A lamp hangs 5.4 m above a level road. A man 1.8 m tall walks straight away from the point below the lamp at 1.2 m/s. How fast is his shadow lengthening?

Explanation

1. 01

   Let the man be $x$ m from the point below the lamp and his shadow $s$ m long. Similar triangles: $\frac{s}{1.8} = \frac{x + s}{5.4}$, so $3.6s = 1.8x$ and $s = \frac{x}{2}$.

2. 02

   $\frac{ds}{dt} = \frac{1}{2}\frac{dx}{dt} = \frac{1}{2} \times 1.2 = 0.6$ m/s.