## Calculus (3rd Edition)

The particle passes through the origin at: $t = 0, 3\sqrt{2}\ seconds$. The particle is instantaneously motionless at $t = 0, 3\ seconds$
s(t) gives us the position, so to find when the particle passes through the origin, we have to find out when the function is equal to 0. So we set the equation to 0, and solve for t: $0 = t^4 - 18t^2$ Now we can factor: $t^2(t^2-18) = 0$ $t^2(t+3\sqrt{2})(t-3\sqrt{2}) = 0$ $t = 0, 3\sqrt{2}, -3\sqrt{2}$ The time cannot be negative, so $t = 0\ and\ 3\sqrt{2}\ seconds$ To find where the particle is motionless, we have to first find the equation for the velocity. To do so, we differentiate the position formula using the product rule: $v(t) = s'(t) = 4t^3 - 36t$ Now, we set this equal to 0 and solve for t: $4t^3 - 36t = 0$ $4t(t^2 - 9) = 0$ $t = 0, 3, -3$ Once again, time cannot be negative, so $t = 0\ and\ 3\ seconds$