(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.
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