Problem 2.35: Faster than a speeding bullet! The Beretta Model 92S (the standard

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Problem 2.35: Faster than a speeding bullet!

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 339m/s.

Part A

What is the acceleration of the bullet while it is in the barrel, assuming it to be constant? Express your answer in m/s2.

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Part B

What is the acceleration of the bullet while it is in the barrel, assuming it to be constant? Express your answer in g's.

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Part C

For how long is the bullet in the barrel?

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Problem 2.35: Faster than a speeding bullet!

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 339m/s.

Part A

What is the acceleration of the bullet while it is in the barrel, assuming it to be constant? Express your answer in m/s2.

a =   m/s2  

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Incorrect; Try Again

Part B

What is the acceleration of the bullet while it is in the barrel, assuming it to be constant? Express your answer in g's.

a =   g  

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Part C

For how long is the bullet in the barrel?

t =   s  

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Explanation / Answer

Distance traveled,?x, in m, is 0.127 m
Initial position is 0 m
Initial velocity,v0, is 0 m/s
Final velocity,v, is 338 m/s
You're solving for acceleration.

Use this kinematic equation:

v = v0 + 2a?x

a = (v-v0)/2?x
a = (v/2?x)
a = 449779.5276 m/s

To express your answer in g's, just divide the acceleration by the gravitational constant, 9.8m/s:
45895.87 g's

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