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Date May 2013 Marks available 3 Reference code 13M.2.hl.TZ1.12
Level HL only Paper 2 Time zone TZ1
Command term Sketch Question number 12 Adapted from N/A

Question

A particle, A, is moving along a straight line. The velocity, vA ms1, of A t seconds after its motion begins is given by

vA=t35t2+6t.

Sketch the graph of vA=t35t2+6t for t, with {v_A} on the vertical axis and t on the horizontal. Show on your sketch the local maximum and minimum points, and the intercepts with the t-axis.

[3]
a.

Write down the times for which the velocity of the particle is increasing.

[2]
b.

Write down the times for which the magnitude of the velocity of the particle is increasing.

[3]
c.

At t = 0 the particle is at point O on the line.

Find an expression for the particle’s displacement, {x_A}{\text{m}}, from O at time t.

[3]
d.

A second particle, B, moving along the same line, has position {x_B}{\text{ m}}, velocity {v_B}{\text{ m}}{{\text{s}}^{ - 1}} and acceleration, {a_B}{\text{ m}}{{\text{s}}^{ - 2}}, where {a_B} = - 2{v_B} for t \geqslant 0. At t = 0,{\text{ }}{x_B} = 20 and {v_B} = - 20.

Find an expression for {v_B} in terms of t.

[4]
e.

Find the value of t when the two particles meet.

[6]
f.

Markscheme

     A1A1A1

Note: Award A1 for general shape, A1 for correct maximum and minimum, A1 for intercepts.

 

Note: Follow through applies to (b) and (c).

 

[3 marks]

a.

0 \leqslant t < 0.785,{\text{ }}\left( {{\text{or }}0 \leqslant t < \frac{{5 - \sqrt 7 }}{3}} \right)     A1

(allow t < 0.785)

and t > 2.55{\text{ }}\left( {{\text{or }}t > \frac{{5 + \sqrt 7 }}{3}} \right)     A1

[2 marks]

b.

0 \leqslant t < 0.785,{\text{ }}\left( {{\text{or }}0 \leqslant t < \frac{{5 - \sqrt 7 }}{3}} \right)     A1

(allow t < 0.785)

2 < t < 2.55,{\text{ }}\left( {{\text{or }}2 < t < \frac{{5 + \sqrt 7 }}{3}} \right)     A1

t > 3     A1

[3 marks]

c.

position of A: {x_A} = \int {{t^3} - 5{t^2} + 6t{\text{ d}}t}     (M1)

{x_A} = \frac{1}{4}{t^4} - \frac{5}{3}{t^3} + 3{t^2}\,\,\,\,\,( + c)     A1

when t = 0,{\text{ }}{x_A} = 0, so c = 0     R1

[3 marks]

d.

\frac{{{\text{d}}{v_B}}}{{{\text{d}}t}} = - 2{v_B} \Rightarrow \int {\frac{1}{{{v_B}}}{\text{d}}{v_B} = \int { - 2{\text{d}}t} }     (M1)

\ln \left| {{v_B}} \right| = - 2t + c     (A1)

{v_B} = A{e^{ - 2t}}     (M1)

{v_B} = - 20 when t = 0 so {v_B} = - 20{e^{ - 2t}}     A1

[4 marks]

e.

{x_B} = 10{e^{ - 2t}}( + c)     (M1)(A1)

{x_B} = 20{\text{ when }}t = 0{\text{ so }}{x_B} = 10{e^{ - 2t}} + 10     (M1)A1

meet when \frac{1}{4}{t^4} - \frac{5}{3}{t^3} + 3{t^2} = 10{e^{ - 2t}} + 10     (M1)

t = 4.41(290 \ldots )     A1

[6 marks]

f.

Examiners report

Part (a) was generally well done, although correct accuracy was often a problem.

a.

Parts (b) and (c) were also generally quite well done.

b.

Parts (b) and (c) were also generally quite well done.

c.

A variety of approaches were seen in part (d) and many candidates were able to obtain at least 2 out of 3. A number missed to consider the +c , thereby losing the last mark.

d.

Surprisingly few candidates were able to solve part (e) correctly. Very few could recognise the easy variable separable differential equation. As a consequence part (f) was frequently left.

e.

Surprisingly few candidates were able to solve part (e) correctly. Very few could recognise the easy variable separable differential equation. As a consequence part (f) was frequently left.

f.

Syllabus sections

Topic 2 - Core: Functions and equations » 2.2 » The graph of a function; its equation y = f\left( x \right) .
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