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.