Explanation:
The particle will be at rest when its velocity [tex]v(t)[/tex] is equal to zero. Recall that the velocity is simply the derivative of the position [tex]x(t)[/tex] with respect to time:
[tex]v(t) = \dfrac{dx}{dt}[/tex]
Since [tex]x(t) = t^2 - 2t + 2[/tex]
then
[tex]v(t) = \dfrac{d}{dt}(t^2 - 2t + 2) = 2t - 2 = 0[/tex]
Solving for t, we find that the particle will be at rest at
[tex]t = 1\:\text{s}[/tex]
