Question

A particle moves in the $x-y$ plane and at time $t$, it is at the point whose coordinates are $(t^2,t^3-2t)$. Then at what instant of time will its velocity and acceleration vectors be perpendicular to each other?

• A. $\frac{1}{3}\text{sec}$
• B. $\frac{2}{3}\text{sec}$
• C. $\frac{3}{2}\text{sec}$
• D. $\text{It will never happen}$

Answer 1

The correct option is B $\frac{2}{3}\text{sec}$

As per the given question we can write the position vector at time $t$ as
$\overrightarrow{r}=t^2\hat{i}+(t^3-2t)\hat{j}$,
$\therefore$ $\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=2t\hat{i}+(3t^2-2)\hat{j}$
and, $\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=2\hat{i}+\left(6t\right)\hat{j}$

If the velocity and acceleration vectors will be perpendicular to each other, their dot product will be zero

$\Rightarrow \overrightarrow{a}.\overrightarrow{v}=[2\hat{i}+6t\hat{j}].[2t\hat{i}+(3t^2-2\left)\hat{j}\right]$
$=4t+6t(3t^2-2)=0$
$4t+18t^3-12t=0$
$\Rightarrow 18t^3-8t=0$
$\Rightarrow 2t[9t^2-4]=0$
So, the possible values of $t$ are
$t=0,9t^2=4$,
$\Rightarrow t^2=\frac{4}{9},\text{or}t=\pm \frac{2}{3}s$
Thus, we get $t=\frac{2}{3}\text{s}$