Answer:
There are no positive times for which the object stops moving.
Step-by-step explanation:
We are given that the location of an object moving along the x-axis is given by the function:
[tex]\displaystyle f(t) = 10t + 5t^2 + 0.8t^3[/tex]
And we want to determine for which positive time(s) does the object stop moving.
If the object stops moving, its velocity will be zero.
Recall that the velocity is the first derivative of the position function. Find velocity:
[tex]\displaystyle \begin{aligned} f'(t) = v(t) & = \frac{d}{dt}\left[10t + 5t^2 + 0.8t^3\right] \\ \\ & = 10 + 10t + 2.4t^2 \end{aligned}[/tex]
Set the velocity equal to zero and solve for t:
[tex]\displaystyle 0 = 10 + 10t + 2.4t^2 \\ \\ \begin{aligned} t & = \frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ & = \frac{-10\pm\sqrt{4}}{4.8} \\ \\ & = \frac{-10+2}{4.8} \text{ or } -\frac{10+2}{4.8} \\ \\ & = -\frac{5}{3} \text{ or } -\frac{5}{2} \end{aligned}[/tex]
Since t ≥ 0, there are no times for which the object stops moving (i.e., the object is always in motion).
