Answer:
A. [tex]\gamma + 3\beta t^2 rad/s[/tex]
B. 0.4 rad/s
C. 1.3 rad/s
D. 0.7 rad/s
Explanation:
Part A. We can take the derivative of the motion function in order to get the angular velocity function with respect to time
[tex]\omega(t) = \theta^{'}(t) = (\gamma t + \beta t^3)^{'} = \gamma + 3\beta t^2 rad/s[/tex]
Part B. The initial value of angular velocity is when t = 0
[tex]\omega_0 = \omega(0) = \gamma + 3\beta 0^2 = \gamma = 0.4 rad/s[/tex]
Part C. The instantaneous value of the angular velocity at t = 5s
[tex]\omega(5) = \gamma + 3\beta 5^2 = \gamma + 75\beta = 0.4 + 75*0.012 = 1.3 rad/s[/tex]
Part D. The angle distance it travels during the first t = 5 seconds is
[tex]\theta(5) = \gamma * 5 + \beta * 5^3 = 5\gamma + 125\beta = 5*0.4 + 125*0.012 = 3.5rad[/tex]
So the average angular velocity during this 5 seconds would be the total angle traveled divided by the time
[tex]\omega_a = \theta(5) / t = 3.5 / 5 = 0.7 rad/s[/tex]
The correct answers to the given questions are:
To find the angular velocity, we would use the derivatives which is:
w(t) = θ ^i(t) = (yt + βt^3)^i = y + βt^2rad/s
The initial value of angular velocity when t = 0 is
w0= w(0) = y + 3β0^2 = y
=0.4 rad/s.
The instantaneous value of the angular velocity at t = 5s
w(5)= y + 33β5^2
= y + 75β
=0.4 + 75 x 0.012
=1.3 rad/s.
The angular distance it travels during the first t = 5 seconds is
θ(5)= y x 5 + β x 5^3
=3.5 rad/s.
Therefore, the average angular velocity during these 5 seconds would be the total angle traveled divided by the time
wα= θ(5)/t
= 3.5/5
=0.7 rad/s
Read more about angular velocity here:
https://brainly.com/question/6860269
