A door starts from rest and swings with an angular acceleration, alpha, α = 6 rad/s2, where t is in seconds. Determine the angle in radians through which the door turns in the first 3 s.
Hint: kinematics equation for angular position (i.e., angle) is exactly the same as for (linear) position from translational kinematics (x = x_i + v_i + 1/2at^2) with Greek letters instead of Latin letters: θ = θ_i + omega_i*t + 1/2alpha*t^2.
To determine the angle in radians through which the door turns in the first 3 seconds, we can use the kinematics equation for angular position:
θ = θ_i + ω_i*t + (1/2)α*t^2
where:
θ is the final angle in radians,
θ_i is the initial angle (assumed to be zero, since the door starts from rest),
ω_i is the initial angular velocity (also assumed to be zero),
α is the angular acceleration (given as 6 rad/s^2), and
t is the time (given as 3 seconds).
Plugging in the given values:
θ = 0 + 0*3 + (1/2)*(6)*(3^2)
= 0 + 0 + 9
= 9 radians
Therefore, the door turns through an angle of 9 radians in the first 3 seconds.