a 5 m long ladder rests against a vertical wall with its feet on horizontal ground. the feet on the ground slip, and at the instant when they are 3 m from the wall, they are moving at 10 ms^-1. at what speed is the other end of the ladder moving at this instant?

Answer 1

The classic introduction question to "rates of change"

At a given time of t seconds, let the foot of the ladder be x m from the wall
and let the top of the ladder be y m above the ground

we know:
x^2 + y^2 = 5^2
2x dx/dt + 2y dy/dt = 0
when x = 3, y = 4 , (using Pythagoras in my above equation), and dx/dt = 10 m/s

2(3) (10) + 2(4)(dy/dt) = 0
dy/dt = -60/8 m/s = -7.5 m/s

the negative tells me the top of the ladder is moving down, or the y value is decreasing.

Answer 2

shouldn't it be -0.0075 m/s because the question has 10 ms?