a student five feet tall is ten feet away from a lamppost fifteen feet tall. she is walking away at a rate of 2 feet per second. how fast is the distance between the end of her shadow and the lamppost growing?
2007-07-05 06:37:37 · 3 answers · asked by kellnugent 1 in Science & Mathematics ➔ Mathematics
If y is the length of the shadow, and x is the distance of the student from the lamppost, then from the similar triangles:
y / 5 = (x + y) / 15
3y = x + y
2y = x
y = x/2
When x = 10ft, y = 5ft.
If x is increasing at 2ft/sec, the shadow is increasing at 1ft/sec.
2007-07-05 07:06:07 · answer #1 · answered by Anonymous · 0⤊ 0⤋
As far as I understand the question if the student moves x feet
from her initial position and calling d the distance of the shadow to the lamppost
d/15 = [d-(10+x)]/5
so d=3d-3(10+x)
and d= 3/2(10+x)
Calling d(d)/dt = d(d)/dx*dx/dt) = 3/2*2 =3feet/s
2007-07-05 14:14:29 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋
3 feets per second
2007-07-05 14:02:53 · answer #3 · answered by oregfiu 7 · 0⤊ 0⤋