Answer
$s(1.3)\approx4$
After $t=1.3$ seconds, the weight is about 4 inches above the equilibrium position.
Work Step by Step
We calculate $s(1.3)$ by substituting $t=1.3$ into the equation and solving:
$s(t)=-5\cos 4\pi t$
$s(1.3)=-5\cos (4\pi\times1.3)$
$s(1.3)=-5\cos (16.34)$
$s(1.3)=-5(-0.81)$
$s(1.3)\approx4$
We know that the motion starts from $-5$ inches. Since the value of $s$ is positive, this means that the weight is moving upwards and has passed the equilibrium position. Therefore, after $t=1.3$ seconds, the weight is about 4 inches above the equilibrium position.
