Question

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

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

Answer 1

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

$v_A=5\hat{i}$
$v_B=5\hat{j}$
Change in velocity $=5\hat{j}-5\hat{i}$

$Avg.ac{c}^n=\frac{changeinvelocity}{time}=\frac{5\hat{j}-5\hat{i}}{10}=\frac{\hat{j}-\hat{i}}{2}$

Magnitude of this vector $=\sqrt{\frac{1^2+1^2}{4}}$$=\frac{1}{\sqrt{2}}$

Direction of this vector is $tan\theta =\frac{-1}{1}⟹\theta ={135}^o$, i.e, N-W direction