16. The angular position of a swinging door is deseribed by 0(t) = 5+ 10t + 2t², where 0

Question

is in radians and t is in seconds. Determine the angular position, angular speed, and angular
acceleration of the door (a) at =0 and (b) at t = 3s.

Answer 1

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

θ(t=0) = 5 + 10(0) + 2(0)^2
θ(t=0) = 5 + 0 + 0
θ(t=0) = 5

Therefore, the angular position at t=0 is 5 radians.

To find the angular speed at t=0, we take the derivative of the equation with respect to time:

ω(t) = dθ/dt = 10 + 4t

Substituting t=0 into the angular speed equation:

ω(t=0) = 10 + 4(0)
ω(t=0) = 10 + 0
ω(t=0) = 10

Therefore, the angular speed at t=0 is 10 radians per second.

To find the angular acceleration at t=0, we take the second derivative of the equation with respect to time:

α(t) = d²θ/dt² = d/dt(10 + 4t) = 4

Therefore, the angular acceleration at t=0 is 4 radians per second squared.

To find the angular position at t=3s, we substitute t=3 into the equation:

θ(t=3) = 5 + 10(3) + 2(3)^2
θ(t=3) = 5 + 30 + 18
θ(t=3) = 53

Therefore, the angular position at t=3s is 53 radians.

To find the angular speed at t=3s, we substitute t=3 into the angular speed equation:

ω(t=3) = 10 + 4(3)
ω(t=3) = 10 + 12
ω(t=3) = 22

Therefore, the angular speed at t=3s is 22 radians per second.

To find the angular acceleration at t=3s, we substitute t=3 into the angular acceleration equation:

α(t=3) = 4

Therefore, the angular acceleration at t=3s is 4 radians per second squared.