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(B) Given the position coordinates as functions of time $t$:$x = 3 \cos 4t$$y = 3(1 - \sin 4t)$To find the velocity components,we differentiate the po

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$24$

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(B) Given the position coordinates as functions of time $t$:$x = 3 \cos 4t$$y = 3(1 - \sin 4t)$To find the velocity components,we differentiate the position with respect to time $t$:$v_x = \frac{dx}{dt} = 3(-4 \sin 4t) = -12 \sin 4t$$v_y = \frac{dy}{dt} = 3(0 - 4 \cos 4t) = -12 \cos 4t$The magnitude of the velocity $v$ is given by:$v = \sqrt{v_x^2 + v_y^2} = \sqrt{(-12 \sin 4t)^2 + (-12 \cos 4t)^2}$$v = \sqrt{144 \sin^2 4t + 144 \cos^2 4t} = \sqrt{144(\sin^2 4t + \cos^2 4t)}$Since $\sin^2 	heta + \cos^2 	heta = 1$,we get:$v = \sqrt{144} = 12 	ext{ m/s}$Since the speed is constant,the distance travelled in time $t = 2 	ext{ s}$ is:$	ext{Distance} = 	ext{speed} 	imes 	ext{time} = 12 	ext{ m/s} 	imes 2 	ext{ s} = 24 	ext{ m}$.

$A$ point moves in the $x-y$ plane according to the law $x = 3 \cos 4t$ and $y = 3(1 - \sin 4t)$. The distance travelled by the particle in $2 \text{ s}$ is...........$m$ (where $x$ and $y$ are in meters).

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