A particle starts moving in a plane with constant tangential acceleration $a_{t} = 2 m / s^{2}$ and normal acceleration $a_{n} = t^{4}$ .Then choose the correct options(s)

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The acceleration of the particle is $2 \sqrt[]{5} m / s^{2}$ after having covered a distance S=2m

The acceleration of the particle is $2 \sqrt[]{2} m / s^{2}$ after having covered a distance S=2m

The radius of curvature of its trajectory is 4m after having covered a distance S=1m

The radius of curvature of its trajectory is 2m after having covered a distance S=1m

answer is A, C.

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Detailed Solution

$a_{t} = 2 m / s^{2} S = u t + \frac{1}{2} a_{1} t^{2} = 0 + \frac{1}{2} \times 2 t^{2} = t^{2} ∴ S = t^{2} v^{2} = u^{2} + 2 a_{1} S = 2 \times 2 S = 4 S ∴ a_{n} = t^{4} = S^{2} a_{n} = S^{2} ∴ a = \sqrt{a_{1}^{2} + a_{n}^{2}} = \sqrt{4 + S^{4}}$

At $S = 2 m$
$a = \sqrt{4 + 16} = 2 \sqrt{5} m / s^{2}$
Radius of curvature of its trajectory
$r = \frac{V^{2}}{a_{n}} = \frac{4 S}{S^{2}} = \frac{4}{S}$
At $S = 1 m, r = 4 m$
Use $g_{e f f} = g \sin 37 = \frac{3 g}{5}$ instead of g in all standard vertical circle cases