$A$ particle moves along a circle with a constant angular speed $\omega$. Its displacement,with respect to the initial position of the particle at time $t = 0$,is plotted against time. The graph would look like:

$A$ particle is moving in the $xy$-plane in a circular path with its center at the origin. If at an instant the position of the particle is given by $\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$,then the velocity of the particle is along .......

$A$ bullet fired at an angle of $30^{\circ}$ with the horizontal hits the ground $3.0 \; km$ away. By adjusting its angle of projection,can one hope to hit a target $5.0 \; km$ away? Assume the muzzle speed to be fixed,and neglect air resistance.

The second's hand of a watch has a length of $6\,cm$. The speed of the end point and the magnitude of the difference of velocities at two perpendicular positions will be:

Read each statement below carefully and state,with reasons,if it is true or false:
$(a)$ The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre.
$(b)$ The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point.
$(c)$ The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.

The position vector of a particle is given as $\vec{r} = (t^2 - 4t + 6)\hat{i} + (t^2)\hat{j}$. The time after which the velocity vector and acceleration vector become perpendicular to each other is equal to ....... $sec$.