**Toolbox:**

- If $s=f(t)$ is the distance function, representing the distance 's' travelled by a particle in time t, then the velocity and acceleration functions are $v=\large\frac{ds}{dt}$$=t'(t)$ and $a=\large\frac{d^2s}{dt^2}$$=f''(t)$

$x=3 \cos (2t-4)\; $-----(i) $mass=1$

Step 1:

Acceleration is given by $\large\frac{d^2x}{dt^2}$

Step 2:

Differentiating (i) twice, we get

$\large\frac{dx}{dt}$$=3(-2s \sin (2t-4))=-6 \sin (2t-4)$----(i)

$\large\frac{d^2x}{dt^2}$$=-6 \times 2 \cos (2t-4)=-12 \cos (2t-4)$----(ii)

Step 3:

When $t=2$

Acceleration $=-12\; \cos 0=-12$

The velocity after 2 sec is got by substituting t=2 in (i)

Velocity $v=-6 \sin 0=0$

Therefore the $K.E= \large\frac{1}{2}$$mv^2=0$