The displacement ‘x’ (in m) of a particle at the instant when its velocity ‘v’ is given by $v=\sqrt{9x+25}m/s$ . Find its acceleration.

Given that  $v=\sqrt{9x+25}$
 $\Rightarrow v^2=9x+25$
 $\Rightarrow v^2-25=9x\mathrm{............}\left(i\right)$
$\therefore$ Comparing this equation with equation
 $v^2-u^2=2as\mathrm{............}\left(ii\right)$
We get  $2a=9$
$\Rightarrow a=4.5m/s^2$