Question

A particle moves towards east with velocity $5m/s$. After $10$ seconds its direction changes towards north with same velocity. The average acceleration of the particle is

• A. Zero
• B. $1\sqrt{2}m/s^{2}N-E$
  
  
• C. $1\sqrt{2}m/s^{2}N-W$
  
  
• D. $1\sqrt{2}m/s^{2}S-W$
  

Answer 1

The correct option is C $1\sqrt{2}m/s^{2}N-W$



Find change in velocity.
Initial velocity, ${\overrightarrow{v}}_i=5\hat{i}m/s$
Final velocity, ${\overrightarrow{v}}_f=5\hat{j}m/s$
$\Delta \overrightarrow{v}=(5\hat{j}-5\hat{i}m/s)$
$\Delta \overrightarrow{v}=(\sqrt{5^2+5^2}=5\sqrt{2}m/s$

Calculate acceleration of the particle.
Formula Used: $\overrightarrow{a}=\frac{△\overrightarrow{v}}{\Delta t}=\frac{(5\hat{j}-5\hat{i})m/s^2}{2}=(-\frac{1}{2}\hat{i}+\frac{1}{2}\hat{j})\frac{m}{s^2}$
$\left|\overrightarrow{a}\right|=\frac{\left|\Delta \overrightarrow{v}\right|}{\Delta t}=\frac{5\sqrt{2}}{10}=\frac{1}{\sqrt{2}}m/s^2$
Final answer: (B)