Hello, jmailloux!

You did it again . . . you made $\displaystyle s$ the wrong part.

A 1.7-m tall man walks away from a 9-m high streetlight at 0.8 m/s.

How fast is *the end of his shadow* moving when he is 20 m from the lamppost? Code:

*
| *
| *
9 | | *
| 1.7| *
| x | s-x *
----*-----------+-----------------------*----
: - - - - - - - - s - - - - - - - - :

This time, we make $\displaystyle s$ be the *entire* distance,

. . from the lampost to the end of his shadow.

From the similar right triangles, we have: .$\displaystyle \frac{s - x}{1.7} \:=\:\frac{s}{9}$

Then: .$\displaystyle 9s - 9x\:=\:1.7s\quad\Rightarrow\quad s\:=\:\frac{9}{7.3}x$

Differentiate with respect to time: .$\displaystyle \frac{ds}{dt}\:=\:\frac{9}{7.3}\left(\frac{dx}{dt} \right)$

Since $\displaystyle \frac{dx}{dt} = 0.8\text{ m/s}:\;\frac{dx}{dt} \:=\:\frac{9}{7.3}(0.8) \:\approx\:0.986\text{ m/s}$