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
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?
5 answers
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?
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?
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.
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.
Reiny is correct.
Can you explain me about similar triangle concept.?
3 answers