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A particle is moing with a velocity of `10m//s` towards east. After 10s its velocity changes to `10m//s` towards north. Its average acceleration is:-A. zeroB. `sqrt(2) m//s^(2)` towards N-WC. `(2)/sqrt(2) m//s^(2)` towards N-WD. `(1)/sqrt(2)m//s^(2)` towards N-W |
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Answer» Correct Answer - B Acceleration `veca=(Deltavecv)/(Deltat)=(10hatj-10hati)/(10)` `=-hati+hatj` (west north) `|veca|=sqrt(1+1)=sqrt(2)` (north-west) |
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