If the velocity of a particle moving along a straight line with uniform acceleration, is given by $v=\left(\sqrt{196-16X}\right)m{s}^{-1},$ then its acceleration is (X is displacement of the particle)

Given,
 $v=\sqrt{196-16X}m/s$
Acceleration is given by
 $a=\frac{dv}{dt}=\frac{dX}{dt}\times \frac{dv}{dX}$
 $\Rightarrow a=v\frac{dv}{dX} \mathrm{...}\left(i\right)$
 $v=\sqrt{196-16X}$
$\Rightarrow v^2=196-16X$ 
Differentiate the above equation w.r.t X
 $2v\frac{dv}{dX}=-16$
 $v\frac{dv}{dX}=-8 \mathrm{....}\left(ii\right)$
Comparing Eq. (i) and (ii), a =$-8m/s^2$