## Fundamentals of Physics Extended (10th Edition)

$t = \sqrt (4/3)$ and t = $-\sqrt(4/3)$
The position is given by $x = 20t - 5t^3$ To find the velocity, we need to take the derivative of the position equation or $\frac{d}{dt} (20t - 5t^3)$ The position is given by $x = 20t - 5t^3$ Since the velocity equation is the derivative of the position equation we need to find $\frac{d}{dt} (20t - 5t^3)$ Since $\frac{d}{dt} t^n = n*t^(n-1)$ then $v = 20 - 5(3)t^2 = 20 - 15t^2$ To find when the velocity is 0, we plug in 0 for v and solve for t This gives us $0 = 20 - 15t^2$ so then $20 = 15t^2$ or $(4/3) = t^2$ Taking the square root of both sides we find that $t = \sqrt (4/3)$ and $-\sqrt(4/3)$