To find the acceleration of the bullet while in the barrel, we can use the kinematic equation:

$v^2 = u^2 + 2as$

Given:

• Final velocity ($v$) = $600.0 \, \text{m/s}$
• Initial velocity ($u$) = $0 \, \text{m/s}$ (assuming the bullet starts from rest)
• Distance traveled ($s$) = $0.90 \, \text{m}$
• Acceleration ($a$) = ?

Rearrange the equation to solve for $a$:

$a = \frac{v^2 - u^2}{2s}$

Substitute the values: $a = \frac{600.0^2 - 0^2}{2 \times 0.90}$

$a = \frac{360000}{1.8}$

$a = 200000 \, \text{m/s}^2$

Answer:

The acceleration of the bullet while in the barrel is $200,000 \, \text{m/s}^2$.

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Would you like a detailed step-by-step breakdown of how the equation works, or do you have any further questions?

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Related Questions:

1. How would the acceleration change if the barrel length was doubled?
2. What other forces might influence the bullet's motion in the barrel?
3. How does the bullet's acceleration compare to that of a car's in a drag race?
4. How does the muzzle velocity affect the bullet's kinetic energy?
5. What would happen if there were air resistance in the barrel?

Tip:

Always double-check units in kinematics problems to ensure consistency, especially when working with distances, speeds, and accelerations.