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Date	November 2019	Marks available	4	Reference code	19N.2.SL.TZ0.S_10
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Find	Question number	S_10	Adapted from	N/A

Question

A rocket is travelling in a straight line, with an initial velocity of $140$ m s⁻¹. It accelerates to a new velocity of $500$ m s⁻¹ in two stages.

During the first stage its acceleration, $a$ m s⁻², after $t$ seconds is given by $a\left( t \right) = 240\,{\text{sin}}\left( {2t} \right)$ , where $0 \leqslant t \leqslant k$ .

The first stage continues for $k$ seconds until the velocity of the rocket reaches $375$ m s⁻¹.

Find an expression for the velocity, $v$ m s⁻¹, of the rocket during the first stage.

[4]

a.

Find the distance that the rocket travels during the first stage.

[4]

b.

During the second stage, the rocket accelerates at a constant rate. The distance which the rocket travels during the second stage is the same as the distance it travels during the first stage.

Find the total time taken for the two stages.

[6]

c.

Markscheme

recognizing that $v = \int a$ (M1)

correct integration A1

eg $- 120\,{\text{cos}}\left( {2t} \right) + c$

attempt to find $c$ using their $v\left( t \right)$ (M1)

eg $- 120\,{\text{cos}}\left( 0 \right) + c = 140$

$v\left( t \right) = - 120\,{\text{cos}}\left( {2t} \right) + 260$ A1 N3

[4 marks]

a.

evidence of valid approach to find time taken in first stage (M1)

eg graph, $- 120\,{\text{cos}}\left( {2t} \right) + 260 = 375$

$k = 1.42595$ A1

attempt to substitute their $v$ and/or their limits into distance formula (M1)

eg $\int_0^{1.42595} {\left| v \right|}$ , $\int {260 - 120} \,{\text{cos}}\left( {2t} \right)$ , $\int_0^k {\left( {260 - 120\,{\text{cos}}\left( {2t} \right)} \right)} \,{\text{d}}t$

$353.608$

distance is $354$ (m) A1 N3

[4 marks]

b.

recognizing velocity of second stage is linear (seen anywhere) R1

eg graph, $s = \frac{1}{2}h\left( {a + b} \right)$ , $v = mt + c$

valid approach (M1)

eg $\int {v = 353.608}$

correct equation (A1)

eg $\frac{1}{2}h\left( {375 + 500} \right) = 353.608$

time for stage two $=0.808248$ ( $0.809142$ from 3 sf) A2

$2.23420$ ( $2.23914$ from 3 sf)

$2.23$ seconds ( $2.24$ from 3 sf) A1 N3

[6 marks]

c.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.8—Solving trig equations

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