2022HSCShort answer

Q12 (c) (i)

1 mark

Question

A particle with mass $1\,\mathrm{kg}$ is moving along the $x$ -axis. Initially, the particle is at the origin and has speed $u\,\mathrm{m\,s^{-1}}$ to the right. The particle experiences a resistive force of magnitude $v+3v^2$ newtons, where $v\,\mathrm{m\,s^{-1}}$ is the speed of the particle after $t$ seconds. The particle is never at rest.

Show that
$\frac{dv}{dx}=-(1+3v).$

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	1

Sample answer:
$F=ma=1\times a=a$
$a=-(v+3v^2)$
$v\frac{dv}{dx}=-v(1+3v)\quad\text{and }v\ne0$
$\frac{dv}{dx}=-(1+3v)$

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Syllabus alignment

Mechanics

Resisted motionDerive and solve motion models with resistance on horizontal and vertical paths, including terminal velocity.

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