Question

A particle is moving eastwards with a velocity of $5{\mathrm{ms}}^{-1}$. In $10\mathrm{s}$ the velocity changes to $5{\mathrm{ms}}^{-1}$ northwards. The average acceleration of particle for this time duration is

• A. Zero
• B. $\frac{1}{\sqrt{2}}{\mathrm{ms}}^{-2}$ towards north-east
• C. $\frac{1}{\sqrt{2}}{\mathrm{ms}}^{-2}$ towards north-west
• D. $\frac{1}{2}{\mathrm{ms}}^{-2}$ towards north

Answer 1

The correct option is C

$\frac{1}{\sqrt{2}}{\mathrm{ms}}^{-2}$ towards north-west

Step 1: Given data

The initial velocity, $v_i=5{\hat{i}\mathrm{ms}}^{-1}$(eastwards)

The final velocity, $v_f=5{\hat{j}\mathrm{ms}}^{-1}$(northwards)

Time, $t=10\mathrm{s}$

Step 2: Find the change in the velocity in $\mathbf{10}\mathbf{}\mathbf{s}$

Consider the directions as given in the diagram below:

Change in velocity, $∆v=v_f-v_i$$=\left(5\hat{j}-5\hat{i}\right)m{s}^{-1}$

$\hat{j}$ is the north direction and $-\hat{i}$ is the west direction. Hence, the direction of the change in velocity is north-west.

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

$$\begin{gathered} \left|∆v\right|=\sqrt{{\left(-5\right)}^2+5^2} \\ \left|∆v\right|=\sqrt{5^2+5^2} \\ \left|∆v\right|=5\sqrt{2}m{s}^{-1} \end{gathered}$$

Step 3: Find the average acceleration of the particle for $\mathbf{10}\mathbf{}\mathbf{s}$

The average acceleration of the particle, $∆a=\frac{\left|∆v\right|}{t}$

$$\begin{gathered} =\frac{5\sqrt{2}}{10} \\ =\frac{1}{\sqrt{2}}m{s}^{-2} \end{gathered}$$

The average acceleration of particle for this time duration is $\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{}\mathit{m}{\mathit{s}}^{\mathbf{-}\mathbf{2}}$ towards north-west.

Hence, option C is correct.