Your 64-cm-diameter car tire is rotating at 3.5 rev/s when suddenly you press down hard on the accelerator. After traveling 200 m, the tire’s rotation has increased to 6.0 rev/s. What was the tire’s angular acceleration? Give your answer in rad/s2.

angular acceleration = 0.748rad/s²

Explanation: according to the question, our answer needs to be in rad/s², thus all units in rev/s will be converted to rad/s

Assuming the motion of the object is of a constant angular acceleration, then newton's laws of motion is applicable.

The formulae below is used

v² = u² + 2αθ

v = final angular speed =6rev/s = 6*2π = 12π rad/s

u =initial angular speed =3.5rev/s = 3.5 *2π = 7π rad/s

Note 1 rev = 2π rad.

α = angular acceleration.

θ = angular displacement.

Diameter = 64cm = 0.64m, radius = 64/2 = 32cm = 0.32m

The angular displacement can be gotten using the formulae below

S = rθ, where s= linear distance covered = 200m, r = radius = 0.32m

θ = S/r = 200/0.32=625 rad.

By substituting the parameter we have that

(12π)² = (7π)² + 2α(625)

1421.22 = 486.31 + 1250α

1421.22 - 486.31 = 1250α

934.91 = 1250α

α = 934.91/1250

α= 0.748 rad/s²

wavelength variability is the right one

the answer would be c