A car starts at time $t=0$ from an initial speed of $10 \mathrm{~m} / \mathrm{s}$ and accelerates uniformly with $2 \mathrm{~m} / \mathrm{s}^2$ on a straight road for time $0 \leq t \leq 10 \mathrm{~s}$. Let $s_1$ and $s_2$ be the distance covered by the car in time $3 \leq t \leq 4 \mathrm{~s}$ and $4 \leq t \leq 5 \mathrm{~s}$, respectively. The ratio $\frac{s_2}{s_1}$ is

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, $g=10 \mathrm{~m} / \mathrm{s}^2$ )