Answer:
[tex]-1.99\:\mathrm{m/s}[/tex]
Explanation:
Assuming that the equation is intended to be [tex]\displaystyle x=\cos\left(\frac{2\pi}{3}t\right)[/tex], we can find the velocity vs. time equation by taking the first derivative with respect to [tex]t[/tex]:
[tex]\displaystyle \frac{dx}{dt}=\frac{d}{dt}\left(\cos\left(\frac{2\pi}{3}t\right)\right)[/tex]
Recall the chain rule:
[tex]\displaystyle f(g(x))'=f'(g(x))\cdot g'(x)[/tex]
Therefore,
[tex]\displaystyle \frac{d}{dt}\left(\cos\left(\frac{2\pi}{3}t\right)\right)=-\sin\left(\frac{2\pi}{3}t\right)\cdot \frac{2\pi}{3}[/tex]
Therefore, the velocity vs. time equation of the object is [tex]\displaystyle v=-\sin\left(\frac{2\pi}{3}t\right)\cdot \frac{2\pi}{3}[/tex].
Substitute [tex]t=0.6\text{ s}[/tex] into this equation to find the velocity at that given time:
[tex]\displaystyle v=-\sin\left(\frac{2\pi}{3}(0.6)\right)\cdot \frac{2\pi}{3}\approx \boxed{-1.99\text{ m/s}}[/tex]
