(III) A particle revolves in a horizontal circle of radius 1.95 m. At a particular instant, its acceleration is 1.05m/s^2 in a direction that makes an angle of 25.0^circ to its direction of motion . Determine its speed (a) at this moment, and (b) 2.00 s later ,assumin g constant tangential acceleration.

To solve this problem, we need to use the relationship between centripetal acceleration, tangential acceleration, and the velocity of the particle.Given:- Radius of the circle, - Centripetal acceleration, - Tangential acceleration, - Angle, (a) To find the speed of the particle at the given instant, we can use the formula for centripetal acceleration: We also know that the total acceleration is given by: Since and , we can solve for : Substituting the given values: (b) To find the speed of the particle 2.00 s later, we need to consider the tangential acceleration. The tangential acceleration is given by: Substituting the given values: The tangential velocity after 2.00 s is given by: Substituting the given values: Therefore, the speed of the particle at the given instant is approximately , and the speed of the particle 2.00 s later is approximately .