Final answer:
The velocity of the bullet as it leaves the barrel is 200 m/s, calculated using kinematic equations with the given acceleration and distance the bullet travels within the barrel.
Step-by-step explanation:
The velocity of the bullet as it leaves the barrel can be calculated using the kinematic equation for motion, which states that the final velocity (v) is equal to the initial velocity (u) plus the product of acceleration (a) and time (t). Since the bullet starts from rest, the initial velocity is 0 m/s. The acceleration is given as 40,000 m/s2, and the time can be derived using the equation s = ut + 0.5at2, where s is the distance traveled. First, solve for time using the distance s = 0.5 m: s = ut + 0.5at2, 0.5 = 0 + 0.5(40,000)t2, t = sqrt(0.5 / (0.5 * 40,000)), t = 0.005 s. Next, calculate the final velocity using v = u + at: v = 0 + (40,000)(0.005), v = 200 m/s. The correct answer is thus (a) 200.0 m/s.