If the velocity of a particle is given by $v=\sqrt {180-16x} $ m/s, its acceleration is:

(A) variable
(B) 8 m/s²
(C) -8 m/s²
(D) 0

Solution

$a=\frac {dv}{dt}=\frac {dv}{dx}.\frac {dx}{dt} =v\frac {dv}{dx} $

Now, $v^2 = 180-16x$

$\therefore 2v\frac {dv}{dx}=0-16$

$\therefore v\frac {dv}{dx}=a=-8$ m/s²

Hence, (C)