Chapter 4 - Graphs of the Circular Functions - Section 4.5 Harmonic Motion - 4.5 Exercises - Page 185: 7b

$3.46$ units

To determine the position at $t=1.25$, we need to substitute $t=1.25$ into the equation and solve: $s(t)=-4\cos\frac{2\pi}{3}t$ $s(t)=-4\cos(\frac{2\pi}{3}\times1.25)$ $s(t)=-4\cos(\frac{2\pi}{3}\times1.25)$ $s(t)=-4\cos(2.618)$ $s(t)=-4(-\frac{\sqrt 3}{2})$ $s(t)=2\sqrt 3=3.46$ units