The top of a 20-foot ladder leans against a wall, and the bottom of the ladder slides away from the wall at a rate of 5 ft/sec. Find the velocity of the top of the ladder at time t=1 sec if the bottom of the ladder is 7 feet away from the wall at time t=0

vy = - 3.75 ft/sec (slides down)

Explanation:

denoting y as the coordinate of the top of the ladder, x as the coordinate of the bottom, L as the length of the ladder, then

x²+y²=L² = constant

deriving by t

2*x*dx/dt + 2*y*dy/dt = 0

since the velocity is vx=dx/dt and vy=dy/dt

x*vx+y*vy=0

then since the velocity of x is constant x=x₀+vx*t

at t=1 sec x=7 ft + 5 ft/sec*1 sec = 12 ft

then x²+y²=L² → y = √ (L²-x²) = √[ (20 ft) ² - (12 ft) ²] = 16 ft

after that

x*vx+y*vy=0 → vy = - (x/y) * vx = - (12 ft/16 ft) * 5 ft/sec = - 3.75 ft/sec

vy = - 3.75 ft/sec (slides down since vx is positive, that means that slides away)

Note

we cannot assume that vy is constant (and do y=y₀+vy*t) when vx is constant, since y = √ (L²-x²) = √ (L² - (x₀+vx*t) ²), that is not linear with t. Nevertheless, vy could be found deriving y with respect to t, but that would be more complicated than the other way.