The velocity $\vec{v}$ of a particle moving in the $x-y$ plane is given by $\vec{v}=\left(6 t-4 t^2\

Question

The velocity $\vec{v}$ of a particle moving in the $x-y$ plane is given by $\vec{v}=\left(6 t-4 t^2\right) \hat{i}+8 \hat{j}$, with $\vec{v}$ in $\mathrm{m} \mathrm{s}^{-1}$ and $t(>0)$ in second. Match the following columns.

Professor PC Thomas Classes · Answer

(a) : (A) → (R); (B) → (P); (C) → (Q); (D) → (S) $$ \begin{aligned} & (\mathrm{A}) ightarrow(\mathrm{R}): \vec{v}=\left(6 t-4 t^2ight) \hat{i}+8 \hat{j} \ & \vec{a}=\frac{d \vec{v}}{d t}=(6-8 t) \hat{i} \ & |a|=\sqrt{6^2=(8 t)^2}=10 \Rightarrow 6^2+(8 t)^2=(10)^2 \end{aligned} $$ or $\quad t=1 \mathrm{~s}$. $\begin{aligned} & (\mathrm{B}) ightarrow(\mathrm{P}): \vec{a}=0 	ext { when } 6-8 t=0 \ & 	ext { or } t=\frac{3}{4} \mathrm{~s} \ & (\mathrm{C}) ightarrow(\mathrm{Q}): \vec{v}=0 	ext { is not possible at any time. } \ & (\mathrm{D}) ightarrow(\mathrm{S}):|v|=\sqrt{\left(6 t-4 t^2ight)+(8)^2}=8 \ & \Rightarrow\left(6 t-\left(4 t^2ight)ight)+8^2=8^2 	ext { or }\left(6 t-(4 t)^2=0ight. \ & 	herefore t=0 	ext { or } \frac{3}{2} \mathrm{~s}=1.5 \mathrm{~s}\end{aligned}$