Question

A particle is moving along the $x-$axis, with its coordinate with time ${}^{\prime }{t}^{\prime }$ 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 $t=1\text{s}$, the speed of the second particle as measured in the frame of the first particle is given as $\sqrt{v}$. Then $v$ $\text{(in m/s)}$ is

Answer 1

$\text{For particle}A,$
$x_A=-3t^2+8t+10$
$\overrightarrow{V_A}=(8-6t)\hat{i}$
For particle $B,$
$y_B=5-8t^3$
$\overrightarrow{V_B}=-24t^2\hat{j}$
At $t=1\text{s}$
${\overrightarrow{V}}_A=2\hat{i}\text{and}{\overrightarrow{V}}_B=-24\hat{j}$
$\therefore {\overrightarrow{V}}_{B/A}=-{\overrightarrow{V}}_A+{\overrightarrow{V}}_B=-2\hat{i}-24\hat{j}$
$\therefore$ Speed of $B$ w.r.t. $A$ is,

${\overrightarrow{V}}_{B/A}=\sqrt{2^2+{24}^2}=\sqrt{580}{\text{ms}}^{-1}=\sqrt{v}$
$\therefore v=580$

Hence, $580$ is the correct answer.