Answer:
(a) α = -0.16 rad/s²
(b) t = 33.2 s
Explanation:
(a)
Applying 3rd equation of motion on the circular motion of the tire:
2αθ = ωf² - ωi²
where,
α = angular acceleration = ?
ωf = final angular velocity = 0 rad/s (tire finally stops)
ωi = initial angular velocity = 5.45 rad/s
θ = Angular Displacement = (14.4 rev)(2π rad/1 rev) = 28.8π rad
Therefore,
2(α)(28.8π rad) = (0 rad/s)² - (5.45 rad/s)²
α = -(29.7 rad²/s²)/(57.6π rad)
<u>α = -0.16 rad/s²</u>
<u>Negative sign shows deceleration</u>
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(b)
Now, we apply 1st equation of motion:
ωf = ωi + αt
0 rad/s = 5.45 rad/s + (-0.16 rad/s²)t
t = (5.45 rad/s)/(0.16 rad/s²)
<u>t = 33.2 s</u>
Answer:
<u>Foot per second. Foot-pound-second system. Frames per second, the frequency (rate) at which consecutive images (frames) appear on a display.</u>
Explanation:
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