Question

A particle is moving eastwards with a velocity of 5 m/s. In 10 s the velocity changes to 5 m/s northwards. The average acceleration in this time is

• A. zero
• B. $\frac{1}{\sqrt{2}}m/s^2$ towards north-west
• C. $\frac{1}{\sqrt{2}}m/s^2$ towards north-east
• D. $\frac{1}{2}m/s^2$ towards north-west

Answer 1

The correct option is B $\frac{1}{\sqrt{2}}m/s^2$ towards north-west

OA = Intial velocity ${\overrightarrow{v}}_1=5m{s}^{-1},$ due east

$\overrightarrow{O}B$ = Final velocity ${\overrightarrow{v}}_2=5m{s}^{-1},$ due north

By $\Delta$ law of vector addition,

$\overrightarrow{OA}+\overrightarrow{AB}=\overrightarrow{OB}$

$\therefore \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}={\overrightarrow{v}}_2-{\overrightarrow{v}}_1$

= change in velocity

$\left|\begin{array}{c}{\overrightarrow{v}}_2-{\overrightarrow{v}}_1\end{array}\right|=AB=\sqrt{O{A}^2+O{B}^2}$

=$\sqrt{5^2+5^2}=5\sqrt{2}m{s}^{-1}$

Average acceleration

$=\frac{\left|\begin{array}{c}{\overrightarrow{v}}_2-{\overrightarrow{v}}_1\end{array}\right|}{t}=\frac{5\sqrt{2}}{10}$

$\frac{1}{\sqrt{2}}m{s}^{-2},$ along N-W direction.