A scooter going due east at $10 m / s$ turns right through an angle of $90^{\circ}$ . If the speed of the scooter remains unchanged in taking this turn, the change in the velocity of the scooter is

20.0 m / s

in south-east direction

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17.5 m / s

in a south-west direction

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10.0 m / s

in south-east direction

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14.14 m / s

in south-west direction

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Solution

The correct option is D $14.14 m / s$ in south-west direction
Let the east be positive x -axis and south be -ve y-axis
Therefore initial velocity is $v_{i} = 10 i$
After turn it becomes $v_{y} = - 10 j$
Therefore the change in velocity is $v_{f i n a l} - v_{i n i t i a l} = - 10 j - 10 i = 10 \sqrt{2}$ in SW direction.

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A scooter going due east at 10 m/s turns right through an angle of 90∘. If the speed of the scooter remains unchanged in taking this turn, the change in the velocity of the scooter is

The correct option is D 14.14 m/s in south-west directionLet the east be positive x -axis and south be -ve y-axisTherefore initial velocity is vi=10iAfter turn it becomes vy=−10jTherefore the change in velocity is vfinal−vinitial=−10j−10i=10√2 in SW direction.

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