1.) a bullet leaves a rifle with a muzzle velocity of 500 m/s. while acceleration through the barrel
of the rifle, the bullet moves a distance of 0.830 m. determine the acceleration of the bullet
(assume uniform acceleration )
a.) 125,500 m/s
b.) 250,500 m/s2
c.) 161,572 m/s
d.) 150,602 m/s2

 (Answer d)

Explanation:

Since the problem doesn't include information on he time during which the bullet was accelerated, we can use the kinematic equation for motion under acceleration, that doesn't include time and relates initial and final velocities () with the acceleration exerted (), and the distance traveled (D):

So, in this case, considering that the bullet starts at rest inside the muzzle (initial velocity = zero), and reaches a final velocity 500 m/s, and the distance the bullet travels inside the muzzle is 0.830 meters, we have the following equation from which we can solve for the acceleration:

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft