question_answer

A particle is moving eastwards with velocity of 5 m/s. In 10 sec the velocity changes to 5 m/s northwards. The average acceleration in this time is      [IIT 1982; AFMC 1999; Pb PET 2000; JIPMER 2001, 02]

A)             Zero

B)             \[\frac{1}{\sqrt{2}}\,\,m\text{/}{{s}^{\text{2}}}\] toward north-west

C)             \[\frac{1}{\sqrt{2}}\,\,m\text{/}{{s}^{\text{2}}}\] toward north-east

D)             \[\frac{1}{2}\,\,m\text{/}{{s}^{\text{2}}}\]toward north-west

Correct Answer: B

Solution :

              \[\Delta \vec{\upsilon }={{\vec{\upsilon }}_{2}}-{{\vec{\upsilon }}_{1}}\]\[=\sqrt{\upsilon _{1}^{2}+\upsilon _{2}^{2}-2{{\upsilon }_{1}}{{\upsilon }_{2}}\,\,\cos {{90}^{o}}}\]            \[=\sqrt{{{5}^{2}}+{{5}^{2}}}=5\sqrt{2}\]                                   Average acceleration             \[=\frac{\Delta \upsilon }{\Delta t}=\frac{5\sqrt{2}}{10}=\frac{1}{\sqrt{2}}\,\,\text{m/}{{\text{s}}^{\text{2}}}\]             Directed toward north-west             (As clear from the figure).