Question

A street lamp is 6m above a straight road.A
Man 2m tall.walks along the road away from
the lamp at a constant speed of 1.5m/s.At what
rate is his shadow lengthening?

Answer 1

Well, it depends where he is I think
at the beginning
x = man from light pole= Xi
then
x = Xi + 1.5 t
if he starts at the pole at t = 0 then Xi = 0 but we can not assume that

y = shadow length=(2/6)(x+y)
or
y = x/2

so
y=(1/2)(Xi + 1.5 t)
and
dy/dt = (1/2)(1.5) = 3/4 m/s

Answer 2

Draw a diagram. Using similar triangles, if

x = the distance from the lamp post
s = length of shadow

s/1.5 = (x+s)/6
s = x/3

so,

ds/dt = 1/3 dx/dt
...

Extra credit: how fast is the tip of the shadow moving?

Answer 3

Using Steve's definitions
by similar triangles:
2/s = 6/(s+x)
6s = 2s + 2x
4s = 2x
2s = x
2 ds/dt = dx/dt
ds/dt = 1.5/2 = 3/4 or .75 m/s

which is what Damon had. I think Steve misread the man's speed as his distance.

Notice the speed of the shadow is independent of where the man is walking.

Answer 4

Reiny is correct.

Answer 5

Can you explain me about similar triangle concept.?