Question #7563

The angular position of a swinging door is described by θ(t) = (5.00 + 2.00t²) rad. Determine the angular position, angular speed, and angular acceleration of the door at t = 3.0 s.

Expert's answer

Angular position of the door at t = 3.0 s:

θ(3) = (5 + 2*3²) = 23 rad;

Angular speed of the door:

θ'(t) = (5 + 2t²)' = 4t

Angular speed of the door at t = 3.0 s:

θ'(3) = 4*3 = 12 rad/s;

Angular acceleration of the door:

θ"(t) = (5 + 2t²)" = 4 rad/s² = const.

So, angular acceleration of the door at t = 3.0 s is 4 4 rad/s².

θ(3) = (5 + 2*3²) = 23 rad;

Angular speed of the door:

θ'(t) = (5 + 2t²)' = 4t

Angular speed of the door at t = 3.0 s:

θ'(3) = 4*3 = 12 rad/s;

Angular acceleration of the door:

θ"(t) = (5 + 2t²)" = 4 rad/s² = const.

So, angular acceleration of the door at t = 3.0 s is 4 4 rad/s².

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