Question

The position vector of a particle is given by $\overrightarrow{r}=\overrightarrow{{{r}_{0}}}\left( 1-at \right)t$, where t is time and a as well as $\overrightarrow{{{r}_{0}}}$ are constant. After what time does the particle return to the starting point?
$\text{A}\text{. }a$
$\text{B}\text{. }\dfrac{1}{a}$
$\text{C}\text{. }{{a}^{2}}$
$\text{D}\text{. }\dfrac{1}{{{a}^{2}}}$

Answer 1

Hint: The starting point of the particle will be at time t=0. Substitute t = 0 in the given equation for the position vector and find the position of vector of the particle at the starting point. Then substitute this value in the same equation again and find the value of t.
Formula used:
$\overrightarrow{v}=\dfrac{d\overrightarrow{r}}{dt}$

Complete answer:
It is given that the position vector of the particle with respect to time is $\overrightarrow{r}=\overrightarrow{{{r}_{0}}}\left( 1-at \right)t$ …. (i).
The starting point of the particle will be at time t=0.
Substitute t=0 in equation (i).
$\Rightarrow \overrightarrow{r}=\overrightarrow{{{r}_{0}}}\left( 1-a(0) \right)(0)=0$
This means that the starting point of the particle is at the origin.
When the particle returns to the starting point, its position vector will be $\overrightarrow{r}=0$.
Substitute $\overrightarrow{r}=0$ in equation (i).
$\Rightarrow 0=\overrightarrow{{{r}_{0}}}\left( 1-at \right)t$
$\Rightarrow t=0$ or $(1-at)=0$.
We know that the time taken for the particle to return at the starting point can not be equal to zero.
This means that $(1-at)=0$.
$\Rightarrow 1=at$
$\Rightarrow t=\dfrac{1}{a}$.
This means that the particle returns at the starting point after a time interval equal to $\dfrac{1}{a}$.

So, the correct answer is “Option B”.

Note:
Suppose the velocity of the particle when it returns to the starting point was asked.
Velocity of a particle is given as $\overrightarrow{v}=\dfrac{d\overrightarrow{r}}{dt}$.
$\Rightarrow \overrightarrow{v}=\dfrac{d}{dt}\left( \overrightarrow{{{r}_{0}}}\left( 1-at \right)t \right)$
Since $\overrightarrow{{{r}_{0}}}$ is a constant, it will come out of the derivative.
$\Rightarrow \overrightarrow{v}=\overrightarrow{{{r}_{0}}}\dfrac{d}{dt}\left( \left( 1-at \right)t \right)$
$\Rightarrow \overrightarrow{v}=\overrightarrow{{{r}_{0}}}\dfrac{d}{dt}\left( t-a{{t}^{2}} \right)$
$\Rightarrow \overrightarrow{v}=\overrightarrow{{{r}_{0}}}\left[ \dfrac{d}{dt}(t)-\dfrac{d}{dt}\left( a{{t}^{2}} \right) \right]$
$\Rightarrow \overrightarrow{v}=\overrightarrow{{{r}_{0}}}\left[ 1-2at \right]$ ….(iii)
We found that when the particle returns at the starting point, $t=\dfrac{1}{a}$.
Substitute the value of t in equation (iii).
$\Rightarrow \overrightarrow{v}=\overrightarrow{{{r}_{0}}}\left[ 1-2a\left( \dfrac{1}{a} \right) \right]$
$\Rightarrow \overrightarrow{v}=\overrightarrow{{{r}_{0}}}\left[ 1-2 \right]=-\overrightarrow{{{r}_{0}}}$.