Question

A particle's velocity changes from $10{\mathrm{ms}}^{-1}$ north to $20{\mathrm{ms}}^{-1}$ east in a time-interval of $5\mathrm{s}$. Find the average acceleration (magnitude and direction both) of the particle for this $5\mathrm{s}$ duration.

Answer 1

Step 1: Given data

The initial velocity, $u=10{\mathrm{ms}}^{-1}$ northwards.

The final velocity, $v=20{\mathrm{ms}}^{-1}$ eastwards.

Time, $t=5s$

Step 2: Find the magnitude of the change in the velocity of a particle

Consider the North direction as $\hat{j}$, South direction as $-\hat{j}$, East direction as $\hat{i}$ and West direction as $-\hat{i}$ on $XYplane$.

The change in the velocity, $∆v=v_f-v_i$

$=\left(20\hat{j}-10\hat{i}\right)m{s}^{-1}$

The magnitude of the change in the velocity, $\left|∆v\right|=\sqrt{v^2+u^2}$

$$\begin{gathered} =\sqrt{{20}^2+{10}^2} \\ =\sqrt{400+100} \\ =\sqrt{500} \\ =10\sqrt{5}m{s}^{-1} \end{gathered}$$

Step 3: Find the magnitude of the average acceleration of a particle

Average acceleration, $a=\frac{|∆v|}{t}$

$$\begin{gathered} =\frac{10\sqrt{5}}{5} \\ =2\sqrt{5}m{s}^{-2} \end{gathered}$$

Step 4: Find the direction of the average acceleration of a particle

The direction of $\mathit{∆}v$ is given by $\tan \theta =\frac{Perpendicular}{Base}$

$\tan \theta =\frac{u}{v}$

$\tan \theta =\frac{10}{20}$

$\tan \theta =\left(\frac{1}{2}\right)$

$\theta ={\tan }^{-1}\left(\frac{1}{2}\right)southofeast$

Hence, the average acceleration is $\mathbf{2}\sqrt{\mathbf{5}}\mathbf{}\mathit{m}{\mathit{s}}^{\mathbf{-}\mathbf{1}}$ and its direction is $\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{1}{2}\right)\mathbf{}\mathit{s}\mathit{o}\mathit{u}\mathit{t}\mathit{h}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{e}\mathit{a}\mathit{s}\mathit{t}$..