Asked by hey
A man 5.5 ft tall walks away from a lamp post 10 ft high at the rate of 8 ft/s. (a.) How fast does his shadow lengthen?
(b.) how fast does the tip of the shadow move?
(b.) how fast does the tip of the shadow move?
Answers
Answered by
Reiny
make a sketch
let the distance of the man from the lamp-post be x
let the length of his shadow be y
by ratios:
5.5/y = 10/(x+y)
5.5x + 5.5y = 10y
5.5x = 4.5y
times 10
55x = 45y
11x = 9y
11 dx/dt = 9 dy/dt
11(8) = 9 dy/dt
dy/dt = 88/9 ft/s or 9 7/9 ft/s --> rate at which the shadow is lengthening.
How fast is the shadow moving??
d(x+y)/dt = dx/dt + dy/dt
= 8 + 88/9 = 160/9 ft/s or 17 7/9 ft/s
let the distance of the man from the lamp-post be x
let the length of his shadow be y
by ratios:
5.5/y = 10/(x+y)
5.5x + 5.5y = 10y
5.5x = 4.5y
times 10
55x = 45y
11x = 9y
11 dx/dt = 9 dy/dt
11(8) = 9 dy/dt
dy/dt = 88/9 ft/s or 9 7/9 ft/s --> rate at which the shadow is lengthening.
How fast is the shadow moving??
d(x+y)/dt = dx/dt + dy/dt
= 8 + 88/9 = 160/9 ft/s or 17 7/9 ft/s
Answered by
hey
Thank You very much !! :D
Answered by
Jean
why was it multiplied to 10?
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