Let a be the acceleration.
Let t be the total time traveled.
Basic calculus tells us that the velocity at time t is a*t and the total distance traveled is a*t2/2.
So we know:
(1) 1=a*t2/2 and
(2) 0.75 = a*(t-3)2/2
Solving both sides for a we have :
(3) 2/t2 = 3/(2*(t-3)2)
4*(t-3)2 = 3*t2
t2 - 24t + 36 = 0
t = (24 +/- 4321/2)/2 (11) t = 6*31/2 + 12
Add 3 for the total distance of 12 + 6*31/2 =~ 22.39230485
Michael Shackleford, ASA - December 6, 2000