Answer:
[tex]v(t_{0})=\frac{-200t_{0} }{(t_{0}^{2}+12)^{2}} ft/s[/tex]
Step-by-step explanation:
Let's remember we could write the speed as a function of position taking the derivative whit respect to time.
[tex] v(t)=\frac{ds}{dt}= \frac{-200t}{(t^{2}+12)^{2}} [/tex]
Now, evaluating the speed at t₀, we have:
[tex] v(t_{0})=\frac{-200t_{0}}{(t_{0}^{2}+12)^{2}} ft/s [/tex]
It would be the maximun speed at that time.
Have a nice day!
