The peripheral speed of the tooth of a 10-in.-diameter circular saw blade is 230 ft/s when the power to the saw is turned off. The speed of the tooth decreases at a constant rate, and the blade comes to rest in 17 s. Determine the time at which the total acceleration of the tooth is 130 ft/s2.

The time is    

Explanation:

From the question we are told that

   The  diameter of the circular saw  is

   The peripheral speed is

    The time taken for the blade to come to rest is t = 17 s

    The total  acceleration of the tooth considered is    

Generally the radius of the blade is mathematically represented as

Generally the tangential acceleration of the blade is mathematically represented as

Here v is the final  velocity of the tooth of the blade which is  zero  since the blade came to  rest

so  

=>      

Generally the total acceleration of the tooth of the blade is mathematically represented as

Here is the radial acceleration , now making the subject of the formula we have that

=>    

=>    

Generally radial  acceleration is mathematically represented as

Here   is the velocity at which the total acceleration is 130 ft/s2.

=>    

=>  

=>  

Generally the time at which the total acceleration is  130 ft/s2. is mathematically represented as

=>      

=>          

we need to know the time interval during which the force acts. impulse of force = momentum change