"A particle P starts from rest from the origin O and moves along the positive x-axis. Its displacement, in metres, from O is given by $x(t)$ where t is the time in seconds. The acceleration a of the particle is given by $a(t) = 25-16v^2$ where $v(t)$ is the velocity of the particle in metres per second. Find v in terms of x."
I decided to utilise the formula $a=v\frac{dv}{dx}$ to calculate this, which ultimately led me to $x=\frac{16v^2+25}{(16v^2-25)^2}+c$ however I cannot find this for v in terms of x. Could someone please provide a different view of the problem?
Thanks.
Are you sure you did the calculation correctly? What you should have is that: $$25-16v^2 = v \frac{dv}{dx}$$ Separating variables, we have that: $$\int \frac{v}{25-16v^2} \ dv = \int \ dx$$
On the left side, let $u = 25-16v^2$. Then, $du = -32v \ dv$. So: $$x = \int -\frac{du}{32u} = -\frac{1}{32} \log(|25-16v^2|) + C$$ In particular, it follows that: $$|25-16v^2| = A \exp(-32x)$$ With a bit more work, you should be able to find $v$ in terms of $x$.