The barrel of a rifle has a length of 0.97m. A bullet leaves the muzzle of a rifle with a speed of 601m/s. What is the acceleration of the bullet while in the barrel?
vf^2=2ad solve for a
To calculate the acceleration of the bullet while in the barrel, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity of the bullet (601 m/s)
u = initial velocity of the bullet (0 m/s, since it starts from rest in the barrel)
s = displacement of the bullet (length of the barrel, 0.97 m)
a = acceleration of the bullet (unknown)
Rearranging the equation, we can solve for the acceleration:
a = (v^2 - u^2) / (2s)
Substituting the given values:
a = (601^2 - 0^2) / (2 * 0.97)
a = (361201 - 0) / 1.94
a ≈ 186497.42 m/s^2
Therefore, the acceleration of the bullet while in the barrel is approximately 186497.42 m/s^2.
To find the acceleration of the bullet while in the barrel of the rifle, we need to use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (speed of the bullet) = 601 m/s
u = initial velocity (at the start of the barrel) = 0 (as the bullet is at rest initially)
a = acceleration
s = displacement (length of the barrel) = 0.97 m
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
Substituting the values we have:
a = (601^2 - 0^2) / (2 * 0.97)
a = 361201 / 1.94
a ≈ 186456.19 m/s^2
Therefore, the acceleration of the bullet while in the barrel is approximately 186456.19 m/s^2.