Question

A $5$ meter long ladder is resting against a vertical wall. At the moment when foot of the ladder is $3$ meter from the wall it starts sliding away from the wall at the rate of $0.5\text{m}/\text{sec}$. The rate at which the angle between the floor and the ladder is decreasing is

• A. $-\frac{1}{8}\text{rad}/\text{sec}$
• B. $-\frac{1}{4}\text{rad}/\text{sec}$
• C. $\frac{1}{6}\text{rad}/\text{sec}$
• D. $\frac{1}{16}\text{rad}/\text{sec}$

Answer 1

Answer: (A)

Length of the ladder is $5\text{m}$

Let $\theta$ be the angle between the floor and the ladder.

$\cos \theta =\frac{x}{5}$

$\Rightarrow -\sin \theta \frac{d\theta }{dt}=\frac{1}{5}\times \frac{dx}{dt}...\left(1\right)$

Given $\frac{dx}{dt}=0.5\text{m/sec}$ $.......\left(2\right)$

Also $\sin \theta =\frac{y}{5}$

Given that the ladder is $3\text{m}$ away from the wall.

As per pythagoras theorem, when the ladder of length $5\text{m}$ is $\text{3 m}$ away from the wall then $y=4\text{m}$

Therefore when $y=4,\sin \theta =\frac{4}{5}$ $.......\left(3\right)$

Substitue $\left(2\right)$ and $\left(3\right)$ in $\left(1\right)$ we get

$-\frac{4}{5}\frac{d\theta }{dt}=\frac{1}{5}\times 0.5$

$\frac{d\theta }{dt}=-\frac{0.5}{4}=-\frac{1}{8}\text{rad/sec}$

Negative sign indicates that $\theta$ is decreasing