A car goes from point A to point B exactly in three hours with $x(t)={59\over 3}t^2-t^3$ .Find the average velocity A car goes from point A to point B exactly in three hours.We assume that the car is moving in a straight line and that the distance of $x(t)$ from A at time $t$ ( where time is measured in hours and the distance in kilometers ) is given by the function $x(t)={59\over 3}t^2-t^3$.

$i)$what is the average velocity on this route? 

$ii)$what is the instantaneous velocity of the car at the time of reaching to point B ($t=3$)

Could anyone give me any ideas ir hints for these two questions?
i) Average velocity is simply $$\frac{\Delta x}{\Delta t}=\frac{x(3)-x(0)}{3-0}$$

ii) The instantaneous velocity $v(t)$ is the derivative of $x(t)$ w.r.t. to $t$. Then the instantaneous velocity at point B (where $t=3$) is just $v(3)$.
First of
$$v(t)=\frac{dx(t)}{dt}$$
Which means in your case:
$$v(t)=\frac{118}{3}t -3t^2$$

I am assuming x(t) is given in units of $[x]=m$ and that means $v(t)$ will be in $[v]=\frac{m}{sec}$. If you want to find $v(3_{[hours]})$ you have to translate $t$ to units of $[t]=sec$ and then assert $t=\text{Translated Value}=t'$.

For the Average speed:
$$v_{Average}=\frac{1}{t_1-t_0}\int_{t_0}^{t_1} v(t)dt$$

In this case:
$$v_{Average}=\frac{1}{t'-0}\int_{0}^{t'} v(t)dt$$
$$\int v(t)dt=x(t)+c \implies v_{Average}=\frac{1}{t'-0}\cdot \left[x(t')-x(0)\right]$$