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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 \(v{\text{ cm}}\,{{\text{s}}^{ - 1}}\) after \(t\) seconds is given by

\[v(t) = \left\{ {\begin{array}{*{20}{l}} { - 2t + 2,}&{{\text{for }}0 \leqslant t \leqslant 1} \\ {3\sqrt t + \frac{4}{{{t^2}}} - 7,}&{{\text{for }}1 \leqslant t \leqslant 12} \end{array}} \right.\]

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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