In given figure, a = 15 ,m s^{- 2} represents total acceleration of a particle moving in clockw

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3-2.Motion in Plane

medium

In the given figure, $a = 15 \,m s^{- 2}$ represents the total acceleration of a particle moving in the clockwise direction in a circle of radius $R = 2.5\, m$ at a given instant of time. The speed of the particle is ........ $m/s$

A$5.7$

B$6.2$

C$4.5$

D$5.0$

(NEET-2016)

Solution

$\begin{gathered}
\,\,\,\,\,Here,a = 15\,m{s^{ – 2}} \hfill \\
\,\,\,\,\,R = 2.5\,m \hfill \\
From\,figure, \hfill \\
{a_c} = a\,\cos \,{30^ \circ } = 15 \times \frac{{\sqrt 3 }}{2}\,m{s^{ – 2}} \hfill \\
As\,we\,know,\,{a_c} = \frac{{{v^2}}}{R} \Rightarrow v = \sqrt {{a_c}R} \hfill \\
\therefore \,\,\,\,\,v = \sqrt {15 \times \frac{{\sqrt 3 }}{2} \times 2.5} = 5.69 = 5.7m\,{s^{ – 1}} \hfill \\
\end{gathered} $

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In the given figure, $a = 15 \,m s^{- 2}$ represents the total acceleration of a particle moving in the clockwise direction in a circle of radius $R = 2.5\, m$ at a given instant of time. The speed of the particle is ........ $m/s$

Solution

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