(a) Differentiate [tex]s(t)[/tex] to get the velocity.
[tex]s(t) = \cos(t^2 - 1)[/tex]
Use the chain rule.
[tex]u = t^2 - 1 \implies \dfrac{du}{dt} = 2t[/tex]
[tex]s(u) = \cos(u) \implies \dfrac{ds}{du} = -\sin(u)[/tex]
[tex]\implies \dfrac{ds}{dt} = \dfrac{ds}{du} \dfrac{du}{dt} = \boxed{-2t\sin(t^2-1)}[/tex]
(b) Evaluate the derivative from (a) at [tex]t=0[/tex].
[tex]\dfrac{ds}{dt}\bigg|_{t=0} = -2\cdot0\cdot\sin(0^2-1) = \boxed{0}[/tex]
(c) The object is stationary when the derivative is zero. This happens for
[tex]-2t \sin(t^2 - 1) = 0[/tex]
[tex]-2t = 0 \text{ or } \sin(t^2 - 1) = 0[/tex]
[tex]t = 0 \text{ or } t^2 - 1 = n\pi[/tex]
(where [tex]n[/tex] is any integer)
[tex]t = 0 \text{ or } t = \pm\sqrt{n\pi + 1}[/tex]
We omit the first case. In the second case, we must have [tex]n\ge0[/tex] for the square root to be defined. Then in the given interval, we have two solutions when [tex]n=\in\{0,1\}[/tex], so the times are
[tex]t_1 = \sqrt{1} = \boxed{1}[/tex]
[tex]t_2 = \boxed{\sqrt{\pi + 1}} \approx 2.035[/tex]
(d) A picture is worth a thousand words. See the attached plot. If you're looking for a verbal description, you can list as many features of the plot as are relevant, such as
• intercepts (solve [tex]s(t)=0[/tex] to find [tex]t[/tex]-intercepts and evaluate [tex]s(0)[/tex] to find the [tex]s[/tex]-intercept)
• intervals where [tex]s(t)[/tex] is increasing or decreasing (first derivative test)
• intervals where [tex]s(t)[/tex] is concave upward or concave downward (second derivative test; at the same time you can determine any local extrema of [tex]s(t)[/tex], which you can see in the plot agrees with the critical points found in (c))
