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The figure shows a circular path taken b...

The figure shows a circular path taken by a particle. If the instantaneous velocity of the particle is `vecv=(2m//s)hati-(2m//s)hatj`, through which quadrant is the particle moving at that instant if it is traveling (a) clockwise and (b) counterclockwise around the circle? For both cases, draw `vecv` on the figure.

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(draw `vecv` tangent to path, tail on path) (a) first, (b) third
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Knowledge Check

  • The instantaneous velocity of a particle moving in xy-plane is vecV=(ay)hati+(V_(0))hatj where y is the instantaneous y co-ordinat of the particle and V_(0) is the particle starts from origin then its trajectory is

    A
    B
    C
    D

  • A particle is projected with a velocity vecv=8hati+6hatj m//s . The time after which it will starts moving perpendicular to its initial direction of motions is

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    `0.5s`
    B
    `1.25s`
    C
    `1s`
    D
    `5//3s`

  • The variation of velocity of a particle moving along straight line is shown in the figure. The distance travelled by the particle in 4 s is

    A
    25m
    B
    30m
    C
    55m
    D
    60m

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