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"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.

bio
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1 Answers1

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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 $x$.

Abhijeet Vats
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  • I believe I did - I simply divided by v between the first and second line. – bio Aug 21 '22 at 14:16
  • Right, you can't quite do that when you're integrating by separating variables. That procedure tells you that you need to get "variables" of each kind on two different sides and then integrate. – Abhijeet Vats Aug 21 '22 at 14:17