A particle is moving along the $x$-axis with its coordinate with time $t$ given by $x(t)=10+8 t-3 t^

Question

A particle is moving along the $x$-axis with its coordinate with time $t$ given by $x(t)=10+8 t-3 t^{2}$. Another particle is moving along the $y$-axis with its coordinate as a function of time given by $y(t)=5-8 t^{3}$. At $t=1 \mathrm{~s}$, the speed of the second particle measured in the frame of the first particle is given as $\sqrt{v}$. Then $v$ (in $\mathrm{m} / \mathrm{s}$ ) is $\qquad$ .\

Professor PC Thomas Classes · Answer

(580) : $x=10+8 t-3 t^{2}, y=5-8 t^{3}$\ $v_{x}=8-6 t, v_{y}=-24 t^{2}$;\ $\vec{v}_{x(t=1 \mathrm{~s})}=2 \hat{i}, \quad \vec{v}_{y(t=1 \mathrm{~s})}=-24 \hat{j}$\ So, the required speed\ $v_{	ext {net }}=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{4+576}=\sqrt{580}$\ $	herefore \quad v=580 \mathrm{~m} \mathrm{~s}^{-1}$\