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A particle is moving in XY-plane such that $x = b (2 t + \sin 2 t), y = b (1 - \cos 2 t)$

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The acceleration of particle is constant

The acceleration of the particle is variable

The direction of motion of the particle makes $\frac{π}{4}$ angle with X-axis at $t = \frac{π}{4}$

The direction of motion of the particle make $\frac{π}{4}$ angle with X-axis at = 1s

answer is A, C.

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Detailed Solution

(1) $∵ x = b (2 t + \sin 2 t)$

or $v_{x} = \frac{dx}{dt} = 2 b + 2 bcos 2 t$

or $a_{x} = \frac{dv}{dt} = - 4 bsin 2 t$

$∵ y = b (1 - \cos 2 t)$

or $v_{y} = 2 bsin 2 t$

or $a_{y} = \frac{{dv}_{y}}{dt} = 4 bcos 2 t$

$∴ a = \sqrt{a_{x}^{2} + a_{y}^{2}} = 4 b$

(3) $\tan θ = \frac{v_{y}}{v_{x}} = \frac{2 bsin 2 t}{2 b + 2 bcos 2 t}$

$= \frac{\sin 2 t}{2 \cos^{2} t} = \frac{2 \sin t \cos t}{2 \cos^{2} t} = \tan t$

$At t = \frac{π}{4}, θ = \frac{π}{4}$

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