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
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
- A
$20.0\, ms^{-1}$ south eastern direction
- B
Zero
- C
$10.0\, ms^{-1}$ in southern direction
- D
$14.14\, ms^{-1}$ in south-west direction
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${\vec F_1} = 5\hat i - 5\hat j + 5\hat k$ ${\vec F_2} = 2\hat i + 8\hat j + 6\hat k$
${\vec F_3} = - 6\hat i + 4\hat j - 7\hat k$ ${\vec F_4} = - \hat i - 3\hat j - 2\hat k$
Then the particle will move
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- [AIIMS 2019]
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