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Distance

A point moves...

Question

A point moves in $x - y$ plane according to the law $x = 4 sin 6 t$ and $y = 4 (1 - cos 6 t)$ . 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 D $96 m$

$x = 4 s i n 6 t$ (Given)

$v_{x} = \frac{d x}{d t} = 24 c o s 6 t$

$y = 4 (1 - c o s 6 t)$ (Given)

$v_{y} = \frac{d y}{d t} = 24 s i n 6 t$

The velocity $v = \sqrt{v_{x}^{2} + v_{y}^{2}}$

$v = 24 m / s$

The distance traversed by the particle in $4 s e c$

$s = v t = 24 \times 4$

$s = 96 m$

The correct option is A.

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A point moves in x−y plane according to the law x=4sin6t and y=4(1−cos6t). The distance traversed by the particle in 4 seconds is :(x and y are in meters)

The correct option is D 96 mx=4sin6t (Given)vx=dxdt=24cos6ty=4(1−cos6t) (Given)vy=dydt=24sin6tThe velocity v=√v2x+v2yv=24m/sThe distance traversed by the particle in 4secs=vt=24×4s=96mThe correct option is A.

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