A particle is moving along the x-axis with its coordinate with time ‘t’ given by $x (t) = 10 + 8 t - 3 t^{2} .$ Another particle is moving along the y-axis with its coordinate as a function of time given by $y (t) = 5 - 8 t^{3} .$ At t=1s, the speed of the second particle as measured in the frame of the first particle is given as $\sqrt{v},$ then $v$ (in m/s) is

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answer is 580.

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Detailed Solution

$\begin{array}{l} F o r p a r i t i c l e' A' F o r p a r t i c l e' B' \\ X_{A} = - 3 t^{2} + 8 t + 10; Y_{B} = 5 - 8 t^{3} \\ \vec{V_{A}} = (8 - 6 t) \hat{i}; \vec{V_{B}} = - 24 t^{2} \hat{j} \\ \vec{a_{A}} = - 6 \hat{i}; \vec{a_{B}} = - 48 t \hat{i} \\ A t t = 1 \sec \\ \vec{V_{A}} = (8 - 6 t) \hat{i} = 2 \hat{i} a n d \vec{V_{B}} = - 24 \hat{j} \\ ∴ \vec{V_{B A}} = - \vec{V_{A}} + \vec{V_{B}} = - 2 \hat{i} - 24 \hat{j} \\ ∴ S p e e d o f B w . r . t A, \sqrt{V} = \sqrt{2^{2} + 24^{2}} \\ = \sqrt{4 + 576} = \sqrt{580} \\ ∴ V = 580 (m / s) \end{array}$