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Date May 2014 Marks available 6 Reference code 14M.1.hl.TZ1.8
Level HL only Paper 1 Time zone TZ1
Command term Find Question number 8 Adapted from N/A

Question

A body is moving in a straight line. When it is \(s\) metres from a fixed point O on the line its velocity, \(v\), is given by \(v =  - \frac{1}{{{s^2}}},{\text{ }}s > 0\).

Find the acceleration of the body when it is 50 cm from O.

Markscheme

\(\frac{{{\text{d}}v}}{{{\text{d}}s}} = 2{s^{ - 3}}\)     M1A1

 

Note:     Award M1 for \(2{s^{ - 3}}\) and A1 for the whole expression.

 

\(a = v\frac{{{\text{d}}v}}{{{\text{d}}s}}\)     (M1)

\(a =  - \frac{1}{{{s^2}}} \times \frac{2}{{{s^3}}}\left( { =  - \frac{2}{{{s^5}}}} \right)\)     (A1)

when \(s = \frac{1}{2},{\text{ }}a =  - \frac{2}{{{{(0.5)}^5}}}{\text{ }}( =  - 64){\text{ (m}}{{\text{s}}^{ - 2}})\)     M1A1

 

Note:     M1 is for the substitution of 0.5 into their equation for acceleration.

Award M1A0 if \(s = 50\) is substituted into the correct equation.

 

[6 marks]

Examiners report

[N/A]

Syllabus sections

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