The string is passing over small smooth pulley rigidly attached to trolley A. If speed of trolley is constant and equal to $V_{A}$ . Speed and magnitude of acceleration of block B at the instant shown in figure is

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$u_{B} = u_{A}, a_{B} = 0$

$a_{B} = 0$

$| u_{B} | = \frac{3}{5} u_{A}$

$| a_{B} | = \frac{16 {υ^{2}}_{A}}{125}$

answer is D.

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Detailed Solution

$(Y - h) + \sqrt{x^{2} + h^{2}} = 1 \frac{d y}{d t} + \frac{x}{\sqrt{x^{2} + h^{2}}} \times \frac{d x}{d t} = 0 \frac{d y}{d t} = - \frac{x}{\sqrt{x^{2} + h^{2}}} \frac{d x}{d t} \frac{d y}{d t} = \frac{- 3}{5} V_{A} \begin{array}{l} v_{B} = \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} > \frac{d y}{d t} \\ \Rightarrow |v_{B}| > \frac{3}{5} v_{A} \end{array} \Rightarrow \frac{d^{2} y}{d t^{2}} = - \frac{{(x^{2} + h^{2})}^{1 / 2} \cdot 1 \cdot \frac{d x}{d t} - x \cdot \frac{1}{2} {(x^{2} + h^{2})}^{- 1 / 2} \cdot 2 x \cdot \frac{d x}{d t}}{(x^{2} + h^{2})} \cdot V_{A} \frac{d y^{2}}{d t^{2}} = {V_{A}}^{} \frac{h^{2}}{{(x^{2} + h^{2})}^{3 / 2}} V_{A} = {V_{A}}^{2} \times \frac{4^{2}}{5^{3}} a_{B} = \frac{16}{125} {V_{A}}^{2}$