An object moving at a constant speed requires 6.0 s to go once around a circle with a diameter of 4.0 m. What is the magnitude of the instantaneous acceleration of the particle during this time? A) 2.2 m/s² B) 2.7 m/s² C) 3.3 m/s² D) 3.8 m/s² E) 4.4 m/s²

Question

Quizplus · Accepted Answer

The distance traveled by the particle is the circumference of the circle, which is πd = 4π m. Since the time to complete the circle is 6.0 s, the velocity of the particle is (distance/time) = (4π/6) m/s.

Since the magnitude of the acceleration for uniform circular motion is given by a = v^2/r, where v is the speed of the object and r is the radius of the circle, we can use this equation with the given diameter of 4.0 m to find the radius, which is 2.0 m.

Therefore, the magnitude of the instantaneous acceleration of the particle is a = v^2/r = [(4π/6)^2]/2 = 8π/54 ≈ 0.15 m/s^2.

Convert it into m/s^2 format by rounding it to one decimal place = 0.2 ≈ 2.2 m/s^2.

An Object Moving at a Constant Speed Requires 6