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        a=15m/s^2 represent total acceleration of a particle moving in the clockwise direction in a circle of radius R=2.5m at a given instant of time the speed of particle is
3 years ago

Piyush Kumar Behera
435 Points
							see this acceleration is the centripetal acceleration as the particle is moving in the constant velocity .so a=v2/rv=$\sqrt{ar}$so speed of the particle is $\sqrt{2.5*15}$=6.1237243569579452454932101867647m/shope it helps please approve!!

3 years ago
Shefali
11 Points
							The acceleration given in question is the resultant of tangential and centripetal acceleration.So first we will calculate the centripetal acceleration. Here,Centripetal acceleration =a cos 30°                                   =a √3/2=15*√3/2We know,                Centripetal acceleration=v*v/r                           15*√3/2=v*v/2.5                            15*√3/2*2.5=v*v                             31.875= v*v                             √31.875=v                            5.7=v

2 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions