The co-ordinates of a moving particle at a ti | vedclass

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Question

$v_{x}=\frac{d x}{d t}=50 \cos 10 t$

$v_{y}=\frac{d y}{d t}=-50 \sin 10 t$

Speed $=\sqrt{v_{x}^{2}+v_{y}^{2}}$

$=\sqrt{(50 \cos 10 t)^{2}+(-5

Vedclass · Accepted Answer

$50$

Vedclass · Answer

$v_{x}=\frac{d x}{d t}=50 \cos 10 t$

$v_{y}=\frac{d y}{d t}=-50 \sin 10 t$

Speed $=\sqrt{v_{x}^{2}+v_{y}^{2}}$

$=\sqrt{(50 \cos 10 t)^{2}+(-50 \sin 10 t)^{2}}=50$

The co-ordinates of a moving particle at a time $t$, are give by, $x = 5 sin 10 t, y = 5 cos 10t$. The speed of the particle is :

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