A) \[\text{1/2}\,\text{m/}{{\text{s}}^{\text{2}}}\] towards North-West
B) \[\text{1/}\sqrt{2}\,\text{m/}{{\text{s}}^{\text{2}}}\] towards North-East
C) zero
D) \[\text{1/}\sqrt{2}\,\text{m/}{{\text{s}}^{\text{2}}}\] towards North-West
Correct Answer: B
Solution :
Suppose positive X-axis be Eastward and positive Y-axis be along North. In terms of components initial velocity \[{{\upsilon }_{A}}={{\upsilon }_{x}}i\] and final velocity \[{{\upsilon }_{B}}={{\upsilon }_{y}}i\] here, \[{{\upsilon }_{x}}=5={{\upsilon }_{y}}\] The magnitude of the change in velocity \[=\sqrt{{{5}^{2}}+{{5}^{2}}}=5\sqrt{2}\] Since, this change occurs in 10 sec. Hence, the average acceleration \[=\frac{5\sqrt{2}}{10}=\frac{1}{\sqrt{2}}\] The direction will be \[\theta ={{\tan }^{-1}}\frac{{{v}_{y}}}{{{v}_{x}}}={{\tan }^{-1}}1=45{}^\circ \] So, the direction is North-East.
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