Mathematical Methods — Question 16

QCAA 2023, Paper 2 · 6 marks

Q16 · 2023 · Technology-activeComplex familiar6 marks

QUESTION 16 (6 marks)

A particle is moving in a straight line. The velocity (

\text{m s}^{-1}

) of the particle is given by

v(t)=\frac{20\sin(2t)}{6-5\cos(2t)},\qquad t\ge0,

where

t

is time (s) after moving from its initial position. The initial position of the particle is

+6.0

m from the origin.

Use calculus methods to determine an equation for the position of the particle from the origin at any time

t

[3 marks]

Determine the position of the particle relative to the origin when it first reaches maximum velocity.

[3 marks]

Question linkOriginal paper

WORKED SOLUTION

QCAA guide · typeset solution

6 marks

ANSWER

s(t)=2\ln(6-5\cos(2t))+6

; b)

7.21227

Worked solution

a) Position is given by

\int v(t)\,dt=\int\frac{20\sin(2t)}{6-5\cos(2t)}\,dt=2\int\frac{10\sin(2t)}{6-5\cos(2t)}\,dt.

Let

f(t)=6-5\cos(2t)

, so

f'(t)=10\sin(2t)

. Thus

s(t)=2\ln(6-5\cos(2t))+c.

Since

s(0)=6

c=6

, so

s(t)=2\ln(6-5\cos(2t))+6.

b) Using a GDC to graph

v(t)

for

t\ge0

, the first maximum occurs at

t=0.292843

s. The displacement to this time is

\int_0^{0.292843}v(t)\,dt=1.21227\text{ m}.

Since

s(0)=6

, the position relative to the origin is

1.21227+6=7.21227\text{ m}.

correctly manipulates the integrand to obtain a numerator that is the derivative of the denominator

[1 mark]

correctly determines position formula

[1 mark]

determines the position formula relative to the origin, i.e. allows for starting at +6 m

[1 mark]

correctly determines first time when velocity is a maximum

[1 mark]

identifies the required integral (or the formula established in 16a) and substitutes the maximum-velocity time

[1 mark]

determines the position of the particle

[1 mark]

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Mathematical Methods — Question 16

QUESTION 16 (6 marks)

QCAA guide · typeset solution

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