a bullet leaves the barrel of a gun with the speed of 320m/s. the barrel is 80.0cm long. Assuming uniform acceleration, determine

a)the time it took

b)the acceleration of the bullet

(a) Time = (barrel length)/(average speed)

= 2*(barrel length)/(exit speed) = 2L/V

(b) Length = (a/2) T^2
Use T from part A and solve for a

a = 2 L/T^2 = 2L*V^2/(4L^2)
= V^2/(2L)

To determine the time it took for the bullet to leave the barrel and the acceleration of the bullet, we can use the equations of motion for uniformly accelerated motion.

a) The first step is to convert the length of the barrel from centimeters to meters, since the speed of the bullet is given in meters per second.

Length of the barrel = 80.0 cm = 80.0/100 = 0.80 m

Now, we can use the equation:
v = u + at

Where:
v = final velocity (speed of the bullet) = 320 m/s
u = initial velocity (since the bullet starts from rest inside the barrel, the initial velocity is 0 m/s)
a = acceleration of the bullet (to be determined)
t = time taken (to be determined)

Substituting the given values:
320 = 0 + a * t

Since the initial velocity is 0, the equation simplifies to:
320 = a * t

Next, we need to determine the value of acceleration (a). To do this, we can use the equation:
s = ut + (1/2)at^2

Where:
s = displacement (length of the barrel) = 0.80 m
u = initial velocity = 0 m/s
t = time taken (to be determined)

Substituting the given values:
0.80 = 0 + (1/2)*a*t^2

Simplifying the equation:
0.80 = (1/2) * a * t^2

Now we have two equations:
320 = a * t
0.80 = (1/2) * a * t^2

To solve for t, let's isolate t in the first equation:
t = 320/a

Substituting this value of t into the second equation:
0.80 = (1/2) * a * (320/a)^2
0.80 = (1/2) * a * (102400/a^2)
0.80 = 51200/a

Now, solve for a:
a = 51200 / 0.80
a = 64000 m/s^2

b) The acceleration of the bullet is determined to be 64000 m/s^2.