Question

A bullet is moving at a speed of 425.0 m/s when it hits a wall and comes to a stop. The bullet travels for a distance of 0.0780m. Determine the acceleration of the bullet.

Answer 1

The acceleration of the bullet is 1.158 x 10⁶ m/s²

Step-by-step 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