question_answer

A scooter going due east at 10 ms?1 turns right through an angle of 90°. If the speed of the scooter remains unchanged in taking turn, the change is the velocity of the scooter is                                                 [BHU 1994]

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.