Answer
The rate of change of the function $s(t)=-2.3t^2$ is equal to $-4.6t$ and $\displaystyle s'(t)=-6.9$
Work Step by Step
\[s(t)=-2.3t^2\]
The derivative of the function $s(t)=-2.3t^2$ is equal to
\[s'(t)=\lim_{h\rightarrow 0}\frac{s(t+h)-s(t)}{h}\]
\[\Rightarrow s'(t)=\lim_{h\rightarrow 0}\frac{-2.3(t+h)^2-[-2.3t^2]}{h}\]
\[\Rightarrow s'(t)=\lim_{h\rightarrow 0}\frac{-2.3(t^2+h^2+2th)-[-2.3t^2]}{h}\]
\[\Rightarrow s'(t)=\lim_{h\rightarrow 0}\frac{-2.3h^2-4.6th}{h}\]
\[\Rightarrow s'(t)=\lim_{h\rightarrow 0}\frac{(-2.3h-4.6t)h}{h}\]
\[\Rightarrow s'(t)=\lim_{h\rightarrow 0}(-2.3h-4.6t)\]
\[\Rightarrow s'(t)=-2.3(0)-4.6t=-4.6t\]
\[s'(t)=-4.6t\]
Substitute $t=1.5$
\[s'(t)=-4.6(1.5)=-6.9\]
Hence , The rate of change of the function $s(t)=-2.3t^2$ is equal to $-4.6t$ and $\displaystyle s'(t)=-6.9$
