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During a certain time interval, the angular position of a swinging door is described by theta = 4.95 + 10.8t + 2.01t², where theta is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door at the following times.

a)t=0
b)t=2.94s

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Final answer:

At t = 0, the swinging door's angular position is 4.95 radians, angular speed is 10.8 rad/s, and angular acceleration is 4.02 rad/s². At t = 2.94s, the angular position is approximately 54.036 radians, angular speed is approximately 22.7692 rad/s, and angular acceleration remains at 4.02 rad/s².

Step-by-step explanation:

The given problem concerns the motion of a swinging door, which is a classical mechanics problem in physics. The door's motion is described by an angular position equation θ(t) = 4.95 + 10.8t + 2.01t², where θ is in radians and t is in seconds.

a) t=0

To find the angular position at t=0, we simply substitute t=0 into the equation:

θ(0) = 4.95 + 10.8(0) + 2.01(0)²
θ(0) = 4.95 radians

To determine the angular speed, we take the first derivative of the angular position function:

ω(t) = dθ/dt = 10.8 + 2(2.01)t
ω(0) = 10.8 rad/s

For the angular acceleration, we take the second derivative:

α(t) = dω/dt = 2(2.01)
α = 4.02 rad/s²

b) t=2.94s

Repeating the process for t=2.94s:

θ(2.94) = 4.95 + 10.8(2.94) + 2.01(2.94)²

Using a calculator, we can find:

θ(2.94) = 4.95 + 31.752 + 17.3339
θ(2.94) ≈ 54.036 radians

The angular speed at t=2.94s is:

ω(2.94) = 10.8 + 2(2.01)(2.94)
ω(2.94) ≈ 22.7692 rad/s

Angular acceleration remains unchanged as it is constant:

α = 4.02 rad/s²

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