Answer: [tex]\dfrac{16}{9}\ s[/tex]
Step-by-step explanation:
Given
[tex]s(t)=-t^3+2t^2+\frac{3}{2}[/tex]
Average velocity is given by
[tex]v_{avg}=\dfrac{\int vdt}{\int dt}[/tex]
[tex]v_{avg}=\dfrac{-\frac{t^4}{4}+\frac{2}{3}t^3+1.5t}{t}\\\\v_{avg}=-\frac{t^3}{4}+\frac{2}{3}t^2+1.5[/tex]
Now, equate the average and instantaneous velocity
[tex]-\frac{t^3}{4}+\frac{2}{3}t^2+1.5=-t^3+2t^2+1.5\\\\\Rightarrow -\dfrac{3t^3}{4}+\dfrac{4t^2}{3}=0\\\\\Rightarrow t^2\left(-\dfrac{3}{4}t+\dfrac{4}{3}\right)=0\\\\\Rightarrow t=\dfrac{16}{9}\ s[/tex]
