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A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.2 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall? (That is, find the angle's rate of change when the bottom of the ladder is 8 ft from the wall.) rad/s

Respuesta :

Answer:0.2 rad/s

Explanation:

Given data

Velocity of the bottom point of the ladder=1.2Ft/s

Length of ladder=10ft

distance of the bottom most point of ladder from origin=8ft

From the data the angle θ with ladder makes with horizontal surface is

Cosθ=[tex]\frac{8}{10}[/tex]

θ=36.86≈37°

We have to find rate of change of θ

From figure we can say that

[tex]x^{2}[/tex]+[tex]y^{2}[/tex]=[tex]AB^{2}[/tex]

Differentiating above equation we get

[tex]\frac{dx}{dt}[/tex]=-[tex]\frac{dy}{dt}[/tex]

i.e [tex]{V_A}=-{V_B}=1.2ft/s[/tex]

[tex]{at\theta}={37}[/tex]

[tex]Y=6ft[/tex]

[tex]and\ about\ Instantaneous\ centre\ of\ rotation[/tex]

[tex]{\omega r_A}={V_A}[/tex]

[tex]{\omega=\frac{1.2}{6}[/tex]

ω=0.2rad/s

i.e.Rate of change of angle=0.2 rad/s

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