## Fundamentals of Physics Extended (10th Edition)

$a_{(3)}=-30\frac{m}{s^2}$
We first differentiate $v=\frac{dx}{dt}=6t-6t^2$ to find the expression for $a$: Therefore, $a=\frac{dv}{dt}=6-12t$ Now, we substitute $t=3$ in the expression to find the value of $a$ at $t=3s$: $a_{(3)}=6-12(3)=6-36=-30\frac{m}{s^2}$