Question

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 as measured in the frame of the first particle is given as $\sqrt{\upsilon}$. Then (in m/s) is __________ .
Given -

Answer 1

$v_{x}=8-6t$
$\left(V_{x}\right)_{t=1 }=2\hat{i}$
$y=5-8t^{3}$
$V_{y}=-24t^{2}
\left(V_{y}\right)_{t=1}=-24\hat{j}$
Now
$\sqrt{v}=\sqrt{\left(24\right)^{2}+\left(2\right)^{2}}=\sqrt{580}$
$\therefore v=580\,m^{2}/s^{2}$