Question

A $2m$ ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate $25$ cm/sec, then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is $1$m above the ground is?

• A. $25\sqrt{3}$
• B. $25$
• C. $\frac{25}{\sqrt{3}}$
• D. $\frac{25}{3}$

Answer 1

The correct option is C $\frac{25}{\sqrt{3}}$

$x^2+y^2=4(\frac{dy}{dt}=-25)$
$x=\frac{dx}{dy}+y\frac{dy}{dt}=0$
$\sqrt{3}\frac{dx}{dt}-1\left(25\right)=0$
$\frac{dx}{dt}=\frac{25}{\sqrt{3}}$cm/sec.