question_answer

A particle is moving eastwards with a velocity of \[5\text{ }m{{s}^{-1}}\]. In 10 seconds the velocity changes to \[5\text{ }m{{s}^{-1}}\] northwards. The average acceleration in this time is

A) \[\frac{1}{2}m{{s}^{-2}}\] toward north

B) \[\frac{1}{\sqrt{2}}m{{s}^{-2}}\] toward north-east

C) \[\frac{1}{\sqrt{2}}m{{s}^{-2}}\] towards north-west

D) zero

Correct Answer: C

Solution :

\[\text{Average acceleration =}\frac{\text{change in velocity}}{\text{time interval}}\] \[\text{=}\frac{\Delta \,\overrightarrow{v}}{t}\] \[\overrightarrow{{{v}_{1}}}=5\hat{i},\,\,\overrightarrow{{{v}_{2}}}=5\hat{j}\] \[\Delta \,\overrightarrow{v}=\left( \overrightarrow{{{v}_{2}}}-\overrightarrow{{{v}_{1}}} \right)\] \[=\sqrt{v_{1}^{2}+v_{2}^{2}+2{{v}_{1}}{{v}_{2}}\,\cos 90}\] \[=\sqrt{{{5}^{2}}+{{5}^{2}}+0}\] \[\left[ As|{{v}_{1}}|\,=\,|{{v}_{2}}|\,=5\text{ }m/s \right]=5\sqrt{2}\text{ }m/s\] \[Avg.\text{ }acc.\text{ }=\text{ }\frac{\Delta \,}{t}=\frac{5\sqrt{2}}{10}=\frac{1}{\sqrt{2}}\text{ }m/{{s}^{2}}\] \[\Rightarrow \tan \theta =\frac{5}{-5}=-1\] Which means \[\theta \] is in the second quadrant. (Towards north-west)