All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$\int \frac{1}{100-v^2}dv = \frac{1}{20} \mathrm{ln}|\frac{10+v}{10-v}| + C$

(ii)
(a)
$\frac{dv}{dt} = 10 - 0.1 v^2$

$\int \frac{1}{100-v^2} dv = \int \frac{1}{10} dt$

$\frac{1}{20} \mathrm{ln} |\frac{10+v}{10-v}| = \frac{1}{10} t + C$

When $t = 0, v = 0 \Rightarrow c = 0$

$\therefore t = 0.5 \mathrm{ln} |\frac{10+v}{10-v}|$

When $v = 5, t = 0.5\mathrm{ln}3$

(b)
When $t = 1$

$\Rightarrow 2 = \mathrm{ln}|\frac{10+v}{10-v}|$

$\pm e^2 = \frac{10 + v}{10 - v}$ Reject $- e^2$ since $v = 0$ when $t = 0$

$e^2 (10 - v) = 10 + v$

$10 e^2 - 10 = v - e^2v$

$v = 7.62$

(c)

From the graph, As $t \rightarrow \infty, v \rightarrow 10$

(i) can be resolved easily with the MF15 formula.
(ii), students can actually plot the entire graph of t against v into the graphing calculator and calculate the v when t = 1. They should draw the graph out then. Should they do this, (c) will be solved conveniently.

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