Step-by-step explanation:
Let vertical height of ladder from ground be y and
horizontal distance of the base of the ladder from the wall be x respectively.
Length of the ladder = l (constant) = 10 ft
Using Pythagoras theorem:
[tex] {l}^{2} = {y}^{2} + {x}^{2} [/tex]
Differentiate both sides w.r.t time
[tex]0 = 2y \frac{dy}{dt} + 2x \frac{dx}{dt} [/tex]
[tex]y \frac{dy}{dt} + x \frac{dx}{dt} = 0[/tex]
We know that (After 1 sec, y = 6 ft and x = 8 ft ; dy/dt = 2 ft/sec)
[tex]6\times 2 + 8\frac{dx}{dt} = 0[/tex]
[tex] \frac{dx}{dt} = - 1.5 ft \: per \: sec[/tex]
( Ignore - ive sign)
Therefore, bottom of the ladder is sliding away from the wall at a speed of 1.5 ft/sec one second after the ladder starts sliding.
