A particle is moving along the x-axis with its coordinate with the time 't' given be
x(t) = 10 + 8t – 3t². Another particle is moving the y-axis with its coordinate as a function of time given by y(t) = 5 – 8t³.
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 ______.

<p>For a particle ‘A’, its position along the x-axis as a function of time $ t $ is given by:</p> <p>$ x(t) = 10 + 8t - 3t^2 $</p> <p>To find the velocity $ v_A $, we take the derivative of $ x(t) $ with respect to $ t $:</p> <p>$ v_A = \frac{d}{dt}[10 + 8t - 3t^2] = 8 - 6t $</p> <p>At $ t = 1 $ second, the velocity of particle A is:</p> <p>$ \vec{v_A} = (8 - 6 \cdot 1)\hat{i} = 2\hat{i} $</p> <p>For a particle ‘B’, its position along the y-axis as a function of time $ t $ is given by:</p> <p>$ y(t) = 5 - 8t^3 $</p> <p>To find the velocity $ v_B $, we take the derivative of $ y(t) $ with respect to $ t $:</p> <p>$ v_B = \frac{d}{dt}[5 - 8t^3] = -24t^2 $</p> <p>At $ t = 1 $ second, the velocity of particle B is:</p> <p>$ \vec{v_B} = -24 \cdot 1^2 \hat{j} = -24\hat{j} $</p> <p>The velocity of particle B relative to particle A ($ \vec{v_{B/A}} $) is calculated as:</p> <p>$ \vec{v_{B/A}} = \vec{v_B} - \vec{v_A} $</p> <p>$ \vec{v_{B/A}} = -24\hat{j} - 2\hat{i} $</p> <p>To find the magnitude of $ \vec{v_{B/A}} $:</p> <p>$ |\vec{v_{B/A}}| = \sqrt{(-24)^2 + (-2)^2} = \sqrt{576 + 4} = \sqrt{580} $</p> <p>Therefore, $ v $ is:</p> <p>$ v = 580 $</p>