Answer :
Answer:
The angle through which the wheel turned is 947.7 rad.
Explanation:
initial angular velocity, [tex]\omega _i[/tex] = 33.3 rad/s
angular acceleration, α = 2.15 rad/s²
final angular velocity, [tex]\omega_f[/tex] = 72 rad/s
angle the wheel turned, θ = ?
The angle through which the wheel turned can be calculated by applying the following kinematic equation;
[tex]\omega_f^2 = \omega_i^2 + 2\alpha \theta\\\\\theta = \frac{\omega_f^2\ -\ \omega_i^2}{2\alpha } \\\\\theta = \frac{(72)^2\ -\ (33.3)^2}{2(2.15)}\\\\\theta = 947.7 \ rad[/tex]
Therefore, the angle through which the wheel turned is 947.7 rad.