Question

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

• A. $4$ km/hr
• B. $2$ km/hr
• C. $6$ km/hr
• D. None of these

Answer 1

The correct option is C $6$ km/hr

As seen from the Diagram

We can write speed of man $=\frac{dx}{dt}$

and speed of shadow $=\frac{dy}{dt}$

We can see that,

$△AFB\sim △AEC$

So, $\frac{AF}{FB}=\frac{AE}{EC}$

$\Rightarrow \frac{3}{x}=\frac{4.5}{y}$

$\Rightarrow y=\frac{3}{2}x$

$\Rightarrow \frac{dy}{dt}=\frac{3}{2}\frac{dx}{dt}$

$\Rightarrow \frac{dy}{dt}=\frac{3}{2}\times 4$

$\Rightarrow \frac{dy}{dt}=6$