Gujarati
Hindi
3-2.Motion in Plane
normal

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 :

A

$25$

B

$50$

C

$10$

D

None

Solution

$v_{x}=\frac{d x}{d t}=50 \cos 10 t$

$v_{y}=\frac{d y}{d t}=-50 \sin 10 t$

Speed $=\sqrt{v_{x}^{2}+v_{y}^{2}}$

$=\sqrt{(50 \cos 10 t)^{2}+(-50 \sin 10 t)^{2}}=50$

Standard 11
Physics

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