For a particle undergoing rectilinear motion with uniform acceleration, the magnitude of displacement is one third the distance covered in some time interval. The magnitude of final velocity is less than magnitude of initial velocity for this time interval. Then the ratio of initial speed to the final speed for this thime interval is:

Answer 1

it evidently goes forward a distance 3d then back a distance 2d to end up a displacoutup to be negative.

deceleration phase out to 3 d
t1 = time out to 3 d
v = Vi + a t1
0 = Vi + a t1
so
t1 = -Vi/a
3 d = Vi t1 + (1/2) a t1^2
3d = -Vi^2/a + .5 Vi^2/a = -.5Vi^2/a
d a = -(1/6) Vi^2

return phase back from 3 d to one d
Vf = 0 + a t2
so
t2 = Vf/a
-2d = (1/2) a t2^2
-2d = .5 Vf^2/a
--------------------------------
algebra from here on :)
d a = -(1/4) Vf^2
(1/6) Vi^2 = (1/4)Vf^2

Answer 2

This ans is wrong