Final Answer:
The position of the particle is given by x(t) = 2.0 cm sin(2.25πt), the maximum speed is 4.5 cm/s at t = 0.172 s, the maximum acceleration is 15.75 cm/s² at t = 0.091 s, and the total distance traveled between t = 0 and t = 1.0 s is 90.9 cm.
Step-by-step explanation:
The position of a particle in simple harmonic motion is given by x(t) = A sin(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase. The frequency and angular frequency are related by ω = 2πf. In this problem, the amplitude is 2.0 cm, the frequency is 1.50 Hz, and the initial position is the origin, so φ = 0. Thus, the position of the particle is given by x(t) = 2.0 cm sin(2.25πt). The maximum speed of the particle is given by the derivative of the position equation, v(t) = Aω cos(ωt + φ). Thus, the maximum speed of the particle is v(t) = 2.25π cm/s. To find the earliest time at which the particle has this speed, we must solve v(t) = 2.25π cm/s for t. This yields t = 0.172 s, which is the earliest time at which the particle has a speed of 4.5 cm/s.
The maximum acceleration of the particle is given by the derivative of the velocity equation, a(t) = -Aω² sin(ωt + φ). Thus, the maximum acceleration of the particle is a(t) = 15.75 cm/s². To find the earliest time at which the particle has this acceleration, we must solve a(t) = 15.75 cm/s² for t. This yields t = 0.091 s, which is the earliest time at which the particle has an acceleration of 15.75 cm/s².
The total distance traveled between t = 0 and t = 1.0 s can be determined by integrating the position equation from t = 0 to t = 1.0 s. Thus, the total distance traveled between t = 0 and t = 1.0 s is 90.9 cm.