(A) We are given the velocity function `v(t)=9t^3+4` and we are asked to find the average velocity over the interval `1<=t<=4` .

The average value of a function f(t) on the interval [a,b] is given by:

`1/(b-a)int_a^bf(t)dt`

`A=1/(4-1)int_1^4(9t^3+1)dt`

`=1/3[9/4t^4+t|_1^4]`

`=1/3[(576+4)-(9/4+1)]`

`=1/3(2307/4)`

`=769/4=192.25`

So the average velocity is 192.25 units per unit time.

(B) Given the position function `s(t)=3t^2+t` we are asked to find the average velocity of the particle over the interval `1<=t<=4`

Using algebra, the average velocity is the displacement divided by the elapsed time.

s(1)=4 and s(4)=52 so the displacement is 48. The elapsed time is 4-1=3.

The average velocity is `A=48/3=16` units per unit time.

**Further Reading**