Question

A man $1.5m$ tall walks away from a lamp post $4.5m$ high at a rate fo $4km/hr$. How fast is the farther end of shadow moving on the pavement?

• A. $4km/hr$
• B. $2km/hr$
• C. $6km/hr$
• D. $5km/hr$

Answer 1

The correct option is B $2km/hr$

$\frac{dx}{dt}=4km/hr$

In $\Delta ABD,\tan \theta =\frac{4.5}{x+y}$ ... (i)

In $\Delta ECD,\tan \theta =\frac{1.5}{y}$ ...(ii)

$\therefore \frac{4.5}{x+y}=\frac{1.5}{y}$

$3y=x+y$

$2y=x$

$y=\frac{x}{2}$

$\therefore \frac{dy}{dt}=\frac{1}{2}\frac{dx}{dt}$

$=\frac{1}{2}\left(4km/hr\right)$

$=2km/hr$