A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (uniform acceleration)

V^2 = 2ad = (521)^2,

2a*0.84 = 271441,
168a = 271441,
a = 271441 / 1.68 = 161572 m/s^2.

Henry just used the equation V2² = V1² + 2ad where V1 (Initial velocity) is 0. Therefore it is shortened to:

V2² = 2ad
V2² / (2d) = a

Thanks but on topper website it is solved in another way which seems correct. I want to understand that way....Could you explain?

To determine the acceleration of the bullet, we can use the equation of motion:

v^2 = u^2 + 2aS

Where:
v = final velocity of the bullet (521 m/s)
u = initial velocity of the bullet (0 m/s, since it starts from rest)
a = acceleration of the bullet (to be determined)
S = displacement of the bullet (0.840 m)

Substituting the given values into the equation, we get:

(521 m/s)^2 = (0 m/s)^2 + 2a(0.840 m)

Simplifying,

271,441 = 0 + 1.68a

Dividing both sides of the equation by 1.68, we get:

a = 271,441 / 1.68

a ≈ 161,408.33 m/s²

Therefore, the acceleration of the bullet is approximately 161,408.33 m/s².