2020HSCShort answer

Q11 (c)

3 marks

Question

A particle starts at the origin with velocity $1$ and acceleration given by
$a=v^2+v,$
where $v$ is the velocity of the particle. Find an expression for $x$ , the displacement of the particle, in terms of $v$ .

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	3
Correctly separates variables, or equivalent merit	2
Uses $a=v\frac{dv}{dx}$ , or equivalent merit	1

Sample answer:

a=v^2+v

\therefore v\times\frac{dv}{dx}=v^2+v

\therefore \frac{dx}{dv}=\frac1{v+1}

x=\ln(v+1)+C

When $x=0$ , $v=1\Rightarrow C=-\ln2$ .

\therefore x=\ln(v+1)-\ln2

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Syllabus alignment

Working MathematicallyMechanics

Modelling motion without resistanceApply forces, Newton's laws and calculus to model straight-line motion without resistance.

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