Question 1154279
x=-4t^3+3t^2+60
t=0 initial distance is x=60
acceleration is 2nd derivative or x''=-24t+6, and x''(0)=6 m/sec^2

average speed is velocity (2)-velocity(1)/(2-1)
velocity is first derivative of position or -12t+6t+60 
at t=1 it is 54
at t=2 it is 48
(48-64)/1=-16m/sec

{{{graph(300,300,-10,10,-100,100,-4x^3+3x^2+60,-12x^2+6x,-24x+6)}}}