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Date	May 2019	Marks available	5	Reference code	19M.1.SL.TZ1.S_7
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 1
Command term	Find	Question number	S_7	Adapted from	N/A

Question

A particle P starts from point O and moves along a straight line. The graph of its velocity, $v$ ms⁻¹ after $t$ seconds, for 0 ≤ $t$ ≤ 6 , is shown in the following diagram.

The graph of $v$ has $t$ -intercepts when $t$ = 0, 2 and 4.

The function $s\left( t \right)$ represents the displacement of P from O after $t$ seconds.

It is known that P travels a distance of 15 metres in the first 2 seconds. It is also known that $s\left( 2 \right) = s\left( 5 \right)$ and $\int_2^4 {v\,{\text{d}}t} = 9$ .

Find the value of $s\left( 4 \right) - s\left( 2 \right)$ .

[2]

Find the total distance travelled in the first 5 seconds.

[5]

Markscheme

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

recognizing relationship between $v$ and $s$ (M1)

eg $\int {v = s}$ , $s' = v$

$s\left( 4 \right) - s\left( 2 \right) = 9$ A1 N2

[2 marks]

correctly interpreting distance travelled in first 2 seconds (seen anywhere, including part (a) or the area of 15 indicated on diagram) (A1)

eg $\int_0^2 {v = 15}$ , $s\left( 2 \right) = 15$

valid approach to find total distance travelled (M1)

eg sum of 3 areas, $\int_0^4 {v + } \int_4^5 v$ , shaded areas in diagram between 0 and 5

Note: Award M0 if only $\int_0^5 {\left| v \right|}$ is seen.

correct working towards finding distance travelled between 2 and 5 (seen anywhere including within total area expression or on diagram) (A1)

eg $\int_2^4 {v - } \int_4^5 v$ , $\int_2^4 {v = } \int_4^5 {\left| v \right|}$ , $\int_4^5 {v\,{\text{d}}t} = - 9$ , $s\left( 4 \right) - s\left( 2 \right) - \left[ {s\left( 5 \right) - s\left( 4 \right)} \right]$ ,

equal areas

correct working using $s\left( 5 \right) = s\left( 2 \right)$ (A1)

eg $15 + 9 - \left( { - 9} \right)$ , $15 + 2\left[ {s\left( 4 \right) - s\left( 2 \right)} \right]$ , $15 + 2\left( 9 \right)$ , $2 \times s\left( 4 \right) - s\left( 2 \right)$ , $48 - 15$

total distance travelled = 33 (m) A1 N2

[5 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5—Calculus » AHL 5.13—Kinematic problems

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Given that particles A and B start at the same point, find the displacement function ${s_{\text{B}}}$ for particle B.
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17M.2.SL.TZ2.S_7:
Note: In this question, distance is in metres and time is in seconds.

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Find the value of $s\left( 4 \right) - s\left( 2 \right)$ .
17N.1.AHL.TZ0.H_5:
A particle moves in a straight line such that at time $t$ seconds $(t \geqslant 0)$ , its velocity $v$ , in ${\text{m}}{{\text{s}}^{ - 1}}$ , is given by $v = 10t{{\text{e}}^{ - 2t}}$ . Find the exact distance travelled by the particle in the first half-second.
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Find an expression for the acceleration of P at time t.
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Interpret this point in the given context.

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