There are two questions:
A ladder, L[m] long, is leaning against a vertical wall. If the bottom of the ladder is pulled horizontally away from the wall at $v_h[\frac{m}{s}]$, how fast is the top of the ladder sliding down the wall when the bottom is the distance $x_0[m]$ from the wall
Suppose a particular downward vertical speed is given as $v_d[\frac{m}{s}]$ . Determine how long it takes the top of the ladder to reach the speed by expressing the time in terms of $v_h, v_d, and L$
So for the first one I used related rates and set up my triangle such that
Given:
$L [m] \rightarrow$ Length of ladder
$V_h[\frac{m}{s}] \rightarrow$ Speed at which the ladder is sliding away from the wall
$x_0[m] \rightarrow$ Some distance the ladder is away from the wall.
Let us represent our rate at which the ladder is sliding down as
$\frac{\delta d}{\delta t} $ which we will later call $V_d[\frac{m}{s}]$
using the Pythagorean Theorem:
$$x_0^2 + d^2 = L^2\quad$$ where d is the point where the ladder is leaning against the wall (height of ladder)
$$2x_0 \frac{\delta x_0}{\delta t} + 2d\frac{\delta d}{\delta t} = 0\quad$$
$L^2$ goes to zero because it is a constant and never changes (ladder)
So solving for the rate at which the ladder is sliding down the wall when the bottom is a $x_0$ distance we isolate $\frac{\delta d}{\delta t}$ such that:
$$ \frac{\delta d}{\delta t} = - \frac{x_0}{d}\frac{\delta x_0}{\delta t}$$
Here we can replace the rate at which $x_0$ changes with our $V_h[m]$ which gives us
$$ \frac{\delta d}{\delta t} = - \frac{x_0}{d}V_h$$
We do not have a value for $d$, but can express it in terms of $x_0$ and $L$:
$$ d = \sqrt{-x_0^2+L^2} $$
So replacing $d$ with this equation we have
$$ \frac{\delta d}{\delta t} = - \frac{x_0}{\sqrt{-x_0^2+L^2}}V_h $$
Checking our units: $$ \frac{m}{s} = \frac{m}{m} \frac{m}{s} \implies \frac{m}{s} = \frac{m}{s} $$
This solves (1) I believe. Moving onto (2), is where I am confused. How do I find the time it takes the ladder to reach a downward velocity when we are given $V_h, V_d$, and $L$.
I know that the first answer is the velocity at which the ladder slides down, and the integral of velocity is position. So would I have to integrate which would give me the right hand side of (1) * t for some equation $y(t)$ ?