Physics Help!?

Question

The Beretta Model 92S (the standard-issue U.S. army pistol) has a barrel 127 mm long. The bullets leave this barrel with a muzzle velocity of 335 m/s .
What is the acceleration of the bullet while it is in the barrel, assuming it to be constant? Express your answer in m/s2 .
I don't know how to get the answer in m/s2, can anybody help me?

Answer 1

a=v^2/d

=[(335)^2m^2/s^2]/(.127m) = 883000m/s^2