Question

An 8 metre long ladder is positions against a wall 3 metres high. If the bottom end of the ladder slides at a rate of 1.5m/s, how fast will the top end of the ladder slide?

Answer 1

I was hoping someone could explain the steps involved?

Answer 2

i was also expecting same thing.:P

Answer 3

the whole ladder is 8m long.and the distance between it's foot and the foot of the wall is \[\sqrt{sqart8-sqart3}\]=7.4 approx.its speed is 1.5m/s.so it will take 8-7.4/1.5 s=.4s.by this .4s the top end have to cross a distance of 3m.
so it's speed is 3/.4=7.5m/s

Answer 4

Aaah I see now. Except you have square root symbols instead of ^2 symbols.

Answer 5

yeah

Answer 6

That is the average velocity of the ladder in the y-axis. However if you're interested in the instantaneous velocity of the ladder at any point in time you have to use calculus.

The given velocity is \[x = {3t \over 2} + x_{0}\]
where x_0 is initial displacement. \[{dx \over dt} = {3 \over 2} m/s\] which is constant at any point in time during the ladder's decent.
Using Pythagoras theorem \[y^2 + x^2 = 8^2\]\[y^2 = 8^2 - x^2\]\[y = \sqrt{8^2 - x^2} = (8^2 - x^2)^{1 \over 2}\]
We are looking after: \[dy \over dt\]
Using the chain rule:
\[{dy \over dt} = {dy \over dx} * {dx \over dt}\]
dy/dx is the only unknown which remains to be found.
\[{dy \over dx} = {1 \over 2}(8^2 - x^2)^{- {1 \over 2}} * (-2x)\]\[{dy \over dx} = -x(8^2 - x^2)^{-{1 \over 2}}\]\[{dy \over dx} = {-{x \over \sqrt{8^2 - x^2}}}\]
Substituting into result by chain rule to obtain dy/dt is:
\[{dy \over dt} = {-{x \over \sqrt{8^2 - x^2}}} * {3 \over 2}\]\[{dy \over dt} = {-{3x \over 2 \sqrt{8^2 - x^2}}} \]
For example at the point x = sqrt(55) the instantaneous velocity with which the ladder is actually falling in the y-axis is:
\[{dy \over dt} = - {{3 \sqrt{55}} \over {2 \sqrt{8^2 - 55}}} = - {\sqrt{55} \over 2} \approx -3.71 m / s\]