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Date May 2019 Marks available 2 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 A = 15 t 6 t 3 e 0.8 t , 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 B = 8 2 t , 0 ≤ t  ≤ 25.

Find the initial displacement of particle A from point P.

[2]
a.

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

[2]
b.

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

[2]
c.

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

[3]
d.

Given that particles A and B start at the same point, find the displacement function s 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 A ( 0 ) , s ( 0 ) , t = 0

15 (cm)      A1  N2

[2 marks]

a.

valid approach       (M1)

eg  s A = 0 , s = 0 , 6.79321 , 14.8651

2.46941

t = 2.47  (seconds)      A1  N2

[2 marks]

b.

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]

c.

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 ( 3 ) s ( 0 ) ,  −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  | v |        (M1)

eg  0 3 | v | d t ,    | 1 18 t 2 e 0.8 t + 4.8 t 3 e 0.8 t |

17.6963

17.7  (cm)      A1  N2

[3 marks]

d.

recognize the need to integrate velocity       (M1)

eg    v ( t )

8 t 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 ( 0 ) 2 ( 0 ) 2 2 + c ,    c = 15

s B ( t ) = 8 t t 2 + 15        A1  N3

[5 marks]

e.i.

valid approach      (M1)

eg    s A = s B , sketch, (9.30404, 2.86710)

9.30404

t = 9.30 (seconds)     A1  N2

Note: If candidates obtain  s B ( t ) = 8 t 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]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.
[N/A]
e.i.
[N/A]
e.ii.

Syllabus sections

Topic 5 —Calculus » SL 5.5—Integration introduction, areas between curve and x axis
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Topic 5 —Calculus

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