Yr 2000 HSC Q16 Help (1 Viewer)

Azzamaral · Jul 25, 2017

This question is legit so hard. I can't do it. Help please
https://drive.google.com/open?id=0B3seGBG_7ovASzBUN05pQjBTRUk

1729 · Jul 25, 2017

Azzamaral said:
This question is legit so hard. I can't do it. Help please
https://drive.google.com/open?id=0B3seGBG_7ovASzBUN05pQjBTRUk

$$\noindent \textbf{(i)} Since $h$ increases at a constant rate, then $h = at$, for some constant $a$. Also, $v = \frac{dx}{dt}$ so $\frac{dx}{dt} = \frac{A}{h} = \frac{A}{at}$. If we let $k = \frac{A}{a}$ where $k$, $A$ and $a$ are constants then $\frac{dx}{dt} = \frac{k}{t}$

$$\noindent \textbf{(ii)} $x = k\ln{t} + C$ upon integrating the result from part (i), for some constant $C$. Using the data in the question, \\ \begin{align*} \quad \text{For } t &= T + 2, \ x = 1 \\ \text{For } t &= T + 5.5, \ x = 2 \\ \text{For } t &= T, \ x = 0\end{align*} \\ We then form these equations, \\ \begin{align*} \quad 0 &= k\ln{T} + C \implies C = -k\ln{T} \\ 1 &= k\ln{(T+2)} + C \implies k = \frac{1}{\ln{(T+2)} - \ln{T}} \\ 2 &= k\ln{(T + 5.5)} + C \implies k = \frac{2}{\ln{(T+5.5)} - \ln{T}}\end{align*} \\ Therefore $\ln{\left(\frac{T+5.5}{T}\right)} = 2\ln{\left(\frac{T+2}{T}\right)}$. This means $\frac{T+5.5}{T} = \frac{(T+2)^2}{T^2}$. Solving for $T$ gives $T = 2\frac{2}{3}$. Therefore it began snowing $2\frac{2}{3}$ hours before 6 am. That is, at 3:20 am.$

Yr 2000 HSC Q16 Help (1 Viewer)

Azzamaral

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1729

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