# Answer to Question #4490 in Differential Equations for kennedy

Question #4490

I hope someone can help me with this question. Thank you very much.

"A woman 6ft tall walks away from a light 10ft above the ground. If her shadow lenghtens at the rate of 2ft/sec, how fast is she walking?"

"A woman 6ft tall walks away from a light 10ft above the ground. If her shadow lenghtens at the rate of 2ft/sec, how fast is she walking?"

Expert's answer

v-velocity of woman

t=1sec

L=2m

We have from first triangle (x+vt)/10=vt/6

From the second one (x+vt+L)/10=L/6

From first : x=(2/3)vt

From second:

6x+6vt+6L=10L

Set x=(2/3)vt

here: 4vt+6vt+6L=10L

So v=2L/5t=4/5

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