Q.

Velocity and acceleration of a particle are v=(2i^-4j^)m/s and a=(-2i^+4j^)m/s2. Which type of motion is this?

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a

None of the above

b

Parallel

 

c

Both

 

d

Antiparallel

 

answer is B.

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Detailed Solution

Angle between v and a  is given by cos θ =a.va.v=-1 θ= 1800

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