Answer:
The acceleration of the bullet is 1.158 x 10⁶ m/s²
Explanation:
Given;
final velocity of the bullet, v = 425 m/s
distance traveled by the bullet before stopping, s = 0.078 m
the acceleration is given by the following kinematic equation;
v² = u² + 2as
where;
u is the initial velocity of the bullet, = 0
a is the acceleration of the bullet = ?
v² = 2as
a = v² / 2s
a = (425²) / (2 x 0.078)
a = 1.158 x 10⁶ m/s²
Therefore, the acceleration of the bullet is 1.158 x 10⁶ m/s²
