Question

Figure shows a circular path taken by a particle. If the instantaneous acceleration of the particle is $\overrightarrow{a}=(-2\hat{i}-2\hat{j})m/s^2$ through which quadrant is the particle moving when it is travelling counter clockwise around the circle? (Assume that speed is constant)

• A. I
• B. II
• C. III
• D. IV

Answer 1

The correct option is A I

Since speed is constant.

$\therefore$ only radical acceleration is acsing

$\therefore$ only in $1st$ quadrant

$\therefore$ Option $A$ is correct.