You program a dot on a computer screen to have a position given by $\displaystyle \vec{r}=(1.5cm + 2.0cm/s^2t^2)\hat{i}+((3.0m/s)t)\hat{j}$. At time $\displaystyle t_{10} > 0$ the dot's displacement from its t = 0 position has a magnitude 10.0cm a.) Find the magnitude and direction of the average velocity of the dot between t = 0 and $\displaystyle t_{10}$. b.) find the magnitude and direction of the instantaneous velocity at t = 0 and at $\displaystyle t_{10}$

the answer is

a.) 5.0cm/s 37 degrees

b.) t = 0: 3.0cm/s, 90 degrees

t_{10} : 8.5cm/s,21 degrees

im stuck with the letter a what i did was substitute 10 as t and got 201.5 as V2 and i got 19.2cm/s as the average velocity

HELP