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Date November 2016 Marks available 4 Reference code 16N.2.sl.TZ0.9
Level SL only Paper 2 Time zone TZ0
Command term Find and Hence Question number 9 Adapted from N/A

Question

A particle P starts from a point A and moves along a horizontal straight line. Its velocity v cms1 after t seconds is given by

v(t)={2t+2,for 0

The following diagram shows the graph of v.

N16/5/MATME/SP2/ENG/TZ0/09

P is at rest when t = 1 and t = p.

When t = q, the acceleration of P is zero.

Find the initial velocity of P.

[2]
a.

Find the value of p.

[2]
b.

(i)     Find the value of q.

(ii)     Hence, find the speed of P when t = q.

[4]
c.

(i)     Find the total distance travelled by P between t = 1 and t = p.

(ii)     Hence or otherwise, find the displacement of P from A when t = p.

[6]
d.

Markscheme

valid attempt to substitute t = 0 into the correct function     (M1)

eg\,\,\,\,\, - 2(0) + 2

2     A1     N2

[2 marks]

a.

recognizing v = 0 when P is at rest     (M1)

5.21834

p = 5.22{\text{ }}({\text{seconds}})     A1     N2

[2 marks]

b.

(i)     recognizing that a = v'     (M1)

eg\,\,\,\,\,v' = 0, minimum on graph

1.95343

q = 1.95     A1     N2

(ii)     valid approach to find their minimum     (M1)

eg\,\,\,\,\,v(q),{\text{ }} - 1.75879, reference to min on graph

1.75879

speed = 1.76{\text{ }}(c\,{\text{m}}\,{{\text{s}}^{ - 1}})     A1     N2

[4 marks]

c.

(i)     substitution of correct v(t) into distance formula,     (A1)

eg\,\,\,\,\,\int_1^p {\left| {3\sqrt t  + \frac{4}{{{t^2}}} - 7} \right|{\text{d}}t,{\text{ }}\left| {\int {3\sqrt t  + \frac{4}{{{t^2}}} - 7{\text{d}}t} } \right|}

4.45368

distance = 4.45{\text{ }}({\text{cm}})     A1     N2

(ii)     displacement from t = 1 to t = p (seen anywhere)     (A1)

eg\,\,\,\,\, - 4.45368,{\text{ }}\int_1^p {\left( {3\sqrt t  + \frac{4}{{{t^2}}} - 7} \right){\text{d}}t}

displacement from t = 0 to t = 1     (A1)

eg\,\,\,\,\,\int_0^1 {( - 2t + 2){\text{d}}t,{\text{ }}0.5 \times 1 \times 2,{\text{ 1}}}

valid approach to find displacement for 0 \leqslant t \leqslant p     M1

eg\,\,\,\,\,\int_0^1 {( - 2t + 2){\text{d}}t + \int_1^p {\left( {3\sqrt t  + \frac{4}{{{t^2}}} - 7} \right){\text{d}}t,{\text{ }}\int_0^1 {( - 2t + 2){\text{d}}t - 4.45} } }

- 3.45368

displacement =  - 3.45{\text{ }}({\text{cm}})     A1     N2

[6 marks]

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.

Syllabus sections

Topic 6 - Calculus » 6.6 » Kinematic problems involving displacement s, velocity v and acceleration a.
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