Net displacement for a point on a circular saw

[BHFCBabe]

Homework Statement

When a carpenter shuts off his circular saw, the 10.0-inch diameter blade slows from 4440 rpm to zero in 2.5 s.
A.What is the angular acceleration of the blade? In rev/sec²
B.What is the distance traveled by a point on the rim of the blade during the deceleration? In feet.
What is the magnitude of the net displacement of a point on the rim of the blade during the deceleration? in inches

I have solved A and B, but cannot get C. The answers are in the back of the book so I know A and B are correct.

Homework Equations

1) $$\omega$$_f=$$\omega$$_o+2$$\alpha$$t
2) $$\omega$$_f²=$$\omega$$_o²+2$$\alpha\Delta\Theta$$

The Attempt at a Solution

A. 4440rev/1min * 1min/60sec=74 rev/sec
Using equation 1:
0=74+$$\alpha$$*2.5
$$\alpha$$=-29.6 rev/sec²

B. Using equation 2:
0=74²+2*-29.4*$$\Delta\Theta$$
$$\Delta\Theta$$=92.5 Rev

1ft/12in*10pi in./1 rev * 92.5rev = 242.16 feet.

C. I know that it completed 92.5 revolutions. There are .5 revolutions left over.
10pi in/ 1 rev *.5 rev = 15.7 inches.

However, this answer is wrong. Why? The proper answer is 10 inches.

 

[Redbelly98]

Displacement is determined by the straight-line distance between two points, not the distance around the circumference.

p.s since nobody said it to you before: welcome to Physics Forums

 

[kerimek]

The proper answer of 10 inches is most likely due to rounding errors. The equations used to solve for the angular acceleration and distance traveled involve decimal numbers, which can introduce rounding errors. It is possible that your calculations were slightly off, resulting in a different answer. To avoid this, it is important to use the exact values instead of rounded values when solving equations. Additionally, double check your calculations and make sure that you are using the correct units.