Step-by-step explanation:
I assume the initial position of the ladder is fully leaning against the wall (0 distance at the bottom of the wall).
the Pythagoras principle applies, as we have only right-angled triangles with the Hypotenuse (the side opposite of the 90° angle) being always the ladder of 13 meters.
let's define
d = distance at the bottom to the wall
h = height on the wall
t = time passed
13² = 169 = d² + h²
h² = 169 - d²
h = sqrt(169 - d²)
the bottom moves at 2 m/s, so, the given bottom distance from the wall of 5 m means that the movement has been going on for 2.5 seconds (2.5 × 2 = 5m).
in our function we know at that moment:
h = sqrt(169 - d²) = sqrt(169 - 5²) = sqrt(169 - 25) = sqrt(144) =
= 12 m
so, it was sliding down 1 m in these 2.5 seconds. as a mean value, the mean speed of dropping was then 1/2.5 = 0.4 m/s. but the actual speed at the end of these 2.5 seconds will be higher, as it will gradually increase over time.
now, the actual speed at a certain point of the drop is the change rate over time. that means we need the first derivative related to time of our main function :
(169 = d² + h²)' = 169×d169/dt = 2d×dd/dt + 2h×dh/dt
d169/dt = 0 which is logical, as the length of the ladder does not change.
so,
0 = 2d×dd/dt + 2h×dh/dt
d = 5
dd/dt = 2 m/s
h = 12
dh/dt = negative (sliding down)
0 = 2×5×2 - 2×12×dh/dt = 20 - 24×dh/dt
24×dh/dt = 20
dh/dt = 20/24 m/s = 5/6 m/s = 0.833333333... m/s
the top of the ladder is sliding down at the speed of 5/6 or 0.833333333... m/s, when the bottom of the ladder is 5 m from the wall.