(a) The position of the particle as a function of time is given by:
x(t) = A cos(2πft)
where A is the amplitude (2.00 cm), f is the frequency (1.50 Hz), and cos is the cosine function.
Substituting the given values, we get:
x(t) = 2.00 cos(3πt)
(b) The maximum speed of the particle occurs at the equilibrium position, where the displacement is zero. At this point, the velocity is maximum and is given by:
vmax = Aω
where ω is the angular frequency and is equal to 2πf. Substituting the given values, we get:
vmax = 2.00 × 2π × 1.50 = 18.85 cm/s
(c) The earliest time at which the particle has this speed is when it passes through the equilibrium position. This happens at t = 0, so the earliest time is t = 0.
(d) The maximum positive acceleration of the particle occurs at the ends of its motion, where the displacement is maximum. At these points, the acceleration is given by:
amax = Aω^2
Substituting the given values, we get:
amax = 2.00 × (2π × 1.50)^2 = 282.74 cm/s^2
(e) The earliest time at which the particle has this acceleration is when it reaches the maximum displacement. This happens at t = 1/4T, where T is the period of the motion. The period is given by:
T = 1/f = 2/3 s
So, t = 1/4T = 1/4 × 2/3 = 0.33 s
(f) The total distance traveled by the particle between t = 0 and t = 1.00 s is equal to one complete cycle of its motion. The distance traveled in one complete cycle is equal to four times the amplitude, or:
4A = 8.00 cm
Therefore, the total distance traveled is:
8.00