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 $$t = 1s$$, 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 ________.
Correct option is A. 580
Let $$x_A$$ be the $$x-$$coordinate of the particle and $$y_B$$ be the y-coordinate of the second particle$$x_A (t) = 10 + 8t - 3t^2$$
$$\dfrac{dx_A(t)}{dt} = 8 - 6t$$ along the x-axis
$$\vec{V_A} = (8-6t)\hat{i}$$
$$y_B(t) = 5-8t^3$$
$$\dfrac{dy_b(t)}{dt} = -24t^2$$ along the y-axis
$$\vec{V_B} = -24t^2\hat{j}$$
We regular $$\vec{V_{BA}} = \vec{V_{B}} - \vec{V_B} = -24t^2 \hat{j} + (6t - 8)\hat{i}$$
At $$t = 1$$
$$\vec{V_{BA}} = -24\hat{j} + (-8+6)\hat{i} = -24\hat{j} - 2\hat{i}$$
$$|\vec{V_{BA}}|^2 =(24)^2 + 2^2 = 580$$
$$|\vec{V_{BA}}| = \sqrt{580} = \sqrt{V}$$
$$\therefore V = 580$$