Question 39 of 121

A particle is moving eastwards with a velocity of $5 \mathrm{~m} / \mathrm{s}$ in 10 seconds the velocity changes to $5 \mathrm{~m} / \mathrm{s}$ northwards. The average acceleration in this time is

A$\frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^2$ towards north-eastB$\frac{1}{2} \mathrm{~m} / \mathrm{s}^2$ towards north.CzeroD$\frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^2$ towards north-west

Show answer and solution▾

Answer: (D) $\frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^2$ towards north-west

$\overrightarrow{\mathrm{a}}=\frac{\overrightarrow{\mathrm{V}}_{\mathrm{f}}-\overrightarrow{\mathrm{V}}_{\mathrm{i}}}{\mathrm{t}}$
$=\frac{5 \hat{\mathrm{j}}-5 \hat{\mathrm{i}}}{10}=\frac{1}{2}(\hat{\mathrm{j}}-\hat{\mathrm{i}})$
$\therefore \mathrm{a}=\frac{1}{\sqrt{2}} \mathrm{~ms}^{-2}$ towards north west