At $ t = 2.00 \, \text{s} $, a particle in counterclockwise circular motion has a constant speed. What is the total acceleration if $ 6.00 \, \text{m/s}^2 $ is radial and $ 4.00 \, \text{m/s}^2 $ is tangential?

A) $ 10.00 \, \text{m/s}^2 $
B) $ 2.00 \, \text{m/s}^2 $
C) $ 8.00 \, \text{m/s}^2 $
D) $ 4.00 \, \text{m/s}^2 $

The total acceleration of a particle at t = 2.0 s with a radial acceleration of 6.00 m/s² and a tangential acceleration of 4.00 m/s² is found using the Pythagorean theorem, resulting in 7.21 m/s², which rounds to the nearest answer choice (c) 8.00 m/s².

To find the total acceleration of the particle in counterclockwise circular motion, we need to consider both the radial (centripetal) acceleration and the tangential acceleration.

Explanation:

The student is asking about the total acceleration of a particle in circular motion at a specific time where two components of acceleration are given: radial acceleration and tangential acceleration. The radial acceleration is 6.00 m/s² and the tangential acceleration is 4.00 m/s². Since these accelerations are perpendicular to each other, the total acceleration can be found using the Pythagorean theorem.

The formula for the total acceleration (a) is:

a = √(radial acceleration² + tangential acceleration²)

So, by substituting the given values:

a = √(6.00² + 4.00²)

a = √(36 + 16)

a = √52

a = 7.21 m/s²

So, the total acceleration of the particle is approximately

7.21

m/s

2

7.21m/s

2

However, the closest answer choice to this value is (c) 8.00 m/s².