A point moves in $x-y$ plane according to the law $x=3\cos 4t$ and $y=3\left(1-\sin 4t\right)$. Find the distance travelled by the particle in $2s$.(Where $x$ and $y$ are in meters)

Given:

$$\begin{gathered} x=3\cos 4t ; \\ y=3(1-\sin 4t) ; \\ t=2sec \end{gathered}$$ 

As we know velocity is given by:

$$\begin{gathered} v_x=\frac{dx}{dt}=-12\sin 4t ; \\ {v}_y=\frac{dy}{dt}=-12\cos 4t \end{gathered}$$

$$\begin{gathered} v=\sqrt{{v_x}^2+{v_y}^2} \\ =\sqrt{{\left(-12\sin 4t\right)}^2+{\left(-12\cos 4t\right)}^2} \\ =\sqrt{{\left(12\right)}^2\left[{\left(\sin 4t\right)}^2+{\left(\cos 4t\right)}^2\right]} \\ =12 \end{gathered}$$ 

Distance$=2\times 12=24 m$