ss is measured in metres.

A
I.

Write s(t)s(t) as a piecewise-defined function without using the modulus sign.

[2]
II.

s(t)s(t) is continuous but not differentiable at t=2t=2.

[2]
B
I.

the velocity of the particle for 0<t<20<t<2 and for 2<t<52<t<5.

[2]
II.

why the particle changes direction at t=2t=2 even though it is not instantaneously at rest at that time.

[1]
C
I.

the displacement of the particle from t=0t=0 to t=5t=5.

[1]
II.

the total distance travelled by the particle from t=0t=0 to t=5t=5.

[2]
Question 54
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A particle moves in a straight line in the positive direction. Its acceleration a m s−2a\ \text{m s}^{-2} is related to its displacement s ms\ \text{m} from a fixed origin by

a=12−0.5sa=12-0.5s

When s=0s=0, the particle has velocity 6 m s−16\ \text{m s}^{-1}.

A
I.

v2=36+24s−0.5s2v^2=36+24s-0.5s^2.

[3]
II.

the velocity of the particle when s=30s=30.

[2]
B
I.

the displacement at which the speed is greatest.

[2]
II.

the greatest speed of the particle.

[2]
C
I.

the time taken for the particle to move from s=0s=0 to s=10s=10.

[2]
II.

the time taken for the particle to reach its greatest speed.

[1]
Question 55
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A particle moves in the plane with position vector

r(t)=e−0.2tcos⁡ti+e−0.2tsin⁡tj,t≥0\mathbf r(t)=e^{-0.2t}\cos t\mathbf i+e^{-0.2t}\sin t\mathbf j,\qquad t\ge0

where distances are measured in metres and tt is measured in seconds.

A
I.

the velocity vector v(t)\mathbf v(t).

[3]
II.

the speed of the particle is 1.04e−0.2t\sqrt{1.04}e^{-0.2t}.

[2]
B
I.

the distance travelled by the particle from t=0t=0 to t=10t=10.

[2]
II.

the time at which the particle is 0.5 m0.5\ \text{m} from the origin.

[1]
C
I.

the angle between the position vector and the velocity vector is constant.

[3]
II.

this angle in degrees.

[1]
Question 56
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A particle moves along a straight line. Its velocity, in m s−1\text{m s}^{-1}, is given by

v(t)=∣t−3∣−1,0≤t≤8v(t)=|t-3|-1,\qquad 0\le t\le 8

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

A
I.

Write v(t)v(t) as a piecewise function without using the absolute value sign.

[2]
II.

the times at which the particle is instantaneously at rest.

[2]
B
I.

the displacement of the particle from t=0t=0 to t=8t=8.

[2]
II.

the total distance travelled from t=0t=0 to t=8t=8.

[2]
C
I.

the acceleration of the particle for t<3t<3 and for t>3t>3.

[2]
II.

Discuss whether the acceleration is defined at t=3t=3.

[1]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 57
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A particle moves along a straight line with velocity v(t)=3sin⁡t−1v(t)=3\sin t-1, for 0≤t≤2π0\le t\le 2\pi.

A
I.

the particle is instantaneously at rest at t=αt=\alpha and t=π−αt=\pi-\alpha, where α=arcsin⁡(13)\alpha=\arcsin\left(\frac13\right).

[2]
II.

the interval on which the particle moves in the positive direction.

[2]
B

the displacement of the particle from t=0t=0 to t=2πt=2\pi.

[2]
C

the total distance travelled is 82+4α8\sqrt2+4\alpha metres.

[4]
D

the average speed of the particle over the interval 0≤t≤2π0\le t\le 2\pi.

[2]
Question 58
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A particle moves along a straight line so that its acceleration is related to its displacement ss from the origin by a=−9sa=-9s. At t=0t=0, s=2s=2 and v=0v=0. The particle initially moves towards the origin.

A
I.

v2=36−9s2v^2=36-9s^2.

[3]
II.

the velocity of the particle when s=1s=1.

[2]
B

the displacement may be written as s=2cos⁡(3t)s=2\cos(3t).

[3]
C

the first time at which the particle passes through the origin.

[2]
D

the total distance travelled in one complete oscillation.

[3]
Question 59
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A test vehicle starts and finishes at rest during a journey of length 200 m200\ \text{m}. Its velocity is modelled by v(t)=kt2(10−t)2v(t)=kt^2(10-t)^2, for 0≤t≤100\le t\le 10, where k>0k>0.

A
I.

the value of kk.

[3]
II.

the maximum velocity of the vehicle.

[2]
B

A similar journey of length DD metres in TT seconds is modelled by v(t)=kt2(T−t)2v(t)=kt^2(T-t)^2. vmax⁡=15D8Tv_{\max}=\dfrac{15D}{8T}.

[4]
C

For a journey of length 200 m200\ \text{m}, the least journey time TT if the maximum velocity must not exceed 30 m s−130\ \text{m s}^{-1}.

[3]
Question 60
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A particle moves in the plane with position vector r(t)=e−t/2cos⁡t i+e−t/2sin⁡t j\mathbf r(t)=e^{-t/2}\cos t\,\mathbf i+e^{-t/2}\sin t\,\mathbf j, for t≥0t\ge 0, where the coordinates are measured in metres and tt is measured in seconds. Use planar vector kinematics for this question.

A
I.

the velocity vector of the particle.

[2]
II.

the speed is 52e−t/2\frac{\sqrt5}{2}e^{-t/2}.

[2]
B

the total distance travelled during the first complete revolution.

[3]
C

the total distance travelled as t→∞t\to\infty.

[3]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview

Integral Calculus

Maclaurin Series