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A point with deacceleration along the circle of radius R so that at any moment of time its tangential and normal acceleration are equal in magnitude. at the initial moment t=0, the velocity of the point is V0 . The velocity of point will be:(S is the distance travelled)
Dear Sanket,
Ans:- The eq of motion is - dv/dt=v²/R
Solving we get,1/v - 1/vº=t/R ( where vº is the initial velocity)
Or v=vº/(1+vºt/R)
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SOUMYAJIT IIT_KHARAGPUR
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