The figure shows a circular path taken by a particle. If the instantaneous velocity of the particle is $\overrightarrow{\mathit{v}}=\left(2\mathrm{m}/\mathrm{s}\right) \hat{\mathrm{i}}-\left(2 \mathrm{m}/\mathrm{s}\right) \hat{\mathrm{j}}$, through which quadrant is the particle moving at that instant if it is travelling (a) clockwise and (b) counterclockwise around the circle? For both cases, draw $\overrightarrow{v}$ on the figure

Dear Student

The direction of instantaneous velocity vector v of a moving particle at any point on the path of the particle is always along the tangent to the path at that point or position.

when the instantaneous velocity is clockwise, the particle would be travelling through the first quadrant as in this case v_x will be positive and v_y will be negative.

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(b)

when the instantaneous velocity is anticlockwise, the particle would be travelling through the third quadrant as in this case v_x will be positive and v_y will be negative.

Regards