Question

Chris, who is 6 feet tall, is walking away from a street light pole 30 feet high at a rate of 2 feet per second. 1. How fast is his shadow increasing in length when he is 24 feet from the pole? 2. How fast is the tip of his shadow moving when he is 24 feet from the pole? @Physics

Answer 1

im learning to read these better :)

Answer 2

we can relate the whole thing with similar triangles

Answer 3

\[\frac{30}{w+s}=\frac{6}{s}\]
\[30s=6(w+s)\]
derive it with respect to time
\[s'=6(w'+s')\]
\[s'=6w'+6s'\]
\[s'-6s'=6w'\]
\[-5s'=6w'\]
\[s'=\frac{6w'}{-5}\]
did i get that right?

Answer 4

since hes walking, w' = 2 but i gotta dbl chk the calculations i made

Answer 5

\[30s=6(w+s)\]
\[30s=6w+6s\]
\[30s-6s=6w\]
\[24s=6w\]
\[s=\frac{1}{4}w\]
\[s'=\frac{1}{4}w'\]

Answer 6

im missing something i can feel it

Answer 7

i dropped the 30 by mistake the first time, but we should have a w or s in there someplace otherwise the rate is the same regardless of distances

Answer 8

so s'=1/4w', which is 1/4(2)=.5 ft/s?

Answer 9

yes, so far as that goes, but still ... im concerned

Answer 10

the shadow growth is s'; the speed at which it moves is w'+s' tho

Answer 11

so 2.5 ft/s?

Answer 12

yeah, as long as the calculations are good

Answer 13

I see, thank you that was a huge help. What about 2. How fast is the tip of his shadow moving when he is 24 feet from the pole?

Answer 14

let me try this from another vantage point:
\[\frac{30}{6}=\frac{(w+s)}{s}\]derived we get:
\[0=s(w'+s')-s'(w+s)\]
\[0=sw'+ss'-s'w+ss'\]
\[0=sw'-s'w\]
\[s'=\frac{sw'}{w}\]
\[s'=\frac{2s}{24}\]
\[\frac{30}{6}=\frac{(24+s)}{s}\]
\[30s=6(24+s)\]
\[30s=144+6s;s=144/24=6\]

\[s'=\frac{2(6)}{24}=\frac{1}{2}ft/sec\]is how fast the shadow is growing
add 2 feet per sec to that from teh walking speed to calculate the tips speed:

5/2 ft/sec

Answer 15

thank you soo much!

Answer 16

Sorry, one more question: How fast is the tip of his shadow moving when he is 24 feet from the pole? I don't know what I should do?