The question I've been given is $$\begin{array}{c|c}t&v\\\hline3&38\\12&200\end{array}$$ Modelling equation is $$v=k\sqrt{t-a}$$ Calculate $a$ and $k$.
I tried to solve like:
$$38 = k \sqrt{12-a}$$ $$200 = k \sqrt{3-a}$$
$$162 = k \sqrt{12-a} - k \sqrt{3-a}$$
$$\frac{162}k = \sqrt{12-a} - \sqrt{3-a}$$
$$(162/k)^2 = 12-a - 3 + a$$
$$(162/k)^2 = 9$$
$$162/k = 3$$
$$162/3 = k$$
$$k = 54$$
WHICH IS GIVEN AS WRONG. $k$ should be $65.45$
On
First mistake:
When time is $3$, the volume is $38$.
When time is $12$, the volume is $200$.
You have made the mistake of thinking when time is $3$, the volume is $200$. Similary for the other equation.
Second mistake:
Guide:
On
Using your data, you have in fact $$38 = k \sqrt{3-a}\tag 1$$ $$200 = k \sqrt{12-a}\tag 2$$ Make the ratio $$\frac{38}{200}=\frac{\sqrt{3-a} } {\sqrt{12-a} }$$ Square both sides to get $$\frac{361}{10000}=\frac{3-a}{12-a}\implies a=\frac{2852}{1071}$$ Plug in $(1)$ to get $k=6 \sqrt{119}\approx 65.4523$.
It seems to me that in general this question would be easier if you were to square both sides:
$$ v^2=k^2(t-a) $$
With the two $(t, v)$ pairs of $(3, 38)$ and $(12, 200)$, we have
$$ 1444 = 3k^2-ak^2 \\ 40000 = 12k^2-ak^2 $$
Subtracting the former from the latter gives us
$$ 9k^2 = 38556 \\ k^2 = 4284 k = \sqrt{4284} \approx 65.452 $$
Then substituting $k^2 = 4284$ back into either of the above yields
$$ a = 12-\frac{40000}{4284} = 3-\frac{1444}{4284} \approx 2.663 $$
Assuming that this model is supposed to produce real numbers as output, the implication is that it cannot be used for $t < a$.