Final answer:
The velocity of the bullet as it leaves the barrel of the gun is approximately 44.72 m/s. This calculation requires knowing the time it takes for the bullet to travel down the barrel, which is calculated using the distance the bullet travels and the acceleration provided.
Step-by-step explanation:
To find the velocity of the bullet as it leaves the barrel of the gun, we can use the kinematic equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Given that the bullet starts from rest, the initial velocity u is 0 m/s. The acceleration a is given as 4000 m/s². However, to use this equation, we need to also know the time t the bullet is accelerating for, which can be found using the kinematic equation s = ut + ½at², where s is the distance.
The distance the bullet travels down the barrel of the gun is 0.5 m. Plugging the known values into the equation, we get:
0.5 = 0 + ½(4000)(t²)
→ t² = ½ * ²(0.5)/4000 m/s²
→ t² = (0.25)/(2000 s²)
→ t = √(0.25)/(2000 s²)
→ t ≈ 0.01118 s
Now that we have the time, we can calculate the final velocity:
v = 0 m/s + (4000 m/s²)(0.01118 s)
→ v ≈ 44.72 m/s
The velocity of the bullet as it leaves the barrel is approximately 44.72 m/s.