a. 21-12t^2=0 or t= sqrt21/12 which is not 6.5 seconds
b. rework a.
displacement=INT (21-12t^2) dt from o to T above
A)At what time does the motor reverse direction?
I got 6.5s
B) Through what angle does the motor turn between t =0 s and the instant at which it reverses direction?
Can anyone do B
b. rework a.
displacement=INT (21-12t^2) dt from o to T above
The angle turned by the motor can be found by calculating the integral of the angular velocity with respect to time:
θ = ∫(ω) dt
Where ω represents the angular velocity of the motor.
Given that the angular velocity is (21 - 12t^2) rad/s, we can plug this into the integral:
θ = ∫(21 - 12t^2) dt
The integral of (21 - 12t^2) with respect to t will give us the angle turned by the motor.
Evaluating this integral will give us the answer to part B.
The given angular velocity function is (21 - 12t^2) rad/s.
Let's integrate this function with respect to time from t = 0 to the time when the motor reverses direction.
∫[(21 - 12t^2) dt] from 0 to t = t_reversal
To solve this integral, we need to find the antiderivative of the function (21 - 12t^2).
The antiderivative of (21 - 12t^2) with respect to t is found by performing power rule integration:
(21t - 4t^3/3) + C,
where C is a constant of integration.
Now, let's evaluate the definite integral using the limits of integration:
[(21t_reversal - 4t_reversal^3/3) + C] - [(21(0) - 4(0)^3/3) + C]
Simplifying further:
(21t_reversal - 4t_reversal^3/3) - 0 - (0) + C
Cancelling out the zeros:
(21t_reversal - 4t_reversal^3/3) + C.
Therefore, the angle through which the motor turns between t = 0 s and the instant at which it reverses direction is (21t_reversal - 4t_reversal^3/3) + C.