# A particle is moving along the x- axis with its coordinate with time 't' given by $x(t)=10+8t-3t^2$. Another particle is moving along the y-axis with its coordinate as a function of time given by $y(t)=5-8t^3.$ At $t=1$ s, the speed of the second particle measured in the frame of the first particle is given by $\sqrt{v}$. Then $v$ (in m/s) is ____________. Option: 1 580 Option: 2 700 Option: 3 100 Option: 4 300

\begin{aligned} &x_{A}=-3 t^{2}+8 t+10 \\ &\vec{v}_{A}=(-6 t+8) \hat{i} \\ &\text { At time } t=1 \mathrm{~s} \\ &\vec{v}_{A}=2 \hat{i} \\ &Y_{B}=5-8 t^{3} \end{aligned}

$\vec{v}_{B}=-24 t^{2} j$

At time t= 1 s

$\vec{v_{B}}=-24 \hat{j}$

\begin{aligned} &\sqrt{v}=\left|\vec{v}_{B}-\vec{v}_{A}\right|=|-24 \hat{j}-2 \hat{i}| \\ &\sqrt{v}=\sqrt{580} \end{aligned}

So the value of v is 580

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