A particle is moving in a circle of radius R = 1 m with constant speed v = 4 m/s. The ratio of the displacement to acceleration of the foot of the perpendicular drawn from the particle onto the diameter of the circle is

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2s²

$\frac{1}{16} s^{2}$

$\frac{1}{2} s^{2}$

16s²

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{s}{a} = \frac{s}{ω^{2} s} \Rightarrow [a = ω^{2} s] \Rightarrow \frac{s}{a} = \frac{1}{ω^{2}} \\ v = Rω \Rightarrow 4 = 1 ω \Rightarrow ω = 4 rad / \sec \\ \frac{s}{a} = \frac{1}{16} \end{array}$

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