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Date	May 2019	Marks available	5	Reference code	19M.2.SL.TZ2.S_8
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Find	Question number	S_8	Adapted from	N/A

Question

In this question distance is in centimetres and time is in seconds.

Particle A is moving along a straight line such that its displacement from a point P, after $t$ seconds, is given by ${s_{\text{A}}} = 15 - t - 6{t^3}{{\text{e}}^{ - 0.8t}}$ , 0 ≤ $t$ ≤ 25. This is shown in the following diagram.

Another particle, B, moves along the same line, starting at the same time as particle A. The velocity of particle B is given by ${v_{\text{B}}} = 8 - 2t$ , 0 ≤ $t$ ≤ 25.

Find the initial displacement of particle A from point P.

[2]

Find the value of $t$ when particle A first reaches point P.

[2]

Find the value of $t$ when particle A first changes direction.

[2]

Find the total distance travelled by particle A in the first 3 seconds.

[3]

Given that particles A and B start at the same point, find the displacement function ${s_{\text{B}}}$ for particle B.

[5]

e.i.

Find the other value of $t$ when particles A and B meet.

[2]

e.ii.

Markscheme

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

valid approach (M1)

eg ${s_{\text{A}}}\left( 0 \right){\text{,}}\,\,s\left( 0 \right){\text{,}}\,\,t = 0$

15 (cm) A1 N2

[2 marks]

valid approach (M1)

eg ${s_{\text{A}}} = 0{\text{,}}\,\,s = 0{\text{,}}\,\,6.79321{\text{,}}\,\,14.8651$

2.46941

$t$ = 2.47 (seconds) A1 N2

[2 marks]

recognizing when change in direction occurs (M1)

eg slope of $s$ changes sign, $s' = 0$ , minimum point, 10.0144, (4.08, −4.66)

4.07702

$t$ = 4.08 (seconds) A1 N2

[2 marks]

METHOD 1 (using displacement)

correct displacement or distance from P at $t = 3$ (seen anywhere) (A1)

eg −2.69630, 2.69630

valid approach (M1)

eg 15 + 2.69630, $s\left( 3 \right) - s\left( 0 \right)$ , −17.6963

17.6963

17.7 (cm) A1 N2

METHOD 2 (using velocity)

attempt to substitute either limits or the velocity function into distance formula involving $\left| v \right|$ (M1)

eg $\int_0^3 {\left| v \right|{\text{d}}t}$ , $\int {\left| { - 1 - 18{t^2}{{\text{e}}^{ - 0.8t}} + 4.8{t^3}{{\text{e}}^{ - 0.8t}}} \right|}$

17.6963

17.7 (cm) A1 N2

[3 marks]

recognize the need to integrate velocity (M1)

eg $\int {v\left( t \right)}$

$8t - \frac{{2{t^2}}}{2} + c$ (accept $x$ instead of $t$ and missing $c$ ) (A2)

substituting initial condition into their integrated expression (must have $c$ ) (M1)

eg $15 = 8\left( 0 \right) - \frac{{2{{\left( 0 \right)}^2}}}{2} + c$ , $c = 15$

${s_{\text{B}}}\left( t \right) = 8t - {t^2} + 15$ A1 N3

[5 marks]

e.i.

valid approach (M1)

eg ${s_{\text{A}}} = {s_{\text{B}}}$ , sketch, (9.30404, 2.86710)

9.30404

$t = 9.30$ (seconds) A1 N2

Note: If candidates obtain ${s_{\text{B}}}\left( t \right) = 8t - {t^2}$ in part (e)(i), there are 2 solutions for part (e)(ii), 1.32463 and 7.79009. Award the last A1 in part (e)(ii) only if both solutions are given.

[2 marks]

e.ii.

Examiners report

[N/A]

e.i.

[N/A]

e.ii.

Syllabus sections

Topic 5—Calculus » AHL 5.13—Kinematic problems

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