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# The position vector of a particle varies with time as $\overrightarrow r = \overrightarrow {{r_0}} t(1 - \alpha t)$where $\overrightarrow {{r_0}}$ is a constant vector and $\alpha$ is a positive constant. The distance travelled by particle in a time interval in which particle returns to its initial position is $\dfrac{{K{r_0}}}{{16\alpha }}$. Determine the value of K?

Last updated date: 18th Sep 2024
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Hint From the position vector we can conclude that the particle moves in a straight line and also it is mentioned that there will be two instances of time when the particle will be at the initial position. We can use this concept to calculate the distance and compare it to get K

Formula used:
$S = ut + \dfrac{1}{2}a{t^2}$

The position vector which is varying with time is given in the question, this time varying position vector can be treated as displacement vector $\overrightarrow r = \overrightarrow {{r_0}} t(1 - \alpha t)$at $t = 0$, r will be zero. Since it is given that the particle returns to its initial position it means that the displacement is zero and there will be two values of t for which displacement is zero. They are
$\Rightarrow {r_0}t(1 - \alpha t) = 0 \Rightarrow t = 0,\dfrac{1}{\alpha }$
To calculate the distance, we first need to calculate velocity, $v = \dfrac{{dr}}{{dt}} = \dfrac{{d\left[ {{r_0}(t - \alpha {t^2})} \right]}}{{dt}} = {r_0}(t - 2\alpha t)$
When a particle has zero displacement and travels some distance, at the point where the particle changes direction velocity is zero which is at half of the distance travelled. The time at which the velocity will be zero is $0 = {r_0}(t - 2\alpha t) \Rightarrow t = \dfrac{1}{{2\alpha }}$
Let the acceleration of particle be a
then $a = \dfrac{{dv}}{{dt}} = - 2\alpha {r_0}$
at initial point t=0, putting it in velocity equation we get initial velocity $u = {r_0}$and the point where velocity is zero, $t = \dfrac{1}{{2\alpha }}$
now using equation of motion $S = ut + \dfrac{1}{2}a{t^2}$where S is the distance travelled by a particle with initial velocity u and acceleration a in time t
let us assume that the particle travels x distance then, $x = {r_0} \times \dfrac{1}{{2\alpha }} + \dfrac{1}{2} \times ( - 2\alpha {r_0}) \times {\left( {\dfrac{1}{{2\alpha }}} \right)^2} \Rightarrow x = \dfrac{{{r_0}}}{{4\alpha }}$
the total distance travelled will be $2 \times \dfrac{{{r_0}}}{{4\alpha }}$,comparing this with the value given in question $\dfrac{{K{r_0}}}{{16\alpha }}$
$\Rightarrow K = 8$

Hence the value of K is 8.

Note:
The particle was moving in the positive x- direction. The negative sign on acceleration indicates that it is retarding in nature. This means that the velocity of the body in motion is gradually decreasing with time.