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Date May 2021 Marks available 2 Reference code 21M.1.SL.TZ2.9
Level Standard Level Paper Paper 1 (without calculator) Time zone Time zone 2
Command term Find Question number 9 Adapted from N/A

Question

Particle A travels in a straight line such that its displacement, s metres, from a fixed origin after t seconds is given by s(t)=8t-t2, for 0t10, as shown in the following diagram.

Particle A starts at the origin and passes through the origin again when t=p.

Particle A changes direction when t=q.

The total distance travelled by particle A is given by d.

Find the value of p.

[2]
a.

Find the value of q.

[2]
b.i.

Find the displacement of particle A from the origin when t=q.

[2]
b.ii.

Find the distance of particle A from the origin when t=10.

[2]
c.

Find the value of d.

[2]
d.

A second particle, particle B, travels along the same straight line such that its velocity is given by v(t)=14-2t, for t0.

When t=k, the distance travelled by particle B is equal to d.

Find the value of k.

[4]
e.

Markscheme

setting st=0            (M1)

8t-t2=0

t8-t=0

p=8  (accept t=8, 8,0)                 A1

 

Note: Award A0 if the candidate’s final answer includes additional solutions (such as p=0,8).

 

[2 marks]

a.

recognition that when particle changes direction v=0 OR local maximum on graph of s OR vertex of parabola            (M1)

q=4 (accept t=4)            A1

 

[2 marks]

b.i.

substituting their value of q into st OR integrating vt from t=0 to t=4             (M1)

displacement=16(m)         A1

 

[2 marks]

b.ii.

s10=-20  OR  distance=st OR integrating vt from t=0 to t=10             (M1)

distance=20(m)        A1

 

[2 marks]

c.

16 forward + 36 backward  OR  16+16+20  OR  010vtdt             (M1)

d=52m        A1

 

[2 marks]

d.

METHOD 1

graphical method with triangles on vt graph             M1

49+x2x2             (A1)

49+x2=52, x=3             (A1)

k=7+3        A1

 

METHOD 2

recognition that distance =vtdt             M1

0714-2tdt+7k2t-14dt

14t-t207+t2-14t7k             (A1)

147-72+k2-14k-72-147=52             (A1)

k=7+3        A1

 

[4 marks]

e.

Examiners report

[N/A]
a.
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b.i.
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b.ii.
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c.
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d.
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e.

Syllabus sections

Topic 5 —Calculus » SL 5.9—Kinematics problems
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Topic 5 —Calculus

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