Please draw a picture of the situation:
- A horizontal line represents the ground.
- A vertical line 12 feet tall represents the lamppost.
Let B be the base of the lamppost, and L be its top.
- A vertical line 6 feet tall represents the student.
Let F be the point at his feet, and H be the point at the top of his head.
- Let u(t) be the length of line segment BF, which is a function of time t
as the student walks away from the lamppost.
- Draw a horizontal line from H towards the lamppost.
Let D be the point where it strikes the lamppost.
The distance between D and H is also u(t).
- Draw the line LH connecting the points L and H and extending to the ground.
Let S be the intersection between the ground and the line LH.
The point S is the tip of his shadow.
- Let s(t) be distance between B and S.
It is the distance between the fixed lamppost and the tip of the student's shadow,
and it is also a function of time as the student walks away.
- Let α be the angle between the ground and the line LH (at the point S).
(It also happens to be a function of time, but it is not important here.)
- s(t) = BL . cot(α)
BL = 12 feet is the the lampost's hight.
- The angle between DH and LH is also α.
- u(t) = DL . cot(α)
DL is the difference between the height of the lamppost and the student,
which is also 6 ft.
- By eliminating cot(α) from the equations for u(t) and s(t)
u(t) = s(t) . DL / BL = s(t) . 6ft / 12ft = 0.5 s(t)
- Speed is derivative of distance w.r.t time:
du/dt = 0.5 ds/dt
- That means that the student moves at half the speed of the tip of his shadow.
When the shadow moves at 5 ft/s, the student moves at 2.5 ft/s.