2020HSCShort answer

Q14 (b)

4 marks

Question

A particle starts from rest and falls through a resisting medium so that its acceleration, in $\mathrm{m/s^2}$ , is modelled by
$a=10\bigl(1-(kv)^2\bigr),$
where $v$ is the velocity of the particle in $\mathrm{m/s}$ and $k=0.01$ . Find the velocity of the particle after $5$ seconds.

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	4
Integrates to find $v$ as a function of $t$ , evaluating the constant, or equivalent merit	3
Correctly applies partial fractions and attempts to integrate, or equivalent merit	2
Correctly separates variables and attempts to apply partial fractions, or equivalent merit	1

Sample answer:

\frac{dv}{dt}=10(1-(kv)^2)

t=\frac1{10}\int\frac{1}{(1-kv)(1+kv)}\,dv

=\frac1{20}\int\left(\frac1{1-kv}+\frac1{1+kv}\right)dv\quad\text{(partial fractions)}

=\frac1{20}\times\frac1k[\ln(1+kv)-\ln(1-kv)]+C

=\frac1{20k}\ln\left(\frac{1+kv}{1-kv}\right)+C

When $t=0$ , $v=0$ , so $C=0$ .

When $t=5$ ,

\ln\left(\frac{1+kv}{1-kv}\right)=5\times20\times0.01=1

1+0.01v=e(1-0.01v)

v=100\left(\frac{e-1}{e+1}\right)\approx46.2\text{ m/s}

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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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