Answer:
(c) 2.034 s; (d) 8.944 cm
Explanation:
Velocity and acceleration
s = 8cos(t) + 4sin(t)
v = -8sin(t) + 4cos(t)
a = -8cos(t) + 4sin(t)
(c) Time to first equilibrium position
The equilibrium position is where the mass hangs before it is pulled downward, that is, at s = 0.
Set s = 0 and solve for t.

If n = 1,
t = -1.107 + π = 2.034 s
(d) Distance from equilibrium position
The mass will reach its maximum distance when v = 0, that is, when it is at the peak or trough of its oscillation.
Set v = 0 and solve for t.

If n = 0,
t = 0.4636
Then
s = 8cos(0.4636) + 4sin(0.4636) = 8×0.8944 + 4×0.4472 = 7.156 + 1.789 = 8.944 cm
The figure below shows the graphs of s and v vs t. They indicate that the mass first reaches its equilibrium position at 2.034 s, and the amplitude of its vibration is 8.944 cm.