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\alphat
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