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The roller at A is moving upward with a velocity of {v}_{A} = 3 ft/s and has an acceleration of {a}_{A} = 4 ft/{s}^{2} when {s}_{A} = 4 ft . Determine the velocity and acceleration of block B at this instant.

Step-by-step

s_{B}+\sqrt{\left(s_{A}\right)^{2}+3^{2}}=l
\dot{s}_{B}+\frac{1}{2}\left[\left(s_{A}\right)^{2}+3^{2}\right]^{-\frac{1}{2}} \left(2 s_{A}\right)\dot{s}_{A}=0 \dot{s_{B}}+\left[s_{A}^{2}+9\right]^{-\frac{1}{2}}\left(s_{A} \dot{s}_{A}\right)=0
\ddot{s}_{B}\left[\left(s_{A}\right)^{2}+9\right]^{-\frac{3}{1}}\left(s_{A}^{2} \dot{s}_{A}^{2}\right)+\left[s_{A}^{2}+9\right]^{-\frac{1}{2}}\left(\dot{s}_{A}^{2}\right)+\left[s_{A}^{2}+9\right]^{-1}\left(s_{A} \ddot{s}_{A}\right)=0
At s_{A}=4 ft, \dot{s}_{A}=3 ft/s \ddot{s}_{A}=4 ft/{s}^{2}
\dot{s}_{B}+\left(\frac{1}{5}\right)(4)(3)=0
v_{B}=-2.4 ft/s=2.40 ft/s \rightarrow
{s}_{B}\left(\frac{1}{5}\right)^{3}(4)^{2}(3)^{2}+\left(\frac{1}{5}\right)(3)^{2}+\left(\frac{1}{5}\right)(4)(4)=0
a_{B}=-3.85 ft/{s}^{2}=3.85 ft/{s}^{2} \rightarrow

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