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velocity and acceleration of a particle ...

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?

Text Solution

Verified by Experts

The correct Answer is:
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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