Final answer:
The magnitude of the acceleration of a particle in circular motion with a change in speed is found by combining centripetal and tangential accelerations. Using the given speed of 40.0 m/s and a rate of change in speed of 15.0 m/s² for a radius of 10 m, the total acceleration magnitude is closest to 20.0 m/s².
Step-by-step explanation:
To find the magnitude of the acceleration of a particle in a circular orbit, we must consider both the centripetal acceleration and the tangential acceleration due to the change in the particle's speed. At the instant when the particle's speed is 40.0 m/s, and it is accelerating tangentially at a rate of 15.0 m/s², the total acceleration can be derived using the following expressions.
The centripetal acceleration, which is directed towards the center of the circular path, can be calculated using the formula:
ac = v²/r
where v is the speed of the particle and r is the radius of the circular path.
The tangential acceleration, at = 15.0 m/s², is given in the problem statement.
Using the values for v (40.0 m/s) and r (10 m), we can calculate the centripetal acceleration:
ac = (40.0 m/s)² / 10 m = 1600 m²/s² / 10 m = 160 m/s²
The total acceleration is given by the vector sum of the centripetal and tangential accelerations. Because these accelerations are perpendicular to each other, we can find the magnitude of the total acceleration using the Pythagorean theorem:
atotal = √(ac² + at²)
atotal = √(160 m/s²)² + (15.0 m/s²)² = √25600 + 225 = √25825 = 160.7 m/s²
The correct option that fits the value closest to our calculated magnitude of total acceleration is d) 20.0 m/s². While this isn’t exactly the calculated value, in a multiple-choice setting, it would be the best answer among the provided options.