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Date November 2020 Marks available 5 Reference code 20N.2.AHL.TZ0.H_11
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Hence and 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=e3tsin6t, where 0<t<π2.

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

At successive times when the acceleration of P is 0m s2 , the velocities of P form a geometric sequence. The acceleration of P is zero at times t1, t2, t3 where t1<t2<t3 and the respective velocities are v1, v2, v3.

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, tan6t=2.

[2]
e.i.

Hence show that v2v1=v3v2=-e-π2.

[5]
e.ii.

Markscheme

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

π6(=0.524)      A1

π3(=1.05)      A1


[2 marks]

a.

attempt to use integration by parts        M1

s=e-3tsin6t dt


EITHER


=-e-3tsin6t3--2e-3t cos6t dt      A1

=-e-3tsin6t3-(2e-3t cos6t3--4e-3t sin6t dt)      A1

=-e-3tsin6t3-(2e-3t cos6t3+4s)

5s=-3e-3tsin6t-6e-3t cos6t9        M1


OR


=-e-3t cos6t6-12e-3t cos6t dt      A1

=-e-3t cos6t6-(e-3t sin6t12+14e-3t sin6t dt)      A1

=-e-3t cos6t6-(e-3t sin6t12+14s)

54s=-2e-3t cos6t-e-3t sin6t12        M1


THEN


s=-e-3t( sin6t+2cos6t)15(+c)      A1

at t=0, s=00=-215+c        M1

c=215      A1

s=215-e-3t( sin6t+2cos6t)15


[7 marks]

b.

EITHER

substituting t=π6 into their equation for s         (M1)

(s=215-e-π2( sinπ+2cosπ)15)


OR

using GDC to find maximum value         (M1)

OR

evaluating π60vdt         (M1)


THEN


=0.161(=215(1+e-π2))       A1 


[2 marks]

c.

METHOD 1 


EITHER

distance required =1.50|e-3tsin6t|dt       (M1)


OR

distance required =π60e-3tsin6tdt+|π3π6e-3tsin6tdt|+1.5π3e-3tsin6tdt       (M1)

(=0.16105+0.033479+0.006806)


THEN


=0.201 (m)       A1

 

METHOD 2


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

0.16105+(0.16105-0.12757)+(0.13453-0.12757)

=0.201 (m)       A1


[2 marks]

d.

valid attempt to find dvdt using product rule and set dvdt=0       M1

dvdt=e-3t6cos6t-3e-3tsin6t        A1

dvdt=0tan6t=2        AG


[2 marks]

e.i.

attempt to evaluate t1, t2, t3 in exact form         M1

6t1=arctan2(t1=16arctan2)

6t2=π+arctan2(t2=π6+16arctan2)

6t3=2π+arctan2(t3=π3+16arctan2)       A1


Note: The A1 is for any two consecutive correct, or showing that 6t2=π+6t1 or 6t3=π+6t2.


showing that sin6tn+1=-sin6tn

eg  tan6t=2sin6t=±25         M1A1

showing that e-3tn+1e-3tn=e-π2         M1

eg   e-3(π6+k)÷e-3k=e-π2


Note: Award the A1 for any two consecutive terms.


v3v2=v2v1=-e-π2        AG


[5 marks]

e.ii.

Examiners report

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b.
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c.
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d.
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e.i.
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e.ii.

Syllabus sections

Topic 1—Number and algebra » SL 1.3—Geometric sequences and series
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Topic 3— Geometry and trigonometry » SL 3.4—Circle: radians, arcs, sectors
Topic 5 —Calculus » SL 5.9—Kinematics problems
Topic 1—Number and algebra
Topic 3— Geometry and trigonometry
Topic 5 —Calculus

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