Through what angle has the wheel turned when its angular speed is 2.20 rad/s

On a certain game show, contestants spin a wheel when it is their turn. One contestant gives the wheel an initial angular speed of 3.40 rad/s. The angular acceleration of the wheel is -0.736 rad/s^2 .

We can use the following angular displacement equation:
[w(final)]^2 = [w(initial)]^2 + 2aθ.

Solving this for θ, the angular displacement, gives:
θ = {[w(final)^2] - [w(initial)]^2}/(2a).

Given that w(final) = 2.20 rad/s, w(initial) = 3.40 rad/s, and a = -0.736 rad/s^2:
θ = {[w(final)]^2 - [w(initial)]^2}/(2a)
= [(2.20 rad/s)^2 - (3.40 rad/s)^2]/[2(-0.736 rad/s^2)]
= 4.57 rad, to 3 s.f.

I hope this helps!

I am just going to use the variables for linear motion. Just assume I mean rotational.
▲x = vot +.5at²
3.4rad/s - .736rad/s²t = 0, when v=0 x stops changing
t = 4.62s
3.4rad/s(4.62s) - .5(.736rad/s²)(4.62s)² = ▲x
▲x = 7.85rad
450 degrees to put it into visual perspective. Just a little extra :)

ω = 2.2 Hz
ω₀ = 3.4 Hz
α = -0.736 Hz/s
Δθ = ?

Now use this equation:

ω² = ω₀² + 2αΔθ

Δθ = (ω² - ω₀²) / (2α)

Δθ = (2.2² - 3.4²) / (2(-.736)) = 4.565 rad.