A particle moves along a circular path having a radius of 2.0 m. At an instant when the speed of the particle is equal to 3.0 m/s and changing at the rate of 5.0 m/s², what is the magnitude of the total acceleration of the particle? A) 7.5 m/s² B) 6.0 m/s² C) 5.4 m/s² D) 6.7 m/s² E) 4.5 m/s²

Question

Quizplus · Accepted Answer

The total acceleration of the particle consists of two components: centripetal acceleration and tangential acceleration. 
The centripetal acceleration is given by ac = v^2/r, where v is the speed of the particle and r is the radius of the circular path. At the instant when the speed of the particle is 3.0 m/s, the centripetal acceleration is ac = (3.0 m/s)^2/2.0 m = 4.5 m/s^2.
The tangential acceleration is given by at = dv/dt, where v is the speed of the particle and t is time. Since the speed of the particle is changing at a constant rate of 5.0 m/s^2, the tangential acceleration is at = 5.0 m/s^2.
The total acceleration of the particle is given by the vector sum of the centripetal and tangential accelerations: a = (ac^2 + at^2)^0.5 = (4.5 m/s^2)^2 + (5.0 m/s^2)^2)^0.5 = 6.7 m/s^2 (rounded to one decimal place). Therefore, the answer is D.

A Particle Moves Along a Circular Path Having a Radius