Question

The position vector of a particle is given as $\overrightarrow{r}=({{t}^{2}}-4t+6)\overrightarrow{i}+({{t}^{2}})\overrightarrow{j}$. The time after which the velocity vector and the acceleration vector become perpendicular to each other is equal to:
(A). $1\,\sec $
(B). $2\,\sec $
(C). $1.5\,\sec $
(D). Not possible

Answer 1

Hint: The rate of change of position vector of a body is the velocity. Similarly, the rate of change of velocity vector of a body is the acceleration. Therefore, differentiating position with respect to time we get velocity and differentiating velocity with respect to time we get acceleration. The dot product of velocity and acceleration is zero when both vectors are perpendicular to each other.

Complete step by step answer:
Given, the position vector of a particle is $\overrightarrow{r}=({{t}^{2}}-4t+6)\overrightarrow{i}+({{t}^{2}})\overrightarrow{j}$. Here, $\hat{i}$ is along the x-axis, $\hat{j}$ is along the y-axis and $\hat{k}$ is along the z-axis.
Velocity is the change in position per unit time therefore, on differentiating the position vector with respect to time we get the velocity vector, therefore,
$\dfrac{d\overrightarrow{r}}{dt}=\overrightarrow{v}=(2t-4)\overrightarrow{i}+(2t)\overrightarrow{j}$
$\Rightarrow \overrightarrow{v}=(2t-4)\overrightarrow{i}+(2t)\overrightarrow{j}$ - (1)
On differentiating velocity vector with respect to time we get the acceleration vector therefore,
$\dfrac{d\overrightarrow{v}}{dt}=\overrightarrow{a}=(2)\overrightarrow{i}+(2)\overrightarrow{j}$
$\Rightarrow \overrightarrow{a}=(2)\overrightarrow{i}+(2)\overrightarrow{j}$ - (2)
When the velocity and acceleration vectors are perpendicular, their dot product will be zero therefore,
$\overrightarrow{a}\cdot \overrightarrow{v}=0$
Substituting given values from eq (1) and eq (2) in the above equation we get,
$\begin{align}
  & [(2t-4)\overrightarrow{i}+(2t)\overrightarrow{j}]\cdot [(2)\overrightarrow{i}+(2)\overrightarrow{j}]=0 \\
 & \Rightarrow 2(2t-4)+2(2t)=0 \\
 & \Rightarrow 4t-8+4t=0 \\
 & \Rightarrow 8t-8=0 \\
 & \therefore t=1\sec \\
\end{align}$
Therefore, the time at which the acceleration vector and the velocity vector becomes perpendicular to each other is at $1\,\sec $.

So, the correct answer is “Option A”.

Note: The dot product or scalar product between two vectors gives a scalar resultant while the vector product gives a vector as resultant. The dot product is the product of magnitude of the vectors and the cosine of the angle between them. The velocity is the slope of the graph between position and time. Similarly, the acceleration is the slope of the graph between velocity and time.