How Is the Average Force on a Bullet Calculated?

[highc]

Ok, I'm using Schaum's Outline of College Physics to supplement my ultra condensed correspondance physics course. As I work through the "Supplementary Problems" I've come across this one which leaves me puzzled.

Typically a bullet leaves a standard 45 caliber pistol (5.0 inch barrel) at a speed of 262 m/s. If it takes 1 ms to traverse the barrel, determine the average acceleration by the 16.2 g bullet within the gun and then compute the average force exerted on it. The provided answers are: 3.0 x 10^5 m/s, 0.4 x 10 N.

I've had no problem working out the average acceleration to 2.62 x 10^5 m/s (3.0 X 10^5 m/s), but I have no idea how the book has arrived at 40 N for the average force exerted.

Does anybody care to show how this was worked out?

 

[quasar987]

This is obviously an error in the book. With 2.62 x 10^5 m/s² (don't forget to square s) as the average acceleration, the average force, which is given by the proportionality of force and acceleration

$$F_{av}=ma_{av}$$

turns out to be 0.43092 x 10^5 N, which would round down to 0.4 x 10^5 N. So it's probably just a printing error; they forgot the $^5$.