The Robot and the lamp post

At any instant of time t>0, the Robot is at a distance of 20t ft from the lamp. Draw a horizontal line from 5 ft above the base of lamp post, joining the head of the robot. We form two similar triangles (first having vertices at light, robot's head and 5 ft above the base of lamp post; second having vertices at Robot's head, its feet and tip of its shadow). Angles of these similar triangles, which they form with horizontals are the same say x. In the first triangle,

$\tan { x } =\quad \frac { 25 }{ 20t } \quad =\quad \frac { 5 }{ 4t }$

Let us denote the distance between robot and tip of its shadow by s. Then in the second triangle,

$\tan { x } =\quad \frac { 5 }{ s } \quad =\quad \frac { 5 }{ 4t } \quad \Longrightarrow \quad s=4t$

So the distance between lamp post and shadow tip is given by L = 20t + 4t = 24t

Thus the speed of movement of shadow is : $\frac { dL }{ dt } \quad =\quad \frac { d }{ dt } \quad (24t)\quad =\quad 24$

Hence speed of the shadow is 24 ft/sec

After 2 seconds; Robot is 40 ft away from the base of light. So we have a right triangle from 5 ft above the base of the light, to the light, to the top of the robot; with Base=40 ft , height=(30-5)=25 ft. Another right triangle is from base of the Robot, to top of the robot, to the tip of the shadow on ground; with base=x(unknown), and height=5.

But the angle of base with hypotenuse of two triangles is same, so we have:

25/40 = 5/x => x = 8

So the tip of shadow is 40+x=40+8=48 ft away from lamp post, after 2 seconds. Hence, shadow's speed is 24 ft/sec.