The coordinates of a moving particle at a tim | vedclass

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(B) The velocity components are obtained by differentiating the position coordinates with respect to time $t$.$v_{x} = \frac{dx}{dt} = \frac{d}{dt}(5

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(B) The velocity components are obtained by differentiating the position coordinates with respect to time $t$.$v_{x} = \frac{dx}{dt} = \frac{d}{dt}(5 \sin 10t) = 50 \cos 10t$$v_{y} = \frac{dy}{dt} = \frac{d}{dt}(5 \cos 10t) = -50 \sin 10t$The speed of the particle is the magnitude of the velocity vector:$	ext{Speed} = \sqrt{v_{x}^{2} + v_{y}^{2}}$$	ext{Speed} = \sqrt{(50 \cos 10t)^{2} + (-50 \sin 10t)^{2}}$$	ext{Speed} = \sqrt{2500 \cos^{2} 10t + 2500 \sin^{2} 10t}$$	ext{Speed} = \sqrt{2500(\cos^{2} 10t + \sin^{2} 10t)}$Since $\sin^{2} 	heta + \cos^{2} 	heta = 1$,we get:$	ext{Speed} = \sqrt{2500 	imes 1} = 50$

The coordinates of a moving particle at a time $t$ are given by $x = 5 \sin 10t$ and $y = 5 \cos 10t$. The speed of the particle is:

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