A particle is moving along the x-axis with its coordinate with time ‘t’ given by $x\left(t\right)=10+8t-3t^2.$ Another particle is moving along the y-axis with its coordinate as a function of time given by $y\left(t\right)=5-8t^3.$ 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

$\begin{array}{l}Forpariticle\text{'}A\text{'}Forparticle\text{'}B\text{'}\\ X_A=-3t^2+8t+10;Y_B=5-8t^3\\ \overrightarrow{V_A}=(8-6t)\widehat{i};\overrightarrow{V_B}=-24t^2\widehat{j}\\ \overrightarrow{a_A}=-6\widehat{i}; \overrightarrow{a_B}=-48t\widehat{i}\\ Att=1\sec \\ \overrightarrow{V_A}=(8-6t)\widehat{i}=2\widehat{i}and\overrightarrow{V_B}=-24\widehat{j}\\ \therefore \overrightarrow{V_{BA}}=-\overrightarrow{V_A}+\overrightarrow{V_B}=-2\widehat{i}-24\widehat{j}\\ \therefore SpeedofBw.r.tA,\sqrt{V}=\sqrt{2^2+{24}^2}\\ =\sqrt{4+576}=\sqrt{580}\\ \therefore V=580(m/s)\end{array}$