Question

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 (assume a uniform acceleration)

Answer 1

a = 1.62*105 m /s2

Step-by-step explanation:

25.28 m2/s2 = vi2

vi = 5.03 m/s

To find hang time, find the time to the peak and then double it.

vf = vi + a*t

0 m/s = 5.03 m/s + (-9.8 m/s2)*tup

-5.03 m/s = (-9.8 m/s2)*tup

(-5.03 m/s)/(-9.8 m/s2) = tup

tup = 0.513 s

hang time = 1.03 s

Return to Problem 11

Given:

vi = 0 m/s

vf = 521 m/s

d = 0.840 m

Find:

a = ??

vf2 = vi2 + 2*a*d

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

271441 m2/s2 = (0 m/s)2 + (1.68 m)*a

(271441 m2/s2)/(1.68 m) = a

a = 1.62*105 m /s2

Answer 2

`velocity=(distance)/(time)\\\\ velocity*time=distance\\\\ time=(distance)/(velocity)\\\\ time=(0,840m)/(521(m)/(s))=0,0016s\\\\ acceleration=(velocity)/(time)=(521)/(0,0016)=325625(m)/(s^2)`