A bullet is fired through a board 11.0 cm thick in such a way that the bullet's line of motion is perpendicular to the face of the board. The initial speed of the bullet is 350 m/s and it emerges from the other side of the board with a speed of 250 m/s.
(a) Find the acceleration of the bullet as it passes through the board.
_____m/s2
(b) Find the total time the bullet is in contact with the board.
_____s
2007-09-20 20:52:43 · 4 answers · asked by lttlstar7 1 in Science & Mathematics ➔ Physics
Do your own homework.
2007-09-20 20:57:49 · answer #1 · answered by bestonnet_00 7 · 0⤊ 0⤋
vf^2 - vi^2 = 2ax
a = (vf^2 - vi^2) / 2x
a will be negative, which reflects that the bullet decellerates
x = vi t + 1/2 at^2
Solve the quadratic equation for t. You'll get one solution that makes no sense whatsoever. Ignore it.
x = (-vi +- sqrt (vi^2 - 4 (a/2)(-x)) ) / 2a/2
Simplify and plug n chug.
2007-09-21 04:00:08 · answer #2 · answered by Anonymous · 0⤊ 1⤋
a))))
V^2 = Va^2 + 2A x..... x= 0.11m, V = 250 , Va = 350
A = - (122500- 62500) / 0.22 = -272727.27 m/s^2
b)))) t = (V - Va) / A = 0.00036667 (s)
2007-09-21 05:08:45 · answer #3 · answered by Mr.Bingo207 1 · 0⤊ 1⤋
ur given v,u&s
so to find a
use v^2=u^2+2a*s;
to get t
use
v=u+at;
2007-09-21 04:02:02 · answer #4 · answered by Ashwin 2 · 0⤊ 1⤋