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2D Motion as Two 1D Motions

A point moves...

Question

A point moves in x-y plane according to the law x=4 sin 6t and y=4(1-cos 6t). The distance traversed by the particle in 4 seconds is (x and y are in meters)

96 m

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48 m

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24 m

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108 m

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Solution

The correct option is A
96 m

x=4 sin 6t; y=4(1-cos 6t)
$\frac{d x}{d t} = v_{x}$ =4.6 cos 6t; $\frac{d y}{d t} = v_{y}$ =4.6 sin 6t
Now speed = $| \to v | = \sqrt{v_{x}^{2} + v_{y}^{2}}$
$= \sqrt{(24 c o s 6 t)^{2} + (24 s i n 6 t)^{2}}$
Speed = 24 m/s
Oh wow!
Speed is constant through the journey
So now it's easy
I know speed = $\frac{d i s t a n c e}{t i m e}$
24= $\frac{d i s t a n c e}{4}$
Distance = 96 m

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A point moves in x-y plane according to the law x=4 sin 6t and y=4(1-cos 6t). The distance traversed by the particle in 4 seconds is (x and y are in meters)

The correct option is A 96 m x=4 sin 6t; y=4(1-cos 6t) dxdt=vx=4.6 cos 6t; dydt=vy=4.6 sin 6t Now speed = |→v|=√v2x+v2y =√(24 cos 6t)2+(24 sin 6t)2 Speed = 24 m/s Oh wow! Speed is constant through the journey So now it's easy I know speed = distancetime 24=distance4 Distance = 96 m