Question

"During a certain time interval, the angular position of a swinging door is described by θ = 5.00 + 1.0t + 2.00t^2, where θ is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door at t = 0."

Answer 1

At t = 0, the angular position of the door is 5.00 radians, the angular speed is 1.0 rad/s, and the angular acceleration is 4.00 rad/s².

Step-by-step explanation:

To determine the angular position, angular speed, and angular acceleration of the door at t = 0, we can use the given equation θ = 5.00 + 1.0t + 2.00t2, where θ is in radians.

Angular position: To find the angular position at time t = 0, we substitute 0 for t in the equation, which gives us θ = 5.00 radians.

Angular speed: Angular speed is the first derivative of the angular position with respect to time. Taking the derivative of θ with respect to t gives us the angular speed ω = dθ/dt = 1.0 + 4.00t. At t = 0, the angular speed is ω = 1.0 rad/s.

Angular acceleration: Angular acceleration is the second derivative of the angular position with respect to time, which is the derivative of angular speed with respect to time. Taking the derivative of ω with respect to t gives us the angular acceleration α = dω/dt = 4.00 rad/s2. Therefore, the angular acceleration at any time, including t = 0, is a constant α = 4.00 rad/s2.