Question

A particle is moving along a curve in $X-Yplane$. At $t=0$ its position is given by coordinate $(3,4)$ and at $t=5\mathrm{s}$ it is at $(6,3)$. Determine the displacement vector and the magnitude of displacement for this $5\mathrm{s}$ duration.

Answer 1

Step 1: Given data

The coordinate of a particle at $t=0$ on $X-axis$, $x_1=3m$

The coordinate of a particle at $t=0$ on $Y-axis$, $y_1=4m$

The coordinate of a particle at $t=5\mathrm{s}$ on $X-axis$, $x_2=6m$

The coordinate of a particle at $t=5\mathrm{s}$ on $Y-axis$, $y_2=3m$

Step 2: Calculate the displacement vector of the particle

Let us assume the position of a particle at $t=0$ as $r_1$ and the position of a particle at $t=5\mathrm{s}$ is $r_2$:

According to the question, the position of a particle at $t=0$:

$$\begin{gathered} \stackrel{\rightharpoonup }{r_1}=x_1\hat{i}+y_1\hat{j} \\ =\left(3\hat{i}+4\hat{j}\right)m \end{gathered}$$

According to the question, the position of a particle at $t=5\mathrm{s}$:

$$\begin{gathered} \stackrel{\rightharpoonup }{r_2}=x_2\hat{i}+y_2\hat{j} \\ =\left(6\hat{i}+3\hat{j}\right)m \end{gathered}$$

$$\begin{gathered} \therefore Displacementvector,\stackrel{\rightharpoonup }{∆r}=\left(x_2-x_1\right)\hat{i}+\left(y_2-y_1\right)\hat{j} \\ =\left(6-3\right)\hat{i}+\left(3-4\right)\hat{j} \\ =\left(3\hat{i}-\hat{j}\right)m \end{gathered}$$

Step 3: Calculate the magnitude of the displacement vector

As the displacement vector $=\left(3\hat{i}-\hat{j}\right)m$

$$\begin{gathered} \therefore \left|\stackrel{\rightharpoonup }{∆r}\right|=\sqrt{{\left(3\right)}^2+{\left(-1\right)}^2} \\ =\sqrt{9+1} \\ =\sqrt{10}m \end{gathered}$$

Hence, the displacement vector of a particle is $\left(3\hat{i}-1\hat{j}\right)\mathbf{}\mathit{m}$ and the magnitude of the displacement vector of a particle is $\sqrt{\mathbf{10}}\mathbf{}\mathit{m}$.