In the given figure, a = 15 ,m s^{- 2} repres | vedclass

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Question

$\begin{gathered}
  \,\,\,\,\,Here,a = 15\,m{s^{ - 2}} \hfill \
  \,\,\,\,\,R = 2.5\,m \hfill \
  From\,figure, \hfill \

Vedclass · Accepted Answer

$5.7$

Vedclass · Answer

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

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 ........ $m/s$

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