A Year 12 exam features the following problem:
The acceleration of a body moving in a straight line is given by $a = 4v^2.$ We're given that $v = e$ when $x=1$.
The problem is to find the velocity of the body when $x = 2$.
I can't solve it.
We can get from $a=4v^2$ to $v = \frac{1}{A-4t}$ pretty easily, where $A$ is an unknown constant. What now?
Using what you already have, $v(t) = \frac{1}{A-4t}$, we integrate this expression to arrive at $$x(t) = \int\frac{dt}{A-4t} = -\frac{1}{4}\log|A-4t| + B$$ We manipulate this to arrive at an expression for $t(x)$ to plug into $v(t)$ to obtain a relation $v(x)$ which we can then evaluate. After some manipulation, we arrive at
$$t(x)=\frac{A\pm e^{4(B-x)}}{4}$$
Plugging this into $v(t) =\frac{1}{A-4t}$ leads to $A$ cancelling out and we obtain:
$$v(x) =\pm\frac{1}{e^{4(B-x)}} = \pm e^{4(x-B)}$$
With our condition that $v(1)=e$, we have:
$$e = e^{4-4B} \Rightarrow 1 = 4-4B\Rightarrow B=\frac{3}{4}$$
where we have dropped the negative case because $e$ is positive and $e^x$ is always positive as well.
Thus, the complete expression is $$v(x) = e^{4x-3}$$
We plug in $x=2$ and obtain $v(2) = e^5$.