Question

The velocity of a particle is v = {3i + (6 – 21)j} m/s, where t is in seconds. If r = 0 when t r = 0 when t = 0, determine displacement of the particle during the time interval t = 1 s to t = 3 s.

Answer 1

Given :-

Velocity vector of particle,

$\vec{V} = 3 \hat{i} + ( 6 - 2t ) \hat{j} m/s$

So Velocity of particle in x direction,

,

Now, velocity of particle in x direction , $V_{x} = \frac{dx}{dt} = 3 m/s$ where x is the position of particle in x direction.

Now integrating above equation on both side,

  

dc = 3dt

  $\therefore x = 3t + c$ where c= integrating constant

Now at t = 0 second position r = 0 so x = 0,

So,. 0 = 3×0 + c $\therefore c = 0$

So,   $\therefore x = 3t$

Now, velocity of particle in y direction,  

Now, $V_{y} = \frac{dy}{dt} = (6 - 2t)$ where y is the position of particle in y direction.

Now integrating above equation on both sides,

$\int {dy} = \int (6 - 2t) dt$

$\therefore y = 6t - t^{2} + c$ where c = integrating constant

Now at t = 0 second position r = 0 so , y = 0 ,

So, 0 = 0 - 0 + c So, C=0

So, $\therefore y = 6t - t^{2}$

So, position vector ,

$\vec{r} = x\hat{i} +y\hat{j}$

$\vec{r} = 3t\hat{i} + (6t - t^{2})\hat{j}$

for ,  

$\D\vec{r_{1}} = 3(1)\hat{i} + [6(1) - (1)^{2}]\hat{j}$

$\D\vec{r_{1}} = 3\hat{i} + 5\hat{j}$

Now for ,

$\vec{r_{2}} = (3\times 3)\hat{i} + ((6\times 3) - 3^{2})\hat{j}$

$\vec{r_{2}} = 9\hat{i} + 9\hat{j}$

Now displacement vector ,

$\Delta \vec{r} = \vec{r_{2}} - \vec{r_{1}}$

$\Delta \vec{r} = (x_{2}-x_{1})\hat{i} + (y_{2}-y_{1})\hat{j}$

$\Delta \vec{r} = (9-3)\hat{i} + (9-5)\hat{j}$

$\Delta \vec{r} = 6\hat{i} +4\hat{j}$

So, value of displacement of particle,

$\Delta r = \sqrt\left ( 6^{2} + (4)^{2}\right )$

So, $\Delta r = 7.211 m$

So, displacement of the particle is 7.211 meters.