Let f(t) = 2 t 3 represent the position of a particle in feet at time t in seconds on the interval 0 ≤ t ≤ 8.

I'll  do  the first one  in (a) .....you should be able to  do the rest based on this one, GM

Average  velocity   =    [ 2(5)^3  - 2(2)^3 ] / [ 5 - 2]    =  [ 150 - 16 ] / 3   ≈  44.66  ft /s

For the second one....we   can use  the   result  of  the  V part of a

[ 2(5.001)^3  - 2(5)^3 ]  /  [ 5.001  - 5 ]   ≈  150 ft/s

{ BTW.....Calculus  can show that this is true.....the instantaneous velocity at 5 sec  =  150 ft / s   exactly }