Answer:
The velocity is [tex]v(-5)=110[/tex].
Explanation:
We have the position as a function of time: [tex]f (t) = -11t^{2}[/tex]
Instantaneous velocity at [tex]t=t_{0}[/tex] is defined as:
[tex]v(t_{0})= \lim_{\triangle t \to 0} \frac{f(t_{0}+\triangle t)-f(t_{0}) }{\triangle t}[/tex]
If [tex]t_{0}=-5[/tex],
[tex]v(-5)= \lim_{\triangle t \to 0} \frac{f(-5+\triangle t)-f(-5) }{\triangle t}=\lim_{\triangle t \to 0} \frac{11(-5+\triangle t)^{2} -11(-5)^{2} }{\triangle t}[/tex]
[tex]=\lim_{\triangle t \to 0} \frac{11(25+10\triangle t+\triangle t^{2}) -275 }{\triangle t}=\lim_{\triangle t \to 0} \frac{110\triangle t+11\triangle t^{2}}{\triangle t}[/tex]
[tex]\lim_{\triangle t \to 0} \frac{(110+11\triangle t)\triangle t}{\triangle t}=\lim_{\triangle t \to 0} 110+11\triangle t=110[/tex]
So,
[tex]v(-5)=110[/tex].
