Question #337941

A ladder 5 meters long is leaning against the wall. The bottom is initially 3.5 meters away from the

wall. You are pushing it towards the wall at a rate of 0.20 m/s. How fast is the top of the ladder moving

up the wall 12 seconds after you start pushing?


1
Expert's answer
2022-05-08T13:48:47-0400
x2+y2=52,y0x^2+y^2=5^2, y\ge 0

y=25x2y=\sqrt{25-x^2}

x=x0+vxtx=x_0+v_xt

vy=dy/dt=2x25x2vxv_y=dy/dt=-\dfrac{2x}{\sqrt{25-x^2}}\cdot v_x

Given x0=3.5m,vx=0.20m/s.x_0=3.5m, v_x=-0.20m/s.

12 seconds after you start pushing


xt=12=3.5m0.20m/s12s=1.1mx|_{t=12}=3.5m-0.20m/s\cdot12s=1.1m

vyt=12=2(1.1m)25(1.1)2(0.20m/s)v_y|_{t=12}=-\dfrac{2(1.1m)}{\sqrt{25-(1.1)^2}}\cdot (-0.20m/s)

0.09m/s\approx0.09m/s

The top of the ladder is moving up the wall at a rate of 0.09m/s 12 seconds after you start pushing.



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