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velocity and acceleration of a particle at some instant are `v=(3 hati-4 hatj+2 hatk) m//s and a=(2 hati+hatj-2hatk) m//s^2` (a) What is the value of dot product of v and a at the given instant? (b) What is the angle between v and a, acute, obtuse or `90^@`? (c) At the given instant, whether speed of the particle is increasing, decreasing or constant? |
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Answer» Correct Answer - A::B::C::D (a) `v.a=6-4-4=-2m^2//s^3` (b) Since dot product is negative. So angle between v and a is obtuse. (c) As angle between v and a at this instant is obtuse, speed is decreasing. |
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