Changing angular acceleration of a shaft

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The discussion focuses on calculating the angular speed of a shaft given its initial angular velocity of 65 rad/s and a time-dependent angular acceleration defined by the equation a = -10.0 rad/s² - 5.00t rad/s³. To find the angular speed at 3 seconds, participants emphasize the need to integrate the angular acceleration function and apply the constant of integration based on the initial conditions. The correct approach involves deriving the velocity function from the acceleration equation and solving for the constant using the initial speed.

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jmf322
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Hello, I'm stuck on this problem and I think its because I am lacking the calculus to get the correct answer.

A shaft is turning at 65 rad/s at time = 0. After, its angular acceleration is given by:

a= -10.0 rad/s^2 - 5.00t rad/s^3 where t is the elapsed time.

Find its angular speed at 3 seconds.

Now where I am stuck is I can find its acceleration at 3 seconds, but I believe I need to take the deritive of something in order to calculate the small sections of a/t. Am I on the right track? Any good links to explain how to take the deriitive of a angular acceleration? Thanks!
 
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if: a=-10-5t

then the velocity is the integral of this function, so, integrate this to find the velocity,

but remember your constant of integration! so you will have a function something like this,

V= - at - 1/2 bt^2 + C C is a constant of integration,

you already know what this value SHOULD be though, because they told you the speed at time 0 is 65. So you have to solve such that 65= - at - 1/2 bt^2 + C ,when time t is zero, this will tell you what C is.

Now you have your general equation, and you just plug in 3 for time to get your angular speed. But ill leave finding a and b to you. it should look similar though.
 
thanks for the help! Do you or anyone else know of any good calclus with physics applications help site? I understand the Physics in my class, but i struggle with the calculus and calculus web sites are very broad. Thanks again!
 

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