A) \[\text{2}0.0m{{s}^{\text{1}}}\] south eastern direction
B) Zero
C) \[\text{1}0.0m{{s}^{\text{1}}}\] in southern direction
D) \[\text{14}.\text{14}m{{s}^{\text{1}}}\] in south-west direction
Correct Answer: D
Solution :
If the magnitude of vector remains same, only direction change by \[\theta \] then \[\overrightarrow{\Delta v}=\overrightarrow{{{v}_{2}}}-\overrightarrow{{{v}_{1}}}\], \[\overrightarrow{\Delta v}=\overrightarrow{{{v}_{2}}}+(-\overrightarrow{{{v}_{1}})}\] Magnitude of change in vector \[|\overrightarrow{\Delta v}|\,=2v\sin \left( \frac{\theta }{2} \right)\] \[|\overrightarrow{\Delta v}|\,=2\times 10\times \sin \left( \frac{90{}^\circ }{2} \right)\]= \[10\sqrt{2}\]= \[14.14\,m/s\] Direction is south-west as shown in figure.You need to login to perform this action.
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