Answer
$s(1)=-2.5$
Therefore, at t=1, the weight is 2.5 inches below the equilibrium point. at the same position it was at t=0. Since the weight is already past half the oscillation point, it means that it is moving upwards.
Work Step by Step
First, lets find $s(1)$ by substituting $t=1$ in the equation:
$s(t)=5\cos (\frac{4\pi}{3} t)$
$s(1)=5\cos (\frac{4\pi}{3}\times 1)$
$s(1)=5\cos (\frac{4\pi}{3})$
$s(1)=5(-0.5)$
$s(1)=-2.5$
Therefore, at t=1, the weight is 2.5 inches below the equilibrium point. at the same position it was at t=0. Since the weight is already past half the oscillation point, it means that it is moving upwards.
