# A point moves in x-y plane according to vec r=3cos(4t)veci +3(1-sin4t) vecj. What is the distance travelled by the particle in 0.5 sec ?

$5.048 \setminus \setminus \textrm{u n i t}$

#### Explanation:

The position of particle at any time $t$ is given as

$\setminus \vec{r} = 3 \setminus \cos \left(4 t\right) \setminus \vec{i} + 3 \left(1 - \setminus \sin \left(4 t\right)\right) \setminus \vec{j}$

Now, the position of particle at time $t = 0$ is given as

$\setminus {\vec{r}}_{1} = 3 \setminus \cos \left(0\right) \setminus \vec{i} + 3 \left(1 - \setminus \sin \left(0\right)\right) \setminus \vec{j} = 3 \setminus \vec{i} + 3 \setminus \vec{j}$

The position of particle at time $t = 0.5$ is given as

$\setminus {\vec{r}}_{2} = 3 \setminus \cos \left(2\right) \setminus \vec{i} + 3 \left(1 - \setminus \sin \left(2\right)\right) \setminus \vec{j} = - 1.248 \setminus \vec{i} + 0.272 \setminus \vec{j}$

Now, the distance traveled by the particle in $t = 0.5$ sec is given as

$| \setminus {\vec{r}}_{1} - \setminus {\vec{r}}_{2} |$

$= | 3 \setminus \vec{i} + 3 \setminus \vec{j} - \left(- 1.248 \setminus \vec{i} + 0.272 \setminus \vec{j}\right) |$

$= | 4.248 \setminus \vec{i} + 2.728 \setminus \vec{j} |$

$= \setminus \sqrt{{4.248}^{2} + {2.728}^{2}}$

$= 5.048 \setminus \setminus \textrm{u n i t}$

Jul 6, 2018

$\sqrt{3} \text{ units}$

#### Explanation:

Vector quantity position/displacement is:

$\vec{r} = 3 \cos \left(4 t\right) \vec{i} + 3 \left(1 - \sin \left(4 t\right)\right) \vec{j}$

Differentiate to obtain vector quantity velocity:

$\frac{d}{\mathrm{dt}} \vec{r} = \vec{v} = - 12 \sin \left(4 t\right) \vec{i} - 12 \cos \left(4 t\right) \vec{j}$

$= - 12 \left(\sin \left(4 t\right) \vec{i} + \cos \left(4 t\right) \vec{j}\right)$

Scalar quantity speed $\dot{s}$ is:

$\dot{s} = \left\mid \vec{v} \right\mid = \sqrt{\vec{v} \cdot \vec{v}} = \sqrt{12} = 2 \sqrt{3}$

In any time period of $0.5 \text{ sec}$, the particle will move a distance :

$s = \frac{1}{2} \cdot 2 \sqrt{3} = \sqrt{3} \text{ units}$