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in the given figure, alpha=15m//s^(2) re...

in the given figure, `alpha=15m//s^(2)` represents the total accleration of a particle moving in the clockwise direction on a circle radius R = 2.5m aat a given of time The speed of the particle is

A

4.5m/s

B

5.0m/s

C

5.7m/s

D

6.2 m/s

Text Solution

Verified by Experts

The correct Answer is:
C
`Tantheta=(dv//dt)/(V^(2)//R)`
`Tan30^(@)=(1)/(sqrt(3))=(dV//dt)/(V^(2)//R)`
`(dV)/(dt)=(1)/(sqrt(3))xx(V^(2))/(R)`
`a=15=sqrt(((dV)/(dt))^(2)+((V^(2))/(R))^(2))` ltBrgt `15=sqrt((4)/(3)((V^(2))/(R))^(2))`
`15=(2)/(sqrt(3))((V^(2))/(R))=(2)/(sqrt(3))xx(V^(2))/(2.5)impliesV^(2)=(15xxsqrt(3)xx2.5)/(2)=32.476=5.7m//s`
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