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Date	November 2020	Marks available	2	Reference code	20N.2.AHL.TZ0.H_11
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Show that	Question number	H_11	Adapted from	N/A

Question

A particle $P$ moves in a straight line such that after time $t$ seconds, its velocity, $v$ in ${m s}^{- 1}$ , is given by $v = e^{- 3 t} \sin 6 t$ , where $0 < t < \frac{π}{2}$ .

At time $t$ , $P$ has displacement $s (t)$ ; at time $t = 0$ , $s (0) = 0$ .

At successive times when the acceleration of $P$ is $0 {m s}^{- 2}$ , the velocities of $P$ form a geometric sequence. The acceleration of $P$ is zero at times $t_{1}, t_{2}, t_{3}$ where $t_{1} < t_{2} < t_{3}$ and the respective velocities are $v_{1}, v_{2}, v_{3}$ .

Find the times when $P$ comes to instantaneous rest.

[2]

a.

Find an expression for $s$ in terms of $t$ .

[7]

b.

Find the maximum displacement of $P$ , in metres, from its initial position.

[2]

c.

Find the total distance travelled by $P$ in the first $1.5$ seconds of its motion.

[2]

d.

Show that, at these times, $\tan 6 t = 2$ .

[2]

e.i.

Hence show that $\frac{v_{2}}{v_{1}} = \frac{v_{3}}{v_{2}} = - e^{- \frac{π}{2}}$ .

[5]

e.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{π}{6} (= 0.524)$ A1

$\frac{π}{3} (= 1.05)$ A1

[2 marks]

a.

attempt to use integration by parts M1

$s = \int e^{- 3 t} \sin 6 t d t$

EITHER

$= - \frac{e^{- 3 t} \sin 6 t}{3} - \int - 2 e^{- 3 t} cos 6 t d t$ A1

$= - \frac{e^{- 3 t} \sin 6 t}{3} - (\frac{2 e^{- 3 t} cos 6 t}{3} - \int - 4 e^{- 3 t} sin 6 t d t)$ A1

$= - \frac{e^{- 3 t} \sin 6 t}{3} - (\frac{2 e^{- 3 t} cos 6 t}{3} + 4 s)$

$5 s = \frac{-3 e^{- 3 t} \sin 6 t - 6 e^{- 3 t} cos 6 t}{9}$ M1

OR

$= - \frac{e^{- 3 t} cos 6 t}{6} - \int \frac{1}{2} e^{- 3 t} cos 6 t d t$ A1

$= - \frac{e^{- 3 t} cos 6 t}{6} - (\frac{e^{- 3 t} sin 6 t}{12} + \int \frac{1}{4} e^{- 3 t} sin 6 t d t)$ A1

$= - \frac{e^{- 3 t} cos 6 t}{6} - (\frac{e^{- 3 t} sin 6 t}{12} + \frac{1}{4} s)$

$\frac{5}{4} s = \frac{- 2 e^{- 3 t} cos 6 t - e^{- 3 t} sin 6 t}{12}$ M1

THEN

$s = - \frac{e^{- 3 t} (sin 6 t + 2 cos 6 t)}{15} (+ c)$ A1

at $t = 0, s = 0 \Rightarrow 0 = - \frac{2}{15} + c$ M1

$c = \frac{2}{15}$ A1

$s = \frac{2}{15} - \frac{e^{- 3 t} (sin 6 t + 2 cos 6 t)}{15}$

[7 marks]

b.

EITHER

substituting $t = \frac{π}{6}$ into their equation for $s$ (M1)

$(s = \frac{2}{15} - \frac{e^{- \frac{π}{2}} (sin π + 2 cos π)}{15})$

OR

using GDC to find maximum value (M1)

OR

evaluating $\int_{0}^{\frac{π}{6}} v d t$ (M1)

THEN

$= 0.161 (= \frac{2}{15} (1 + e^{- \frac{π}{2}}))$ A1

[2 marks]

c.

METHOD 1

EITHER

distance required $= \int_{0}^{1.5} |e^{- 3 t} \sin 6 t| d t$ (M1)

OR

distance required $= \int_{0}^{\frac{π}{6}} e^{- 3 t} \sin 6 t d t + |\int_{\frac{π}{6}}^{\frac{π}{3}} e^{- 3 t} \sin 6 t d t| + \int_{\frac{π}{3}}^{1.5} e^{- 3 t} \sin 6 t d t$ (M1)

$(= 0.16105 \dots + 0.033479 \dots + 0.006806 \dots)$

THEN

$= 0.201 (m)$ A1

METHOD 2

using successive minimum and maximum values on the displacement graph (M1)

$0.16105 \dots + (0.16105 \dots - 0.12757 \dots) + (0.13453 \dots - 0.12757 \dots)$

$= 0.201 (m)$ A1

[2 marks]

d.

valid attempt to find $\frac{d v}{d t}$ using product rule and set $\frac{d v}{d t} = 0$ M1

$\frac{d v}{d t} = e^{- 3 t} 6 \cos 6 t - 3 e^{- 3 t} \sin 6 t$ A1

$\frac{d v}{d t} = 0 \Rightarrow \tan 6 t = 2$ AG

[2 marks]

e.i.

attempt to evaluate $t_{1}, t_{2}, t_{3}$ in exact form M1

$6 t_{1} = arctan 2 (\Rightarrow t_{1} = \frac{1}{6} arctan 2)$

$6 t_{2} = π + arctan 2 (\Rightarrow t_{2} = \frac{π}{6} + \frac{1}{6} arctan 2)$

$6 t_{3} = 2 π + arctan 2 (\Rightarrow t_{3} = \frac{π}{3} + \frac{1}{6} arctan 2)$ A1

Note: The A1 is for any two consecutive correct, or showing that $6 t_{2} = π + 6 t_{1}$ or $6 t_{3} = π + 6 t_{2}$ .

showing that $\sin 6 t_{n + 1} = - \sin 6 t_{n}$

eg $\tan 6 t = 2 \Rightarrow \sin 6 t = \pm \frac{2}{\sqrt{5}}$ M1A1

showing that $\frac{e^{- 3 t_{n +1}}}{e^{- 3 t_{n}}} = e^{- \frac{π}{2}}$ M1

eg $e^{- 3 (\frac{π}{6} + k)} \div e^{- 3 k} = e^{- \frac{π}{2}}$

Note: Award the A1 for any two consecutive terms.

$\frac{v_{3}}{v_{2}} = \frac{v_{2}}{v_{1}} = - e^{- \frac{π}{2}}$ AG

[5 marks]

e.ii.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

[N/A]

e.i.

[N/A]

e.ii.

Syllabus sections

Topic 1—Number and algebra » SL 1.3—Geometric sequences and series

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