Free worked solutions · 2026 Edition

IB Maths AI HL Chapter 20: Kinematics worked solutions

Worked solutions to the practice questions in Chapter 20 of Photon Guidebook: Mathematics Applications and Interpretation HL for the IB Diploma (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[3 marks]
A lift has position s(t)=t^2-6t+11 metres.
Find its velocity and acceleration at t=4 seconds.
[3]
Show solutionHide solution
v=2t-6, so v(4)=2 metres per second. M1A1
a=2 metres per second squared.
A1

Final answer

v(4)=2 m s^{-1}; a=2 m s^{-2}.

Question 2

[2 marks]
A shuttle has s(t)=t^3-9t^2+15t+8 metres for 0\le t\le6 seconds.
Find v(t) and the times it is at rest.
[2]
Show solutionHide solution
v=3t^2-18t+15=3(t-1)(t-5).
M1
At rest at t=1 and t=5 seconds.
A1

Final answer

v=3(t-1)(t-5); at rest at t=1 and t=5.

Question 3

[4 marks]
A vehicle has acceleration a(t)=0.6t metres per second squared. At t=0, v=4 metres per second and s=3 metres.
Find velocity and position.
[4]
Show solutionHide solution
v=0.3t^2+C and C=4, so v=0.3t^2+4. M1A1
s=0.1t^3+4t+K and K=3, so s=0.1t^3+4t+3. M1A1

Final answer

v=0.3t^2+4; s=0.1t^3+4t+3.

Question 4

[3 marks]
A particle has v(t)=(t-2)(t-6) metres per second for 0\le t\le8.
Find its total distance travelled, showing the intervals used.
[3]
Show solutionHide solution
v is positive on (0,2) and (6,8), negative on (2,6).
M1
D=\int_0^2v\,dt-\int_2^6v\,dt+\int_6^8v\,dt.
M1
An antiderivative is t^3/3-4t^2+12t; D=32 metres.
A1

Final answer

32 m.

Question 5

[3 marks]
A particle has v(t)=2\sin t-1 metres per second on 0\le t\le\pi seconds, in radians.
(a) Find the reversal times.
[1]
(b) Use these times to calculate total distance.
[2]
Show solutionHide solution
(a) \sin t=1/2, so t=\pi/6,5\pi/6.
A1
(b) D=-\int_0^{\pi/6}v\,dt+\int_{\pi/6}^{5\pi/6}v\,dt-\int_{5\pi/6}^{\pi}v\,dt.
M1
D=4\sqrt3-4-\pi/3=1.88 metres to three significant figures.
A1

Final answer

(a) t=\frac\pi6,\ \frac{5\pi}6. (b) 4\sqrt3-4-\frac\pi3=1.88 m.

Question 6

[5 marks]
A surveying drone has position \mathbf r(t)=\begin{pmatrix}t^2+1\\3t+2\end{pmatrix} metres at t seconds, with the first component east and the second north.
(a) Find its velocity vector and the angle its direction makes north of east at t=2.
[3]
(b) Determine when its velocity points northeast at 45^\circ.
[2]
Show solutionHide solution
(a) \mathbf v(t)=\begin{pmatrix}2t\\3\end{pmatrix} metres per second.
A1
At t=2, \tan\theta=3/4, giving \theta=36.9^\circ north of east. M1A1
(b) 2t=3, so t=1.5 seconds. M1A1

Final answer

(a) \mathbf v=(2t,\,3); at t=2 the direction is 36.9^\circ north of east. (b) t=1.5 s.

Question 7

[3 marks]
A launch has velocity \mathbf v(t)=\begin{pmatrix}3\\4-t\end{pmatrix} metres per second and \mathbf r(0)=\begin{pmatrix}5\\-2\end{pmatrix} metres.
Find its position at t=6.
[3]
Show solutionHide solution
\mathbf r(t)=\begin{pmatrix}3t+5\\4t-t^2/2-2\end{pmatrix}. M1A1
\mathbf r(6)=\begin{pmatrix}23\\4\end{pmatrix} metres.
A1

Final answer

\mathbf r(6)=(23,\,4) m.

Question 8

[4 marks]
A projectile is launched from height 2.4 metres with horizontal velocity 9 metres per second and vertical velocity 13 metres per second. Take g=9.8 metres per second squared and neglect air resistance.
(a) Write its position vector for time t after launch.
[2]
(b) Find its horizontal distance when it first reaches ground level.
[2]
Show solutionHide solution
(a) \mathbf r(t)=\begin{pmatrix}9t\\2.4+13t-4.9t^2\end{pmatrix}. A1A1
(b) 2.4+13t-4.9t^2=0 gives positive time t=2.82636\ldots.
M1
Horizontal distance =9t=25.4 metres to three significant figures.
A1

Final answer

(a) \mathbf r=\begin{pmatrix}9t\\2.4+13t-4.9t^2\end{pmatrix}. (b) 25.4 m.

Question 9

[4 marks]
A train begins braking at s=0 with v(0)=17 metres per second. Its acceleration is a(t)=-2-0.4t metres per second squared, used until first rest.
(a) Find stopping time.
[2]
(b) Find stopping distance using the unrounded time.
[2]
Show solutionHide solution
(a) v=17-2t-0.2t^2.
M1
v=0 gives positive root t=5.49 s.
A1
(b) s=17t-t^2-t^3/15 at the stopping time.
M1
Stopping distance =52.2 m to three significant figures.
A1

Final answer

(a) 5.49 s. (b) 52.2 m.

Question 10

[4 marks]
Two boats start from the origin at t=0. Their velocities, in metres per second, are \mathbf v_A=\begin{pmatrix}2\\t\end{pmatrix} and \mathbf v_B=\begin{pmatrix}t\\2\end{pmatrix}, for 0\le t\le6.
(a) Find their position vectors.
[2]
(b) Find when they next meet and their speeds then.
[2]
Show solutionHide solution
(a) \mathbf r_A=\begin{pmatrix}2t\\t^2/2\end{pmatrix}, \mathbf r_B=\begin{pmatrix}t^2/2\\2t\end{pmatrix}. M1A1
(b) 2t=t^2/2 gives next meeting at t=4 s.
M1
Both speeds are \sqrt{20}=4.47 metres per second to three significant figures.
A1

Final answer

(a) \mathbf r_A=(2t,\,\frac{t^2}2), \mathbf r_B=(\frac{t^2}2,\,2t). (b) They meet at t=4 s; both speeds are 4.47 m s^{-1}.

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 11

[8 marks]
Siew is told that a pallet moves along a straight rail with acceleration a(t)=6t-4\ \,\mathrm{m\,s^{-2}}. At t=0 the pallet is 3 m from the origin O and moving with velocity -2\,\mathrm{m\,s^{-1}}. Siew writes v=3t^2-4t,\qquad s=t^3-2t^2.
(a) Identify two errors in Siew's working.
[2]
(b) Find the correct expressions for v(t) and s(t).
[4]
(c) Use your GDC to find the time, for t>0, at which the pallet is momentarily at rest.
[2]
Show solutionHide solution
(a) Siew has left out the constant of integration in v (so ignored v(0)=-2),
[1]
and has left out the constant in s (so ignored s(0)=3).
[1]
(b) v=3t^2-4t+C; v(0)=-2\Rightarrow C=-2.
[1]
v=3t^2-4t-2
[1]
s=t^3-2t^2-2t+D; s(0)=3\Rightarrow D=3.
[1]
s=t^3-2t^2-2t+3
[1]
(c) Solve 3t^2-4t-2=0 on the GDC: t=-0.387 or t=1.7207\ldots
[1]
Since t>0: t=\mathbf{1.72\ s}.
[1]

Final answer

(a) Both constants are missing: v(0)\ne-2 and s(0)\ne3. (b) v=3t^2-4t-2; s=t^3-2t^2-2t+3. (c) 1.72 s.

Question 12

[8 marks]
A camera hangs from a wire above a football pitch and travels along it. Its velocity is v(t)=3\cos(0.7t)-1\ \,\mathrm{m\,s^{-1}} for 0\le t\le 8 (use radians), where t is in seconds and positive means towards the north goal. At t=0 the camera is above the centre spot O.
(a) Use your GDC to find the times at which the camera is at rest.
[2]
(b) Find the displacement of the camera from O at t=8, and the total distance it travels in the 8 seconds.
[4]
(c) Find the greatest speed of the camera and the time at which it occurs.
[2]
Show solutionHide solution
(a) Solve 3\cos(0.7t)-1=0 (graph v and use the zero finder, window 0\le t\le8):
[1]
t=\mathbf{1.76\ s} and t=\mathbf{7.22\ s}.
[1]
(b) Displacement =\displaystyle\int_0^8(3\cos(0.7t)-1)\,dt=-10.705\ldots\approx\mathbf{-10.7\ m} (10.7 m south of O).
[2]
Total distance =\displaystyle\int_0^8\lvert3\cos(0.7t)-1\rvert\,dt=16.37\ldots\approx\mathbf{16.4\ m}.
[2]
(c) a=-2.1\sin(0.7t)=0 at t=0 and t=\pi/0.7=4.49 in the interval, so compare |v| at these and at the endpoint: |v(0)|=2, |v(4.49)|=|-4|=4, |v(8)|=1.33.
[1]
Greatest speed \mathbf{4\,\mathrm{m\,s^{-1}}} at t=\mathbf{4.49\ s}.
[1]

Final answer

(a) 1.76 s and 7.22 s. (b) Displacement −10.7 m; distance 16.4 m. (c) 4 m s^{-1} at t=4.49 s.

Question 13

[10 marks]
A remotely operated vehicle (ROV) descends from a research ship. Its downward velocity is v(t)=6te^{-0.4t}\ \,\mathrm{m\,s^{-1}} for 0\le t\le12, where t is in seconds and s(t) metres is its depth below the ship.
(a) Find an expression for the acceleration a(t), and write it in the form 6e^{-0.4t}(p-qt).
[3]
(b) Find the time at which the ROV is descending fastest, and its greatest speed, justifying that no endpoint gives a greater value.
[3]
(c) Use your GDC to find the depth of the ROV at t=12, given s(0)=0.
[2]
(d) Use your GDC to find the time at which the ROV reaches a depth of 30 m.
[2]
Show solutionHide solution
(a) Product rule: a=6e^{-0.4t}+6t(-0.4)e^{-0.4t}
[2]
a=6e^{-0.4t}(1-0.4t), so p=1, q=0.4.
[1]
(b) a=0\Rightarrow t=2.5\ \mathrm{s}.
[1]
v(2.5)=15e^{-1}=5.518\ldots\approx\mathbf{5.52\,\mathrm{m\,s^{-1}}}.
[1]
Endpoints: v(0)=0 and v(12)=0.593, both smaller, so this is the greatest.
[1]
(c) s(12)=\displaystyle\int_0^{12}6te^{-0.4t}\,dt=35.71\ldots\approx\mathbf{35.7\ m}.
[2]
(d) Solve \displaystyle\int_0^{T}6te^{-0.4t}\,dt=30 (graph the integral as a function of T and find where it meets y=30):
[1]
T=7.4857\ldots\approx\mathbf{7.49\ s}.
[1]

Final answer

(a) 6e^{-0.4t}(1-0.4t), so p=1 and q=0.4. (b) t=2.5 s; 5.52 m s^{-1}. (c) 35.7 m. (d) 7.49 s.

Question 14

[7 marks]
The hook of an overhead crane moves along a straight gantry. Its velocity is v(t)=4\sin(0.8t)\,\mathrm{m\,s^{-1}} for 0\le t\le6 (use radians), where t is in seconds and positive is towards the loading bay. The hook starts at the origin O.
(a) Find the time at which the hook is next at rest after t=0.
[1]
(b) Find the displacement of the hook from O at t=6.
[2]
(c) Find the total distance travelled by the hook in the 6 seconds.
[2]
(d) Find the average speed and the average velocity of the hook over the 6 seconds.
[2]
Show solutionHide solution
(a) \sin(0.8t)=0\Rightarrow0.8t=\pi\Rightarrow t=3.927\ldots\approx\mathbf{3.93\ s}.
[1]
(b) \displaystyle\int_0^6 4\sin(0.8t)\,dt=4.5625\ldots\approx\mathbf{4.56\ m}.
[2]
(c) \displaystyle\int_0^6\lvert4\sin(0.8t)\rvert\,dt=15.44\ldots\approx\mathbf{15.4\ m} (the hook reverses at t=3.93).
[2]
(d) Average speed =\dfrac{15.44}{6}=2.573\ldots\approx\mathbf{2.57\,\mathrm{m\,s^{-1}}}; average velocity =\dfrac{4.5625}{6}=0.7604\ldots\approx\mathbf{0.760\,\mathrm{m\,s^{-1}}}.
[2]

Final answer

(a) 3.93 s. (b) 4.56 m. (c) 15.4 m. (d) Average speed 2.57 m s^{-1}; average velocity 0.760 m s^{-1}.

Question 15

[8 marks]
A test sled runs along a straight track. Its velocity is v(t)=2t for 0\le t\le5 and v(t)=p-0.4t^2 for 5<t\le11, in metres per second, where t is the time in seconds, p is a constant and the velocity is continuous at t=5. The sled starts at the origin.
(a) Find p.
[1]
(b) Find the time at which the sled is at rest.
[2]
(c) Find the greatest displacement of the sled from the origin.
[2]
(d) Find the time at which the sled returns to the origin.
[3]
Show solutionHide solution
(a) 10=p-0.4(25), so p=20.
A1
(b) 20-0.4t^2=0.
M1
t=\sqrt{50}=7.07 s.
A1
(c) The displacement is greatest when the sled stops: \displaystyle\int_0^5 2t\,dt+\int_5^{7.07}(20-0.4t^2)\,dt.
M1
=25+10.95=35.9 m.
A1
(d) 25+\displaystyle\int_5^k(20-0.4t^2)\,dt=0.
M1
25+20(k-5)-\dfrac{0.4}{3}(k^3-125)=0.
A1
k=10.4 s, which lies in 5<k\le11.
A1

Final answer

(a) p=20. (b) 7.07 s. (c) 35.9 m. (d) 10.4 s.

Question 16

[7 marks]
A toy boat moves on a lake with position vector \mathbf r(t)=\begin{pmatrix}\tfrac13t^3-2t^2+6t\\t^2-5t\end{pmatrix} metres relative to the origin O at time t seconds, for 0\le t\le5.
(a) Find the velocity vector \mathbf v(t) and the acceleration vector \mathbf a(t).
[2]
(b) Find the exact speed of the boat when t=2.
[1]
(c) Use your GDC to find the least speed of the boat during the 5 seconds, and the time at which it occurs.
[3]
(d) At the time of least speed, the acceleration is perpendicular to the velocity. Explain why this must be so.
[1]
Show solutionHide solution
(a) \mathbf v=\begin{pmatrix}t^2-4t+6\\2t-5\end{pmatrix} [1] \mathbf a=\begin{pmatrix}2t-4\\2\end{pmatrix}
[1]
(b) \mathbf v(2)=\begin{pmatrix}2\\-1\end{pmatrix}, so speed =\sqrt{4+1}=\mathbf{\sqrt5\,\mathrm{m\,s^{-1}}}.
[1]
(c) Graph |\mathbf v|=\sqrt{(t^2-4t+6)^2+(2t-5)^2} for 0\le t\le5 and find the minimum:
[1]
t=2.25\ \mathrm{s}, [1] least speed =2.12\ldots\approx\mathbf{2.12\,\mathrm{m\,s^{-1}}}.
[1]
(The endpoint speeds are 7.81 and 12.1, larger.)
(d) \dfrac{d}{dt}|\mathbf v|^2=2\,\mathbf v\cdot\mathbf a, and at an interior minimum of the speed this is zero, so \mathbf v\cdot\mathbf a=0.
[1]

Final answer

(a) \mathbf v=(t^2-4t+6,\ 2t-5); \mathbf a=(2t-4,\ 2). (b) \sqrt5 m s^{-1}. (c) 2.12 m s^{-1} at t=2.25 s. (d) \frac d{dt}\lvert\mathbf v\rvert^2=2\,\mathbf v\cdot\mathbf a=0 at a least speed.

Question 17

[8 marks]
A radio-controlled buggy drives across a sandy beach. Its acceleration is \mathbf a(t)=\begin{pmatrix}0.6t\\-1\end{pmatrix}\,\mathrm{m\,s^{-2}}. At t=0 the buggy is at the origin O with velocity \begin{pmatrix}2\\3\end{pmatrix}\,\mathrm{m\,s^{-1}}.
(a) Find the velocity vector \mathbf v(t).
[3]
(b) Find the position vector \mathbf r(t).
[2]
(c) Find the time at which the buggy is moving parallel to the x-axis, and its position at that time.
[3]
Show solutionHide solution
(a) \mathbf v=\displaystyle\int\mathbf a\,dt=\begin{pmatrix}0.3t^2\\-t\end{pmatrix}+\mathbf C
[2]
\mathbf v(0)=\binom23\Rightarrow\mathbf C=\binom23, so \mathbf v=\begin{pmatrix}0.3t^2+2\\3-t\end{pmatrix}.
[1]
(b) \mathbf r=\begin{pmatrix}0.1t^3+2t\\3t-0.5t^2\end{pmatrix} (the constant is \mathbf 0 since \mathbf r(0)=\mathbf 0).
[2]
(c) Parallel to the x-axis means the y-component of velocity is zero: 3-t=0\Rightarrow t=\mathbf{3\ s}.
[1]
\mathbf r(3)=\begin{pmatrix}2.7+6\\9-4.5\end{pmatrix}=\begin{pmatrix}\mathbf{8.7}\\\mathbf{4.5}\end{pmatrix} m.
[2]

Final answer

(a) \mathbf v=(0.3t^2+2,\ 3-t). (b) \mathbf r=(0.1t^3+2t,\ 3t-0.5t^2). (c) t=3 s; (8.7,\,4.5) m.

Question 18

[8 marks]
A runner jogs round an oval track. Her position, in metres relative to the centre O of the track, is \mathbf r(t)=\begin{pmatrix}60\cos(0.1t)\\40\sin(0.1t)\end{pmatrix} at time t seconds (use radians).
(a) Find her velocity vector \mathbf v(t).
[2]
(b) Find her speed when t=5.
[2]
(c) Use your GDC to find the greatest speed of the runner and the first time at which it occurs.
[2]
(d) Find the runner's position at that time.
[2]
Show solutionHide solution
(a) \mathbf v=\begin{pmatrix}-6\sin(0.1t)\\4\cos(0.1t)\end{pmatrix}
[2]
(b) |\mathbf v(5)|=\sqrt{36\sin^2(0.5)+16\cos^2(0.5)}=\sqrt{20.60\ldots}=4.538\ldots\approx\mathbf{4.54\,\mathrm{m\,s^{-1}}}.
[2]
(c) Graph |\mathbf v|=\sqrt{36\sin^2(0.1t)+16\cos^2(0.1t)}: the maximum is \mathbf{6\,\mathrm{m\,s^{-1}}},
[1]
first reached when \sin(0.1t)=1, i.e. t=5\pi=\mathbf{15.7\ s}.
[1]
(d) \mathbf r(5\pi)=\begin{pmatrix}60\cos(\pi/2)\\40\sin(\pi/2)\end{pmatrix}=\begin{pmatrix}\mathbf0\\\mathbf{40}\end{pmatrix} m: the end of the short axis of the oval.
[2]

Final answer

(a) \mathbf v=(-6\sin0.1t,\ 4\cos0.1t). (b) 4.54 m s^{-1}. (c) 6 m s^{-1} at t=15.7 s. (d) (0,\,40) m.

Question 19

[9 marks]
A launcher fires a foam dart at 20\,\mathrm{m\,s^{-1}} from ground level. Take g=9.8\,\mathrm{m\,s^{-2}} and ignore air resistance. The launcher is first set at 25^\circ above the horizontal.
(a) Find the time of flight of the dart and the horizontal distance it travels before landing.
[4]
The launcher is now set at 65^\circ above the horizontal, with the same launch speed.
(b) Show that the dart lands at the same distance from the launcher as in part (a), to 3 significant figures.
[2]
For a launch angle \theta (in degrees) the horizontal range of the dart is R=\dfrac{400\sin 2\theta}{9.8} metres. A target lies on the ground 30 m from the launcher.
(c) Use your GDC to find the two angles \theta, with 0^\circ<\theta<90^\circ, at which the dart hits the target, and comment on how they are related to each other.
[3]
Show solutionHide solution
(a) y=20\sin25^\circ\,t-4.9t^2=0\Rightarrow t=\dfrac{40\sin25^\circ}{9.8}=1.7249\ldots\approx\mathbf{1.72\ s}.
[2]
x=20\cos25^\circ\times1.7249=31.26\ldots\approx\mathbf{31.3\ m}.
[2]
(b) At 65^\circ: t=\dfrac{40\sin65^\circ}{9.8}=3.699\ldots s.
[1]
x=20\cos65^\circ\times3.699=31.26\ldots\approx31.3\ \mathrm{m}, the same as in part (a).
[1]
(c) Solve \dfrac{400\sin2\theta}{9.8}=30 on the GDC (graph R(\theta) in degree mode and the line y=30, intersection):
[1]
\theta=\mathbf{23.7^\circ} and \theta=\mathbf{66.3^\circ}.
[1]
The two angles add up to 90^\circ (complementary), as with 25^\circ and 65^\circ in parts (a) and (b), because \sin2\theta=\sin(180^\circ-2\theta).
[1]

Final answer

(a) 1.72 s; 31.3 m. (b) 31.3 m again. (c) 23.7^\circ and 66.3^\circ; the angles are complementary.

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 20

[11 marks]
A submersible moves vertically in the water. It starts moving downward with velocity -5\text{ m s}^{-1} (downward is negative) and its acceleration, in \text{m s}^{-2}, is a(t)=6-2t, where t is the time in seconds.
(a) Find an expression for the velocity v(t).
[3]
(b) Find the two times at which the submersible is momentarily at rest.
[3]
(c) Find the displacement of the submersible between t=0 and t=5.
[2]
(d) Find the total distance travelled by the submersible between t=0 and t=5.
[3]
Show solutionHide solution
(a)
v(t)=\int (6-2t)\,dt=6t-t^2+c
Using v(0)=-5: c=-5, so v(t)=-t^2+6t-5.
[3]
(b) -t^2+6t-5=-(t-1)(t-5)=0\Rightarrow t=1 or t=5 s.
[3]
(c)
\int_0^5 (-t^2+6t-5)\,dt=\left[-\tfrac{t^3}{3}+3t^2-5t\right]_0^5=-\tfrac{125}{3}+75-25=\tfrac{25}{3}\approx 8.33\text{ m}
[2]
(d) The velocity is negative on [0,1] and positive on [1,5]:
\int_0^1 v\,dt=-\tfrac{7}{3},\qquad \int_1^5 v\,dt=\tfrac{32}{3}
Total distance =\tfrac{7}{3}+\tfrac{32}{3}=\tfrac{39}{3}=13\text{ m}.
[3]

Final answer

(a) v=-t^2+6t-5. (b) t=1 and t=5. (c) \frac{25}3=8.33 m. (d) 13 m.

Question 21

[14 marks]
A kayak races across a lake. Relative to a fixed origin O, in metres, its position at time t seconds is \mathbf r(t)=\begin{pmatrix}6t-0.25t^2\\8+3t\end{pmatrix} for 0\le t\le24. The x-direction is east and the y-direction is north.
(a) Find the velocity \mathbf v(t) and the acceleration \mathbf a(t) of the kayak.
[2]
(b) Find the initial speed of the kayak.
[1]
(c) Show that the kayak is heading due north at t=12, and find its position then.
[3]
A marker buoy is at the point B(30,\,40).
(d) Find the distance of the kayak from B at t=12.
[1]
(e) Use your GDC to find the least distance between the kayak and B, and the time at which it occurs.
[3]
(f) A crew member must call out whenever the kayak is within 6 m of B. Find the time interval over which this happens.
[3]
(g) State the greatest speed of the kayak during the 24 seconds.
[1]
Show solutionHide solution
(a) \mathbf v=\begin{pmatrix}6-0.5t\\3\end{pmatrix} [1] \mathbf a=\begin{pmatrix}-0.5\\0\end{pmatrix}
[1]
(b) \mathbf v(0)=\binom63, so speed =\sqrt{36+9}=\sqrt{45}=\mathbf{6.71\,\mathrm{m\,s^{-1}}}.
[1]
(c) \mathbf v(12)=\begin{pmatrix}0\\3\end{pmatrix}: the x-component is zero, so the kayak is moving due north.
[1]
\mathbf r(12)=\begin{pmatrix}72-36\\8+36\end{pmatrix}=\begin{pmatrix}\mathbf{36}\\\mathbf{44}\end{pmatrix} m.
[2]
(d) Distance =\sqrt{(36-30)^2+(44-40)^2}=\sqrt{52}=\mathbf{7.21\ m}.
[1]
(e) Distance d(t)=\sqrt{(6t-0.25t^2-30)^2+(3t-32)^2}. Graph it and find the minimum:
[1]
t=\mathbf{10.1\ s} [1] least distance =\mathbf{5.37\ m}.
[1]
(f) Solve d(t)=6 (intersect the graph with y=6):
[1]
t=9.18 and t=11.15.
[1]
So the kayak is within 6 m of the buoy for \mathbf{9.18\le t\le11.15} (about 1.97 s).
[1]
(g) |\mathbf v|^2=(6-0.5t)^2+9 is largest at the ends, t=0 and t=24: greatest speed =\mathbf{6.71\,\mathrm{m\,s^{-1}}}.
[1]

Final answer

(a) \mathbf v=(6-0.5t,\ 3); \mathbf a=(-0.5,\ 0). (b) 6.71 m s^{-1}. (c) (36,\,44). (d) 7.21 m. (e) 5.37 m at t=10.1 s. (f) 9.18\le t\le11.2. (g) 6.71 m s^{-1}.

Question 22

[12 marks]
Leila hits a ball from ground level at 28\,\mathrm{m\,s^{-1}} at 35^\circ above the horizontal. Take g=9.8\,\mathrm{m\,s^{-2}} and ignore air resistance. Dev, a fielder, stands on the ground in the same vertical plane as the shot, and will catch the ball 1.0 m above the ground, when it is on its way down.
(a) Write down expressions for the horizontal displacement x(t) and the height y(t) of the ball.
[2]
(b) Use your GDC to find the two times at which the ball is 1.0 m above the ground.
[3]
(c) Find the horizontal distance from Leila at which Dev must catch the ball.
[2]
(d) Dev starts 45 m from Leila and runs directly away from her at a constant speed to make the catch. Find that speed.
[2]
(e) Dev can run at no more than 7\,\mathrm{m\,s^{-1}}. Find the range of starting distances from Leila from which Dev could make the catch, running either towards or away from her.
[3]
Show solutionHide solution
(a) x(t)=28\cos35^\circ\,t=22.94t [1] y(t)=28\sin35^\circ\,t-4.9t^2=16.06t-4.9t^2
[1]
(b) Solve 16.06t-4.9t^2=1 on the GDC:
[2]
t=0.0635 s (rising) and t=3.214 s (falling).
[1]
(c) The catch is on the way down: t=3.214\ldots
[1]
x=22.94(3.214)=73.72\ldots\approx\mathbf{73.7\ m}.
[1]
(d) Dev covers 73.72-45=28.72 m in 3.214 s: speed =\dfrac{28.72}{3.214}=8.935\ldots\approx\mathbf{8.94\,\mathrm{m\,s^{-1}}}.
[2]
(e) In 3.214 s Dev can cover at most 7(3.214)=22.50 m.
[1]
He must start within 22.50 m of 73.72 m: from 73.72-22.50=51.22 m to 73.72+22.50=96.22 m,
[1]
i.e. from 51.2 m to 96.2 m from Leila.
[1]

Final answer

(a) x=22.94t; y=16.06t-4.9t^2. (b) 0.0635 s and 3.21 s. (c) 73.7 m. (d) 8.94 m s^{-1}. (e) From 51.2 m to 96.2 m.

Question 23

[10 marks]
Kemal flies a model aircraft on a control line in a horizontal circle. On a plan of the field, the position of the aircraft relative to a camera at O at time t seconds is \mathbf r(t)=\begin{pmatrix}2+5\cos(0.4t)\\5\sin(0.4t)\end{pmatrix} metres.
(a) Find the velocity and acceleration vectors of the aircraft.
[3]
(b) Show that the aircraft stays at a constant distance from the point C(2,\,0), and state this distance.
[2]
(c) Show that the acceleration of the aircraft is always directed towards C.
[2]
(d) Find the first time t>0 at which the velocity of the aircraft is perpendicular to its position vector \mathbf r, and the distance of the aircraft from O at that time.
[3]
Show solutionHide solution
(a) \mathbf v=\begin{pmatrix}-2\sin(0.4t)\\2\cos(0.4t)\end{pmatrix}.
M1A1
\mathbf a=\begin{pmatrix}-0.8\cos(0.4t)\\-0.8\sin(0.4t)\end{pmatrix}.
A1
(b) \mathbf r-\overrightarrow{OC}=\begin{pmatrix}5\cos(0.4t)\\5\sin(0.4t)\end{pmatrix} has squared length 25\cos^2(0.4t)+25\sin^2(0.4t)=25.
M1
The distance from C is 5 m.
A1
(c) \mathbf a=-0.16(\mathbf r-\overrightarrow{OC}).
M1
A negative multiple of the vector from C, so \mathbf a is directed towards C.
R1
(d) \mathbf r\cdot\mathbf v=-4\sin(0.4t)-10\sin(0.4t)\cos(0.4t)+10\sin(0.4t)\cos(0.4t)=-4\sin(0.4t).
M1
This is zero first when 0.4t=\pi, so t=2.5\pi=7.85 s.
A1
\mathbf r=(-3,\,0), so the aircraft is 3 m from O.
A1

Final answer

(a) \mathbf v=(-2\sin0.4t,\ 2\cos0.4t); \mathbf a=(-0.8\cos0.4t,\ -0.8\sin0.4t). (b) 5 m. (c) \mathbf a=-0.16(\mathbf r-\overrightarrow{OC}). (d) t=2.5\pi=7.85 s; 3 m.

Question 24

[20 marks]
A survey drone flies above a field, drifting with the wind as it circles. Relative to a base station O, its position at time t seconds is \mathbf r(t)=\begin{pmatrix}12\cos(0.2t)+0.5t\\12\sin(0.2t)\end{pmatrix} metres for 0\le t\le60.
(a) Find the velocity vector of the drone.
[2]
(b) Show that the square of the speed of the drone is 6.01-2.4\sin(0.2t).
[3]
(c) Hence find the greatest and least speeds of the drone.
[2]
(d) Find the first time at which the drone moves parallel to the y-axis.
[2]
(e) Find the acceleration vector of the drone, and show that the scalar product of its acceleration and velocity is -0.24\cos(0.2t).
[3]
(f) Explain the connection between your answers to parts (c) and (e).
[1]
(g) Find the total distance travelled by the drone in the 60 seconds, and the distance between its positions at t=0 and t=60.
[4]
A radio mast stands at the point M(20,\,-5).
(h) Find the least distance between the drone and the mast, and the time at which it occurs.
[3]
Show solutionHide solution
(a) \mathbf v=\begin{pmatrix}-2.4\sin(0.2t)+0.5\\2.4\cos(0.2t)\end{pmatrix}.
M1A1
(b) \lvert\mathbf v\rvert^2=5.76\sin^2(0.2t)-2.4\sin(0.2t)+0.25+5.76\cos^2(0.2t).
M1M1
=5.76-2.4\sin(0.2t)+0.25.
A1
=6.01-2.4\sin(0.2t).
AG
(c) Greatest when \sin(0.2t)=-1: \sqrt{8.41}=2.9 m s^{-1}.
A1
Least when \sin(0.2t)=1: \sqrt{3.61}=1.9 m s^{-1}.
A1
(d) -2.4\sin(0.2t)+0.5=0.
M1
t=1.05 s.
A1
(e) \mathbf a=\begin{pmatrix}-0.48\cos(0.2t)\\-0.48\sin(0.2t)\end{pmatrix}.
A1
\mathbf a\cdot\mathbf v=1.152\sin\cos-0.24\cos-1.152\sin\cos.
M1
=-0.24\cos(0.2t).
A1
(f) \mathbf a\cdot\mathbf v=0 when \cos(0.2t)=0, that is \sin(0.2t)=\pm1: exactly where the speed is greatest or least.
R1
(g) Distance =\displaystyle\int_0^{60}\sqrt{6.01-2.4\sin(0.2t)}\,dt.
M1
=145 m.
A1
\mathbf r(60)-\mathbf r(0)=\begin{pmatrix}12\cos12+18\\12\sin12\end{pmatrix}.
M1
Length 28.9 m.
A1
(h) Minimise \sqrt{(12\cos0.2t+0.5t-20)^2+(12\sin0.2t+5)^2} on the GDC.
M1
4.12 m
A1
at t=27.7 s.
A1

Final answer

(a) \mathbf v=(-2.4\sin0.2t+0.5,\ 2.4\cos0.2t). (b) Shown. (c) 2.9 and 1.9 m s^{-1}. (d) 1.05 s. (e) \mathbf a=(-0.48\cos0.2t,\ -0.48\sin0.2t); shown. (f) \mathbf a\cdot\mathbf v=0 exactly when the speed is greatest or least. (g) 145 m; 28.9 m. (h) 4.12 m at t=27.7 s.

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 Motion in a circle

G1 Worked

[4 marks]
Mariam watches a carousel in a park. A horse is 6 m from the centre and the carousel makes one full turn every 20 seconds.
(a) Find the angle turned per second, \omega.
[1]
(b) Find the speed of the horse.
[1]
(c) Find the size and direction of the acceleration of the horse.
[2]
Show solutionHide solution
(a) \omega=\frac{2\pi}{20}=\frac{\pi}{10}=0.314 rad per second (A1). (b) Speed =R\omega=6\times\frac{\pi}{10}=1.88 m s^{-1} (A1). (c) R\omega^2=6\left(\frac{\pi}{10}\right)^2=0.592 m s^{-2} (A1), directed towards the centre of the carousel (A1).

Final answer

0.314 rad s^{-1}; 1.88 m s^{-1}; 0.592 m s^{-2} towards the centre.

G3 Investigation

[18 marks]
A ball-throwing machine for tennis practice has an arm of length 0.9 m that turns in a vertical plane about a hub 1.2 m above level ground, at a constant 10 radians per second. Take the origin on the ground directly below the hub, with x horizontal in the direction of the throw and y vertical. When the arm has turned through an angle \varphi from pointing straight down, the ball is at (0.9\sin\varphi,\ 1.2-0.9\cos\varphi).
(a) Taking \varphi=10t, find the velocity of the ball while it is on the arm and show that its speed is 9 metres per second. Show also that its acceleration is directed towards the hub and has magnitude 90 metres per second squared.
[6]
(b) Show that when the ball is released at angle \varphi its velocity is 9 metres per second at an angle \varphi above the horizontal. After release the ball has acceleration \begin{pmatrix}0\\-9.8\end{pmatrix} metres per second squared. For \varphi=\frac{\pi}{4}, find the time of flight and the horizontal distance from the origin at which the ball lands.
[5]
(c) Use your GDC to find the release angle that gives the greatest distance, and that distance.
[3]
(d) Explain why the best release angle is a little more than \frac{\pi}{4}.
[1]
(e) The coach wants the balls to land between 7.5 m and 8.5 m from the origin. Find the two intervals of release angle that achieve this.
[3]
Show solutionHide solution
(a) \mathbf v=\begin{pmatrix}0.9\cos\varphi\times10\\0.9\sin\varphi\times10\end{pmatrix}=\begin{pmatrix}9\cos10t\\9\sin10t\end{pmatrix} (M1 A1), speed \sqrt{81\cos^2+81\sin^2}=9 m s^{-1} (A1). \mathbf a=\begin{pmatrix}-90\sin10t\\90\cos10t\end{pmatrix} (A1) =-100\begin{pmatrix}0.9\sin10t\\-0.9\cos10t\end{pmatrix}, a negative multiple of the vector from the hub to the ball, so directed towards the hub (M1), with magnitude 90 m s^{-2} (A1). (b) \mathbf v=9(\cos\varphi,\sin\varphi), at angle \varphi above the horizontal (A1). For \varphi=\frac{\pi}{4} the release point is (0.636,\ 0.564) and the velocity is (6.364,\ 6.364) (A1). Then 0.564+6.364t-4.9t^2=0 (M1) gives t=1.38 s (A1), and x=0.636+6.364(1.38)=9.43 m (A1). (c) The landing distance as a function of \varphi is x=0.9\sin\varphi+9\cos\varphi\,t_f, with t_f the positive root of 1.2-0.9\cos\varphi+9\sin\varphi\,t-4.9t^2=0 (M1). The GDC maximum is at \varphi=0.791 rad, 45.3^\circ (A1), with distance 9.43 m (A1); the value at \frac{\pi}{4} is 9.4314 against 9.4320. (d) A larger \varphi moves the release point forward and higher, which outweighs the small loss from the angle (R1). (e) Solve x(\varphi)=7.5 and x(\varphi)=8.5 (M1): \varphi=0.441,\ 0.552,\ 1.03,\ 1.14 (A1), so 0.441\le\varphi\le0.552 or 1.03\le\varphi\le1.14 radians (A1).

Final answer

Shown; 1.38 s and 9.43 m; \varphi=0.791 rad, 9.43 m; a larger \varphi also raises and advances the release point; 0.441\le\varphi\le0.552 or 1.03\le\varphi\le1.14 radians.

From the book

Photon Guidebook: Mathematics Applications and Interpretation HL for the IB Diploma

Chapter 20 teaches kinematics from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book

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