$A$ particle of mass $m$ is moving in the $xy$-plane such that its velocity at a point $(x, y)$ is given as $\vec{v} = \alpha(y \hat{i} + 2x \hat{j})$,where $\alpha$ is a non-zero constant. What is the force $\vec{F}$ acting on the particle?

The coordinates of a moving particle at a time $t$ are given by $x = 5 \sin 10t$ and $y = 5 \cos 10t$. The speed of the particle is:

The displacement vector of a particle of mass $m$ is given by $\vec{r}(t) = A \cos \omega t \hat{i} + B \sin \omega t \hat{j}$.
$(a)$ Show that the trajectory is an ellipse.
$(b)$ Show that $\vec{F} = -m \omega^2 \vec{r}$.

$A$ particle is moving in a circular path. The acceleration and momentum vectors at an instant of time are $\vec{a} = 2 \hat{i} + 3 \hat{j} \ m/s^2$ and $\vec{p} = 6 \hat{i} + 4 \hat{j} \ kg \cdot m/s$. Then the motion of the particle is

The centripetal acceleration of a particle in uniform circular motion is $18 \,m/s^2$. If the radius of the circular path is $50 \,cm$, the change in velocity of the particle in a time of $\frac{\pi}{18} \,s$ is (in $\,m/s$)

$A$ man wants to reach from $A$ to the opposite corner of the square $C$. The sides of the square are $100\, m$. $A$ central square of $50\, m \times 50\, m$ is filled with sand. Outside this square,he can walk at a speed of $1\, m/s$. In the central square,he can walk only at a speed of $v\, m/s$ $(v < 1)$. What is the smallest value of $v$ for which he can reach faster via a straight path through the sand than any path in the square outside the sand?