A bullet is fired through a board 12.0 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 450 m/s and it emerges from the other side of the board with a speed of 350 m/s.
1. Find the acceleration of the bullet as it passes through the board in ( m/s2 )
Find the total time the bullet is in contact with the board in (s ).
2007-09-20 19:28:45 · 2 answers · asked by smilenhope 1 in Science & Mathematics ➔ Physics
hey (Ω)Mistress Bekki i dont get what those symbol or letters mean..in your equation..can u please explain it...thanks
2007-09-20 21:43:54 · update #1
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.
2007-09-20 20:47:32 · answer #1 · answered by Anonymous · 0⤊ 0⤋
This symbolism might be a little clearer:
v(avg) = (v(zero) + v(final))/2
v(avg) = (450 + 350)/2
v(avg) = 400m/s
t = D/r = (0.12meters)/(400) = 0.0003sec
a = (v(final) - v(zero))/t
a = (350 - 450)/0.0003
a = -333333m/s^2
2007-09-21 15:49:42 · answer #2 · answered by jsardi56 7 · 0⤊ 0⤋