Time at a certain reference line

[bearhug]

At t=0, a flywheel has an angular velocity of 4.7 rad/s, a constant angular acceleration of -0.25 rad/s^2, and a reference line at θ(t=0)=0 rad.
Assuming the motion proceeded similary at times before t=0, at what negative time was the reference line at θ=-10.5 rad?

Based on this question I'm assuming that acceleration is still -0.25 rad/s^2 and angular velocity is still 4.7 rad/s.

Θf = θi + ωit + 1/2αt^2 is the equation I was planning on using to find time. However I'm confused as to what I should consider the final radians and final velocity. Should it be at θ=0 with final velocity at 4.7 rad/s?

 

[OlderDan]

At t=0, a flywheel has an angular velocity of 4.7 rad/s, a constant angular acceleration of -0.25 rad/s^2, and a reference line at θ(t=0)=0 rad.
Assuming the motion proceeded similary at times before t=0, at what negative time was the reference line at θ=-10.5 rad?

Based on this question I'm assuming that acceleration is still -0.25 rad/s^2 and angular velocity is still 4.7 rad/s.

Θf = θi + ωit + 1/2αt^2 is the equation I was planning on using to find time. However I'm confused as to what I should consider the final radians and final velocity. Should it be at θ=0 with final velocity at 4.7 rad/s?

When you write
Θ = θi + ωit + 1/2αt^2
you are effectively saying that α is a constant and that
ω = ωi + αt
So
θi is the value of Θ when t = 0
ωi is the value of ω when t = 0
Putting those values into
Θ = θi + ωit + 1/2αt^2
allows you to find Θ at any other time (positve or negative)