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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

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

Grade:12th pass

3 Answers

Piyush Kumar Behera
417 Points
7 years ago
see this acceleration is the centripetal acceleration as the particle is moving in the constant velocity .
so a=v2/r
v=\sqrt{ar}
so speed of the particle is \sqrt{2.5*15}
=6.1237243569579452454932101867647m/s
hope it helps please approve!!
Shefali
11 Points
6 years ago
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
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

Given :
R=2.5 m
a=15 m/s2
centripetal acceleration
ac​ = acos30o
=15×0.866
=13 m/s^2
Using
ac​ = v^2/R​
∴13 = v^2/2.5​
⟹v=5.7 m/s
Thanks and Regards

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