Step-by-step explanation:
The particle is at rest when the derivative of x(t), which is its velocity, is zero, i.e.,
[tex]\dfrac{d}{dt}[x(t)] = \dot{x}(t) = 0[/tex]
Since [tex]x(t) = 2t^3 - 21t^2 + 72t - 53,[/tex] its derivative is
[tex]\dot{x}(t) = 6t^2 - 42t + 72[/tex]
Equating this to zero, we get
[tex]6t^2 - 42t + 72 = 0 \Rightarrow t^2 - 7t + 12 = 0[/tex]
This is a familiar quadratic equation whose roots are
[tex]t = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]\;\;\;= \dfrac{7 \pm \sqrt{(7)^2 - 4(1)(12)}}{2}[/tex]
[tex]\;\;\;= \dfrac{7 \pm 1}{2}[/tex]
[tex]\;\;\;= 3\;\text{and}\;4[/tex]
This means that the particle will be at rest at t = 3 and t= 4.
