Explanation

(c) Given, $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = (180 - 16x)^{1/2}$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v^{2} = 180 - 16x$
Differentiating with respect to $t,$ we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2v\frac{dv}{dt} = 0 - 16\frac{dx}{dt}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2v\frac{dv}{dt} = - 16v$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{dv}{dt} = - 8$
Hence, particle decelerates at the rate of $8\ {ms}^{- 2}.$