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Date May 2016 Marks available 2 Reference code 16M.2.hl.TZ1.3
Level HL only Paper 2 Time zone TZ1
Command term Find Question number 3 Adapted from N/A

Question

The displacement, s, in metres, of a particle t seconds after it passes through the origin is given by the expression s=ln(2et), t.

Find an expression for the velocity, v, of the particle at time t.

[2]
a.

Find an expression for the acceleration, a, of the particle at time t.

[2]
b.

Find the acceleration of the particle at time t = 0.

[1]
c.

Markscheme

v - \frac{{{\text{d}}s}}{{{\text{d}}t}} = \frac{{{{\text{e}}^{ - t}}}}{{2 - {{\text{e}}^{ - t}}}}{\text{ }}\left( { = \frac{1}{{2{{\text{e}}^t} - 1}}{\text{ or }} - 1 + \frac{2}{{2 - {{\text{e}}^{ - t}}}}} \right)     M1A1

[2 marks]

a.

a = \frac{{{{\text{d}}^2}s}}{{{\text{d}}{t^2}}} = \frac{{ - {{\text{e}}^{ - t}}(2 - {{\text{e}}^{ - t}}{\text{)}} - {{\text{e}}^{ - t}} \times {{\text{e}}^{ - t}}}}{{{{(2 - {{\text{e}}^{ - t}})}^2}}}{\text{ }}\left( { = \frac{{ - 2{{\text{e}}^{ - t}}}}{{{{(2 - {{\text{e}}^{ - t}})}^2}}}} \right)     M1A1

Note:     If simplified in part (a) award (M1)A1 for a = \frac{{{{\text{d}}^2}s}}{{{\text{d}}{t^2}}} = \frac{{ - 2{{\text{e}}^t}}}{{{{(2{{\text{e}}^t} - 1)}^2}}}.

Note:     Award M1A1 for a =  - {{\text{e}}^{ - t}}{(2 - {{\text{e}}^{ - t}})^{ - 2}}({{\text{e}}^{ - t}}) - {{\text{e}}^{ - t}}{(2 - {{\text{e}}^{ - t}})^{ - 1}}.

[2 marks]

b.

a =  - 2{\text{ }}({\text{m}}{{\text{s}}^{ - 2}})     A1

[1 mark]

c.

Examiners report

Mostly well done. There were a few sign errors but most candidates were correctly applying the quotient or chain rules.

a.

Mostly well done. There were a few sign errors but most candidates were correctly applying the quotient or chain rules.

b.

Mostly well done. There were a few sign errors but most candidates were correctly applying the quotient or chain rules.

c.

Syllabus sections

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