Speed of moving shadow due to light source

hqjb
Messages
38
Reaction score
0

Homework Statement



A girl 5 feet tall is running at the rate of 12 feet/second and
passes under a street light 20 feet above the ground. Find how
rapidly the length of her shadow is increasing when she is 20
feet past the base of the street light.

The Attempt at a Solution



I let the distance of girl from light be S = 12t
I let the distance of shadow from light be X,

Due to similar triangle,

\frac{x}{x-12t} = \frac{20}{5}

-15x=-240t

x=16t

\frac{dx}{dt}=16

But the ans is 4ft/sec any clues as to where I went wrong?
 
Physics news on Phys.org
The problem does not want dx/dt. Because x is not the length of her shadow. x is the length of her shadow plus her distance from the base of the street light.
 
BruceW said:
The problem does not want dx/dt. Because x is not the length of her shadow. x is the length of her shadow plus her distance from the base of the street light.

Ah...semantics. Thanks. So I suppose that if they were looking for the speed for the tip of the shadow then that would be the right answer?
 
Yes. Your answer of 16 ft/second is the speed of the tip of the shadow. The rate of change of the length of the shadow is different to this because the other end of the shadow (at the girl) is also moving.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top