Particle $A$ (which was located at the origin at time $t=0$ ) is moving along the $X$-axis with a constant speed of $1 \mathrm{~m} / \mathrm{s}$. Location of particle $B$ which is moving along the $Y$-axis is given by $y=c t^2$, where $c=1 \mathrm{~m} / \mathrm{s}^2$. Find the speed of particle $A$ relative to particle $B$ at $t=1 \mathrm{~s}$
A particle is moving in $X Y$-plane as $\mathbf{x}=\left(4 t+t^2\right) \hat{\mathbf{i}}$, $\mathbf{y}=\left(2 t+\frac{t^2}{2}\right) \hat{\mathbf{j}}$, where $\mathbf{x}$ and $\mathbf{y}$ are displacements measured along $X$ and $Y$-axes respectively, in metres and $t$ in seconds, What is the velocity of the particle?
The surface of a hill inclined at an angle $30^{\circ}$ to the horizontal. A stone is thrown from the summit of the hill (point $A$ ) at an initial speed $10 \mathrm{~m} / \mathrm{s}$ at angle $60^{\circ}$ to the vertical. If the stone strikes the hill at point $B$ as shown in the figure, the distance between $A$ and $B$ is (take, $g=10 \mathrm{~m} / \mathrm{s}^2$ )

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