Question

A bullet accelerates the length of the barrel of a gun 0.750 meter long with a magnitude of 5.35×10^5 m/s2. with what speed does the bullet exit the barrel.​

Answer 1

Explanation:

Given velocity , v=750m/s

initial velocity , u=0m/s

Length of barrel ,s=60cm=0.6m

Therefore average velocity V

av

=

2

V+u

=

2

750+0

=375m/s

& from equation of motion ,we have

V

2

=V

2

+2as

a=

2s

v

2

−u

2

=

2×0.6

750

2

=4.68×10

5

m/s

2

Also we know

a=

t

v−u

=>t=

4.68

750×10

−5

=160.25×10

−5