Question 3 40 pts An object is attached to a spring that is stretched and released. The equation d = a (cos(bt)) models the distance, d, of the object in inches above or below the rest position as a function of time, t, in seconds. If a = 5 and b = 5 Approximately when will the object be 2 inches above the rest position? Round to the nearest hundredth, if necessary.

The object will be 2 inches above the rest position when at approximately 0.23 seconds

Step-by-step explanation:

The given parameters are;

The equation that models the distance, d, of the object in inches above or below the rest position = a×(cos(b·t))

If a = 5 and b = 5, we have;

d = 5 × (cos(5 × t))

When the object is approximately 2 inches above the ground, we have;

d = 2 = 5 × (cos(5 × t))

cos(5 × t) = 2/5

5 × t = cos⁻¹(2/5) ≈ 1.16

t =  (cos⁻¹(2/5))/5 ≈ 1.16/5

t ≈ 0.23 s

The object will be 2 inches above the rest position when t ≈ 0.23 s.

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step-by-step explanation: